Explanation:
Okay, so I need to find the largest possible value of |a| + |b| + |c| given that the absolute value of the quadratic function ax² + bx + c is less than or equal to 1 for all x between 0 and 1. Hmm, let me think about how to approach this.

First, since the quadratic is bounded between -1 and 1 on the interval [0, 1], maybe I can use some properties of polynomials or optimization techniques. Maybe Lagrange multipliers? Or perhaps considering specific points in the interval where the maximum absolute value occurs?

Let me recall that for a quadratic function, the maximum or minimum occurs either at the vertex or at the endpoints of the interval. Since the interval is [0, 1], the critical points are at x = 0, x = 1, and the vertex x = -b/(2a). But I need to ensure that the function doesn't exceed 1 or go below -1 anywhere in between.

But the problem is about maximizing |a| + |b| + |c| under the constraint that |ax² + bx + c| ≤ 1 for x in [0, 1]. So maybe I can find expressions for a, b, c that satisfy the constraints and then compute |a| + |b| + |c|?

Alternatively, perhaps I can use some known inequalities or transformations. Let me think about some specific values.

First, let me note that when x = 0, the function evaluates to c, so |c| ≤ 1. Similarly, when x = 1, the function is a + b + c, so |a + b + c| ≤ 1. Also, the vertex of the quadratic is at x = -b/(2a), but if that lies within [0, 1], then the maximum or minimum could be there. However, if the vertex is outside [0, 1], then the maximum and minimum would occur at the endpoints.

But since the quadratic is bounded between -1 and 1 on [0, 1], we need to ensure that the function doesn't exceed those bounds anywhere in the interval.

But how do I relate this to |a| + |b| + |c|? Maybe I can parametrize the quadratic in terms of its values at certain points?

Another idea: perhaps use Chebyshev polynomials? Because Chebyshev polynomials are known for minimizing the maximum deviation, but I need the opposite—maximizing |a| + |b| + |c| under the maximum deviation constraint. Maybe not directly, but perhaps similar techniques can be applied.

Wait, Chebyshev polynomials have extrema at certain points. Let me recall that the Chebyshev polynomial of degree 2 is T₂(x) = 2x² - 1, which oscillates between -1 and 1 on the interval [-1, 1]. But our interval is [0, 1], so maybe we can adjust it?

If I make a substitution, let me map the interval [0, 1] to [-1, 1]. Let t = 2x - 1, so when x = 0, t = -1, and x = 1, t = 1. Then, substituting x = (t + 1)/2 into the quadratic, we get:

ax² + bx + c = a[(t + 1)/2]^2 + b[(t + 1)/2] + c.

Expanding this:

a[(t² + 2t + 1)/4] + b[(t + 1)/2] + c

= (a/4)t² + (a/2)t + a/4 + (b/2)t + b/2 + c

Combine like terms:

= (a/4)t² + [(a/2) + (b/2)]t + [a/4 + b/2 + c]

So if we want this to be similar to the Chebyshev polynomial T₂(t) = 2t² - 1, which has maximum absolute value 1 on [-1, 1], maybe we can set coefficients accordingly?

But our quadratic is bounded by 1 in absolute value on [0, 1], which corresponds to t in [-1, 1]. So maybe if we scale the Chebyshev polynomial appropriately?

But let me think. Suppose we want our quadratic to attain the maximum absolute value 1 at several points in [0, 1]. For Chebyshev polynomials, they alternate between 1 and -1 at n+1 points, which is maximally oscillating. Maybe if our quadratic is constructed similarly, it could achieve the maximal |a| + |b| + |c|?

Alternatively, maybe consider that to maximize |a| + |b| + |c|, we need the quadratic to have maximum coefficients. However, since the quadratic is constrained to be within [-1, 1] on [0,1], maybe we need to find the quadratic that touches the bounds 1 or -1 at several points, such that the coefficients are as large as possible.

For instance, if we can find a quadratic that reaches 1 at x=0, -1 at x=1, and 1 at some point in between, maybe that would create the maximum coefficients? Or some combination like that.

Alternatively, maybe consider specific cases where the maximum of |ax² + bx + c| is achieved at three points, which would be necessary for the maximum coefficients?

Wait, for a quadratic, the maximum on an interval can be at most two points, unless it's constant. Wait, but since it's a parabola, it can have only one extremum. So the maximum absolute value can be achieved at the endpoints and the critical point. Hence, maybe three points? For example, the maximum is achieved at x=0, x=1, and x=-b/(2a) if that's within [0,1].

So perhaps if we have the quadratic reaching 1 at x=0, -1 at x=1, and 1 at the critical point, or some combination like that. Let me try to set up equations.

Suppose that the quadratic reaches maximum absolute value at three points. For example, let me assume that the maximum is achieved at x=0, x=1, and the critical point. Let me assume that:

At x=0: |c| = 1, so c = 1 or -1.

At x=1: |a + b + c| = 1.

At the critical point x = -b/(2a): |ax² + bx + c| = 1.

But this seems a bit vague. Let me think of a specific example. Suppose we have the quadratic that attains 1 at x=0, -1 at x=1, and 1 at the midpoint x=1/2. Then, we can set up equations.

Let me try that.

Let me assume that:

At x=0: ax² + bx + c = 1 => c = 1.

At x=1: ax² + bx + c = -1 => a + b + c = -1.

At x=1/2: a*(1/2)^2 + b*(1/2) + c = 1 => (a/4) + (b/2) + c = 1.

Since c = 1, substitute into the other equations.

From x=1: a + b + 1 = -1 => a + b = -2.

From x=1/2: (a/4) + (b/2) + 1 = 1 => (a/4) + (b/2) = 0 => multiply by 4: a + 2b = 0.

So we have:

a + b = -2,

a + 2b = 0.

Subtract first equation from the second: b = 2. Then from first equation: a = -2 - b = -2 - 2 = -4.

Thus, the quadratic is -4x² + 2x + 1.

Let me check if this quadratic stays within [-1, 1] on [0, 1].

First, at x=0: 1, which is okay.

At x=1: -4 + 2 + 1 = -1, okay.

At x=1/2: -4*(1/4) + 2*(1/2) + 1 = -1 + 1 + 1 = 1, okay.

Now check the vertex. The critical point is at x = -b/(2a) = -2/(2*(-4)) = -2/(-8) = 1/4.

So at x=1/4, the value is:

-4*(1/4)^2 + 2*(1/4) + 1 = -4*(1/16) + 1/2 + 1 = -1/4 + 1/2 + 1 = (-1/4 + 1/2) + 1 = 1/4 + 1 = 5/4 = 1.25.

Wait, that's 1.25, which is greater than 1! That violates the condition. So this quadratic does not satisfy the constraint. So my assumption that it reaches 1 at x=0, -1 at x=1, and 1 at x=1/2 is wrong because it exceeds 1 in between. Hence, this approach is flawed.

Hmm, so maybe I need to ensure that the maximum is not exceeded. So perhaps the maximum is achieved at the endpoints and the critical point, but with alternating signs? Let me think again.

Alternatively, maybe set up the quadratic so that it reaches 1 at two points and -1 at another, such that the maximum is achieved at those points. However, for a quadratic, it can only have one extremum. So if it reaches 1 at two points and -1 at one, or similar. But how?

Alternatively, maybe use the Lagrange interpolation. Suppose the quadratic has maximum absolute value 1 at three points: x=0, x=1/2, x=1. Let me suppose that it alternates between 1 and -1 at these points. For instance, suppose at x=0: 1, x=1/2: -1, x=1: 1. Then we can find the quadratic passing through these points.

Let me try that. Let me denote the quadratic as f(x) = ax² + bx + c.

We have:

f(0) = c = 1.

f(1/2) = a*(1/4) + b*(1/2) + c = -1.

f(1) = a + b + c = 1.

So substituting c = 1 into the other equations:

From f(1/2): (a/4) + (b/2) + 1 = -1 => (a/4) + (b/2) = -2.

Multiply by 4: a + 2b = -8.

From f(1): a + b + 1 = 1 => a + b = 0.

So from a + b = 0, we have a = -b. Substitute into a + 2b = -8: -b + 2b = -8 => b = -8. Then a = 8.

Thus, the quadratic is 8x² - 8x + 1.

Let me check its value at the critical point. The critical point is at x = -b/(2a) = 8/(2*8) = 8/16 = 1/2. Wait, that's the same as the point where we have f(1/2) = -1. So the vertex is at x=1/2, which is already one of the points where the function reaches -1. Therefore, the maximum and minimum are at the endpoints and the vertex. But let's check the value at x=1/2, which is -1, and the other points are 1 at x=0 and x=1. Let's check other points in between to see if the function stays within [-1, 1].

Take x=1/4:

f(1/4) = 8*(1/16) - 8*(1/4) + 1 = 0.5 - 2 + 1 = -0.5, which is between -1 and 1.

Take x=3/4:

f(3/4) = 8*(9/16) -8*(3/4) + 1 = 4.5 - 6 + 1 = -0.5, same as above.

So it seems that this quadratic oscillates between 1 and -1 at x=0, 1/2, 1, and stays within the bounds in between. Therefore, it satisfies the condition |f(x)| ≤ 1 on [0, 1].

Now, compute |a| + |b| + |c| for this quadratic: |8| + |-8| + |1| = 8 + 8 + 1 = 17. Hmm, that's quite large. Is this the maximum?

But wait, let me check if there are other quadratics with higher |a| + |b| + |c|.

Alternatively, maybe considering another configuration where the quadratic reaches 1 at three points or alternates signs differently? But since a quadratic is determined by three points, if we set it up to pass through three points with alternating signs, maybe that's the maximum?

Wait, but maybe there's a quadratic that reaches 1 at x=0, -1 at x=1, and 1 at some other point, but arranged such that the maximum is not exceeded. However, earlier when I tried such a case, the vertex exceeded the bound. Let me try again with different points.

Alternatively, maybe take another approach. Let me use the method of Lagrange multipliers. Let me try to maximize |a| + |b| + |c| subject to |ax² + bx + c| ≤ 1 for all x in [0,1].

But dealing with absolute values and inequalities over an interval is complicated. Maybe instead, consider that the maximum of |ax² + bx + c| over [0, 1] is ≤ 1. Therefore, the maximum of ax² + bx + c is ≤ 1, and the minimum is ≥ -1.

Alternatively, since the quadratic is continuous on [0,1], we can write that:

-1 ≤ ax² + bx + c ≤ 1 for all x in [0,1].

To maximize |a| + |b| + |c|, perhaps we need to find the extremal quadratic which touches the bounds at several points. The previous example gave |a| + |b| + |c| = 17. Let me check if that's indeed the maximum.

Alternatively, consider that when we have the quadratic 8x² - 8x + 1, which we saw satisfies the condition. Also, let's check the value of |a| + |b| + |c| for other quadratics.

Alternatively, suppose that we set up the quadratic to reach maximum at x=0, x=1, and the critical point. Wait, but if we do that, we might have similar equations.

But maybe another approach: express the quadratic in terms of its values at three points. Since a quadratic is uniquely determined by its values at three points. Let me use the three points x=0, x=1/2, x=1, and set up equations such that the quadratic is extremal.

Alternatively, maybe use some linear algebra. Let me think.

Let me consider that the quadratic function is bounded by 1 in absolute value on [0, 1]. Let me denote f(x) = ax² + bx + c.

We have the constraints:

For all x in [0, 1], -1 ≤ f(x) ≤ 1.

We need to maximize |a| + |b| + |c|.

This seems like an optimization problem with infinitely many constraints. However, perhaps the maximum is attained when the function f(x) touches the bounds ±1 at several points, which is similar to the Chebyshev polynomial idea.

Earlier, with the quadratic 8x² - 8x + 1, we saw that it reaches 1 at x=0 and x=1, and -1 at x=1/2, and stays within the bounds. So this is similar to the Chebyshev polynomial oscillating between extremes.

Alternatively, maybe there's a quadratic that oscillates between 1 and -1 three times? But a quadratic can only have one extremum. So maximum two alternations? Wait, on the interval [0, 1], if the function has one extremum, then it can alternate between max and min at most two times. For example, going from 1 at x=0, down to -1 at x=1/2, then back up to 1 at x=1. Which is exactly what the quadratic 8x² - 8x + 1 does. So this is the maximal oscillation for a quadratic on [0, 1], which might lead to the maximal coefficients.

So if this is the case, then |a| + |b| + |c| = 8 + 8 + 1 = 17. But let me check if there are other quadratics with higher values.

Alternatively, consider if we set the quadratic to reach 1 at x=0, -1 at x=1, and then have another extremum? Wait, but the extremum is already at x=1/2.

Alternatively, what if we have maximum at x=0, minimum at x=1, and maximum at some other point? But the quadratic can't have two maxima or minima. So only one extremum.

Alternatively, suppose the quadratic has maximum at x=0 and x=1, and minimum somewhere in between? But then the quadratic would be concave down. For example, like -8x² + 8x - 1? Wait, let me check.

Wait, let me try this quadratic: Let me assume that f(0) = 1, f(1) = 1, and f(1/2) = -1. Wait, that's the same as before but with a sign change. Wait, actually, the previous quadratic was 8x² -8x +1, which gives f(1/2) = -1. If we take the negative of that, -8x² +8x -1, then f(0) = -1, f(1) = -1, and f(1/2)=1. But then |a| + |b| + |c| is same: |-8| + |8| + |-1| = 8 + 8 +1 = 17. So same value.

Alternatively, maybe there's a way to have larger coefficients?

Wait, suppose I set up the quadratic so that it reaches 1 at x=0, -1 at x=1, and also reaches 1 at some other point within the interval. But earlier when I tried that, the vertex exceeded the bound. However, maybe if we adjust the points?

Wait, let me try again. Suppose f(0) = 1, f(1) = -1, and f(1/2) = 1. Then set up equations:

f(0) = c = 1.

f(1) = a + b + c = -1 => a + b = -2.

f(1/2) = a*(1/4) + b*(1/2) + c = 1 => (a/4) + (b/2) + 1 = 1 => (a/4) + (b/2) = 0 => a + 2b = 0.

So solving:

From a + b = -2 and a + 2b = 0, subtract first equation from second:

b = 2. Then a = -2 - b = -4. So quadratic is -4x² + 2x + 1. Earlier, this had a vertex at x=1/4 with value f(1/4) = -4*(1/16) + 2*(1/4) + 1 = -0.25 + 0.5 + 1 = 1.25. Which is over the bound. Therefore, this quadratic violates the condition. Therefore, this approach doesn't work.

Therefore, maybe the previous quadratic that attains 1 at x=0 and x=1, and -1 at x=1/2 is the maximal in some sense. But how to confirm?

Alternatively, maybe consider other extremal points. Let me think about how to maximize |a| + |b| + |c|. Since c is evaluated at x=0, which is bounded by 1 in absolute value, so |c| ≤ 1. Similarly, a + b + c is evaluated at x=1, so |a + b + c| ≤ 1. Also, if we can find other constraints.

Alternatively, let me think in terms of linear algebra. Suppose I want to maximize |a| + |b| + |c|, given that for x in [0, 1], |ax² + bx + c| ≤ 1.

But this is a complex constraint. Maybe we can use some specific points.

For example, the maximum of |a| + |b| + |c| could be achieved when c is maximized or minimized, along with a and b. However, since c is bounded by 1, and similarly a + b + c is bounded by 1, perhaps there's a relationship between these terms.

Alternatively, perhaps consider that the maximum of |a| + |b| + |c| occurs when the quadratic is extremal in some direction.

Alternatively, think of the problem as an optimization problem with variables a, b, c, and constraints that for all x in [0, 1], -1 ≤ ax² + bx + c ≤ 1. To maximize |a| + |b| + |c|.

But solving this requires knowledge of the maximum of the quadratic on [0, 1]. However, since the quadratic is continuous, its maximum and minimum on [0,1] occur either at critical points or endpoints.

Therefore, maybe the constraints can be written as:

-1 ≤ c ≤ 1,

-1 ≤ a + b + c ≤ 1,

and if the critical point x_v = -b/(2a) is in [0, 1], then -1 ≤ a(x_v)^2 + b(x_v) + c ≤ 1.

But this seems complicated due to the dependence on a and b.

Alternatively, maybe consider specific cases where the quadratic takes the maximum absolute value at the endpoints and the critical point. For example, suppose that the maximum is achieved at three points: x=0, x=1, and x= t in (0,1). Then, we can set up equations such that f(0) = 1, f(1) = 1, f(t) = -1 or something like that.

Alternatively, maybe use the previous example where the quadratic reaches 1 at x=0, -1 at x=1/2, and 1 at x=1. Which gives us |a| + |b| + |c| = 17. Let me check if there is a quadratic with larger |a| + |b| + |c|.

Alternatively, maybe consider another quadratic which attains maximum absolute value at different points. For example, suppose it attains 1 at x=0 and x=1, and another value at the critical point. Wait, but we saw that if we set it to attain 1 at x=0, x=1, and -1 at x=1/2, then it gives a quadratic with |a| + |b| + |c| = 17. What if we do something else?

Alternatively, suppose we take the quadratic that attains 1 at x=0, and -1 at x=1, and then another extrema. Wait, but the quadratic can only have one extremum. So maybe the extremum is somewhere else.

Alternatively, think of a quadratic that has maximum slope or something? Not sure.

Alternatively, consider that the maximum of |a| + |b| + |c| could be achieved when a, b, c have the same sign? Or different signs?

Wait, in the previous example, a=8, b=-8, c=1. So |a| + |b| + |c| = 17. If I can find a quadratic with larger coefficients, that would be better.

Alternatively, consider the case where the quadratic reaches 1 at x=0, and -1 at x=1, and also reaches another extremum. But how?

Alternatively, think of the following: if I can write the quadratic in terms of its values at x=0, x=1, and x=1/2, then express a, b, c in terms of those values.

Let me denote:

Let’s say f(0) = c = y₀,

f(1) = a + b + c = y₁,

f(1/2) = a(1/4) + b(1/2) + c = y₂.

Then, solving for a, b, c:

From f(0): c = y₀.

From f(1): a + b = y₁ - y₀.

From f(1/2): (a/4) + (b/2) = y₂ - y₀.

Multiply the third equation by 4: a + 2b = 4(y₂ - y₀).

So subtract the second equation (a + b = y₁ - y₀) from this: (a + 2b) - (a + b) = b = 4(y₂ - y₀) - (y₁ - y₀) = 4y₂ - 4y₀ - y₁ + y₀ = 4y₂ - y₁ - 3y₀.

Then, from a + b = y₁ - y₀, we get a = y₁ - y₀ - b = y₁ - y₀ - [4y₂ - y₁ - 3y₀] = y₁ - y₀ - 4y₂ + y₁ + 3y₀ = 2y₁ + 2y₀ - 4y₂.

Therefore,

a = 2y₁ + 2y₀ - 4y₂,

b = 4y₂ - y₁ - 3y₀,

c = y₀.

Now, if we want |a| + |b| + |c| to be maximized, given that |y₀| ≤ 1, |y₁| ≤ 1, |y₂| ≤ 1, and also the maximum of |ax² + bx + c| on [0,1] is ≤ 1.

But this seems complicated, but maybe we can consider the case where the maximum is achieved at the three points x=0, x=1/2, x=1. So if we set y₀, y₁, y₂ to be ±1, such that the quadratic oscillates between 1 and -1.

For example, if we set y₀ = 1, y₁ = 1, y₂ = -1, which gives us the quadratic we discussed earlier:

a = 2*1 + 2*1 - 4*(-1) = 2 + 2 + 4 = 8,

b = 4*(-1) - 1 - 3*1 = -4 -1 -3 = -8,

c = 1.

So that gives |a| + |b| + |c| = 8 + 8 + 1 = 17.

Alternatively, if we set y₀ = 1, y₁ = -1, y₂ = 1, then:

a = 2*(-1) + 2*1 - 4*1 = -2 + 2 -4 = -4,

b = 4*1 - (-1) - 3*1 = 4 +1 -3 = 2,

c = 1.

Which gives quadratic -4x² + 2x +1, which as we saw earlier has a maximum of 1.25 at x=1/4, which violates the constraint. Therefore, this case is invalid.

Therefore, the previous case with y₀=1, y₁=1, y₂=-1 is valid, giving the maximum of 17.

Alternatively, what if we set y₀ = -1, y₁ = -1, y₂ = 1? Then similarly, a = 2*(-1) + 2*(-1) - 4*1 = -2 -2 -4 = -8, b = 4*1 - (-1) - 3*(-1) = 4 +1 +3 = 8, c = -1. So the quadratic is -8x² +8x -1, which gives |a| + |b| + |c| = 8 +8 +1=17 again.

So that also gives 17. So that seems consistent.

Alternatively, if we set y₀ = -1, y₁ = 1, y₂ = -1? Let's check:

a = 2*1 + 2*(-1) - 4*(-1) = 2 -2 +4 = 4,

b = 4*(-1) -1 -3*(-1) = -4 -1 +3 = -2,

c = -1.

Thus, quadratic is 4x² -2x -1.

Check its maximum. Let's compute the vertex at x = -b/(2a) = 2/(2*4) = 1/4. At x=1/4:

f(1/4) = 4*(1/16) -2*(1/4) -1 = 0.25 -0.5 -1 = -1.25, which is less than -1. So it violates the constraint. Therefore, this is invalid.

So only when we set y₀ and y₁ to the same sign and y₂ to the opposite, we get a valid quadratic. Therefore, 17 might be the maximum.

But let me check if there are other configurations with different y₀, y₁, y₂ that could result in a larger |a| + |b| + |c|.

For example, suppose y₀ = 1, y₁ = 1, y₂ = -1 gives 17. What if we take y₀ = 1, y₁ = something else, and y₂ = something else?

Wait, but if we want to maximize |a| + |b| + |c|, then we need to maximize |2y₁ + 2y₀ - 4y₂| + |4y₂ - y₁ -3y₀| + |y₀|.

This seems complex, but maybe we can consider that y₀, y₁, y₂ are constrained by the fact that the quadratic must not exceed 1 or -1 in between. Therefore, perhaps the maximal |a| + |b| + |c| occurs when y₀, y₁, y₂ are set to ±1, with the appropriate alternations.

Alternatively, let's think of the following: since we already have a quadratic with |a| + |b| + |c| = 17, can we find a quadratic with larger sum?

Suppose we consider another case where f(0) = 1, f(1) = -1, and f(1/2) = something else. Wait, but earlier this led to a violation. However, maybe if we adjust the values?

Wait, perhaps not. Let me think. Suppose we want to maximize |a| + |b| + |c|. If we can have some of a, b, c to be larger in magnitude, maybe by adjusting the points where the function attains the maximum absolute value.

Wait, another approach: Let me consider the general quadratic and try to express |a| + |b| + |c| in terms of some other variables.

Alternatively, think about the maximum of |a| + |b| + |c| given the constraints on f(x). Let me note that at x=0, c is bounded by 1, so |c| ≤ 1. Similarly, at x=1, |a + b + c| ≤ 1. Also, the derivative at x=0 is b, and at x=1 is 2a + b. Maybe not helpful.

Alternatively, let me think about the coefficients in terms of the function's values. For example, if I can relate a, b, c to the function's values at certain points.

But maybe using the concept of the maximum norm. The problem is similar to finding the maximum of the coefficients of a quadratic polynomial bounded by 1 in the maximum norm on [0, 1].

There might be known results for this. For example, in approximation theory, the Chebyshev polynomials are extremal for certain problems.

But since we are dealing with maximum norm on [0, 1], perhaps scaling the Chebyshev polynomial.

The standard Chebyshev polynomial of degree 2 is T₂(t) = 2t² - 1, which has extrema of ±1 on [-1, 1]. If we map the interval [0, 1] to [-1, 1], then t = 2x - 1 as before.

Then, substituting into T₂(t):

T₂(2x - 1) = 2(2x - 1)^2 - 1 = 2(4x² -4x +1) -1 = 8x² -8x +2 -1 = 8x² -8x +1.

Ah! Which is exactly the quadratic we found earlier: 8x² -8x +1. Which has maximum absolute value 1 on [0,1], and |a| + |b| + |c| = 8 + 8 +1 = 17. Therefore, this is the Chebyshev polynomial on [0,1], scaled appropriately. Hence, this might be the extremal polynomial for our problem.

Therefore, it's likely that 17 is the maximum value of |a| + |b| + |c|.

But to confirm, let me see if there's a way to get a larger sum. Suppose someone claims that there is a quadratic with |a| + |b| + |c| > 17, but still satisfying |ax² + bx + c| ≤ 1 on [0,1].

Assume such a quadratic exists. Let's suppose |a| + |b| + |c| = S > 17.

But since |c| ≤ 1, and |a + b + c| ≤ 1, then |a + b| ≤ |a + b + c| + |c| ≤ 1 + 1 = 2. Also, from the Chebyshev polynomial case, we have a relation between a, b, c.

Alternatively, maybe use some inequalities. Let me think.

Let me consider that for any quadratic f(x) = ax² + bx + c with |f(x)| ≤ 1 on [0, 1], then we can bound |a| + |b| + |c|.

Alternatively, express a, b, c in terms of the values of f at three points. For instance, take x=0, x=1, and x=1/2.

As we did earlier, with:

c = f(0),

a + b + c = f(1),

and (a/4) + (b/2) + c = f(1/2).

Which gives us expressions for a, b, c in terms of f(0), f(1), f(1/2). If we assume that f(0)=1, f(1)=1, f(1/2)=-1, we get a=8, b=-8, c=1, which gives S=17. Suppose we try other values. For example, if f(0)=1, f(1)=1, f(1/2)=k, then:

a = 2*1 + 2*1 - 4k = 4 - 4k,

b = 4k - 1 - 3*1 = 4k -4,

c =1.

Then |a| + |b| + |c| = |4 - 4k| + |4k - 4| + 1.

Let me compute this expression. Note that |4 - 4k| + |4k - 4| = 2*|4 -4k| if we consider that 4k -4 = - (4 -4k). Therefore, this is 2*|4 -4k| + 1.

To maximize this expression, we need to maximize |4 -4k|. Since f(1/2)=k must satisfy |k| ≤ 1 (since |f(x)| ≤1 on [0,1]). Therefore, k ∈ [-1, 1].

Therefore, 4 - 4k ranges from 4 -4*(1)=0 to 4 -4*(-1)=8. Therefore, |4 -4k| ranges from 0 to 8. Therefore, the maximum of 2*|4 -4k| +1 is 2*8 +1=17, which occurs when k=-1. Which corresponds to our previous quadratic. Hence, this case indeed gives the maximum of 17.

Similarly, if we take other combinations of f(0), f(1), f(1/2), we might not get higher values. For example, if we take f(0)=1, f(1)=-1, f(1/2)=k, but then the quadratic may not satisfy the constraints, as seen earlier, leading to violations.

Alternatively, if we take other points, such as x=0, x=1, and some other x in between?

Alternatively, suppose we take three points not equally spaced, but maybe that complicates things. However, since the Chebyshev polynomial gives the maximal alternation, it is likely optimal.

Alternatively, think of the problem in terms of linear algebra. Suppose we want to maximize |a| + |b| + |c| given that -1 ≤ ax² + bx + c ≤1 for all x in [0,1]. This is equivalent to maximizing a linear function over a convex set defined by infinitely many constraints. The maximum should occur at an extreme point of this convex set. The extreme points are likely the polynomials which achieve the maximum absolute value at several points, similar to the Chebyshev polynomials.

Therefore, it's reasonable to conjecture that the maximum is 17.

But to be thorough, let me check if there are other quadratics where |a| + |b| + |c| could be larger. For instance, suppose we consider quadratics that are not passing through the three points x=0, x=1, x=1/2, but maybe have different extrema.

Alternatively, consider the case where the quadratic reaches 1 at x=0, and -1 at some point t in (0,1), and also reaches -1 at x=1. Wait, but that would require the quadratic to have two points where it reaches -1, which would mean it has a minimum at two points, but a quadratic can only have one extremum. Hence, impossible unless it's a constant function. Therefore, that's not possible.

Alternatively, suppose the quadratic reaches 1 at x=0 and x=1, and has a minimum somewhere inside the interval. Which is what we already considered, leading to the quadratic 8x² -8x +1. Similarly, if we have a maximum at x=0 and x=1, and minimum somewhere else.

Alternatively, suppose the quadratic reaches 1 at x=0 and at another point, and -1 at x=1. For example, suppose f(0) = 1, f(1) = -1, and f(t) = 1 for some t in (0,1). Then, solving this would give us a quadratic, but we need to ensure that it doesn't exceed 1 or -1 in between. However, when I tried this earlier with t=1/2, the quadratic exceeded the bound. Perhaps with a different t, it could be adjusted?

Let me try with t= something else. Let me suppose that f(0) = 1, f(1) = -1, and f(t) = 1. Let me find a quadratic that satisfies these three conditions.

So, f(0) = c = 1.

f(1) = a + b + c = -1 ⇒ a + b = -2.

f(t) = a t² + b t + c = 1 ⇒ a t² + b t + 1 = 1 ⇒ a t² + b t = 0 ⇒ a t² = -b t ⇒ a t = -b (assuming t ≠ 0).

So from a t = -b, we have b = -a t. Substitute into a + b = -2:

a - a t = -2 ⇒ a(1 - t) = -2 ⇒ a = -2 / (1 - t).

Therefore, b = -a t = 2 t / (1 - t).

Therefore, the quadratic is f(x) = ax² + bx + c = [-2/(1 - t)]x² + [2t/(1 - t)]x + 1.

Now, we need to ensure that |f(x)| ≤ 1 for all x in [0,1]. Let's analyze this quadratic.

First, note that the critical point is at x = -b/(2a) = [-2t/(1 - t)] / [2*(-2/(1 - t))] = [ -2t/(1 - t) ] / [ -4/(1 - t) ] = ( -2t ) / ( -4 ) = t/2.

Therefore, the critical point is at x = t/2. So, depending on the value of t, this critical point is in [0, 1] as long as t ∈ [0, 2], but since t ∈ (0,1), then x = t/2 ∈ (0, 0.5).

Therefore, we need to check the value of the quadratic at x = t/2 and ensure that it lies within [-1, 1].

So compute f(t/2):

f(t/2) = a(t/2)^2 + b(t/2) + c.

Substitute a, b:

= [ -2/(1 - t) ]*(t² / 4) + [ 2t/(1 - t) ]*(t / 2 ) + 1

= [ -2t² / (4(1 - t)) ] + [ 2t² / (2(1 - t)) ] + 1

Simplify:

= [ -t² / (2(1 - t)) ] + [ t² / (1 - t) ] + 1

= [ (-t² + 2t² ) / (2(1 - t)) ] + 1

= [ t² / (2(1 - t)) ] + 1

We need this value to be between -1 and 1. Let's compute:

t² / [ 2(1 - t) ] + 1 ≤ 1 ⇒ t² / [ 2(1 - t) ] ≤ 0 ⇒ since t ∈ (0,1), denominator is positive, numerator t² is positive ⇒ t² / [ 2(1 - t) ] > 0 ⇒ 0 + 1 > 1 ⇒ which would mean f(t/2) > 1, violating the constraint. Therefore, regardless of t ∈ (0,1), this quadratic will have a value greater than 1 at x = t/2. Therefore, this quadratic does not satisfy the constraint. Therefore, this approach doesn't work.

Therefore, the only valid case where the quadratic touches the bounds at three points is when it alternates between 1 and -1 at three points, which is the Chebyshev polynomial case, leading to |a| + |b| + |c| = 17.

Alternatively, is there a quadratic that attains 1 at x=0 and x=1, and has its maximum at the critical point, but has the maximum absolute value at the same time? Wait, in the case of the Chebyshev polynomial, the critical point is where it reaches -1. So this seems to be the maximal case.

Alternatively, what if we consider quadratics with different signs? For example, suppose c = -1, a + b + c = 1, and the critical point is -1. But that would be similar to the previous case.

Alternatively, if I set c = -1, f(1) = 1, and f(1/2) = 1. Then solving:

c = -1,

a + b + c = 1 ⇒ a + b = 2,

at x=1/2: (a/4) + (b/2) + (-1) = 1 ⇒ (a/4) + (b/2) = 2 ⇒ multiply by 4: a + 2b = 8.

Therefore:

a + b = 2,

a + 2b = 8.

Subtracting gives b = 6, then a = 2 - 6 = -4. So quadratic is -4x² + 6x -1.

Check its value at critical point x = -b/(2a) = -6/(2*(-4)) = 6/8 = 3/4.

At x = 3/4:

f(3/4) = -4*(9/16) + 6*(3/4) -1 = -9/4 + 9/2 -1 = (-9/4 + 18/4 - 4/4) = 5/4 = 1.25 > 1. Again violating the constraint. Therefore invalid.

Hence, this approach also doesn't work.

Therefore, after trying various configurations, it seems that the only valid quadratic that achieves the maximal |a| + |b| + |c| without violating the constraints is the Chebyshev polynomial scaled to [0, 1], which gives 17. Hence, I think that 17 is indeed the maximum value.

Therefore, the answer should be \boxed{17}.

**Final Answer**
\boxed{17}

To find the largest possible value of $ |a| + |b| + |c| $ given the constraint that the quadratic expression $ f(x) = ax^2 + bx + c $ satisfies
$$
|f(x)| \leq 1 \quad \text{for all } x \in [0, 1],
$$
we proceed by analyzing the extremal properties of such quadratics and leveraging the structure of the function over the interval.

---

### Key Observations

- The function $ f(x) = ax^2 + bx + c $ is continuous and differentiable on $ [0, 1] $.
- Since it is a quadratic function, it has at most **one critical point** (i.e., a maximum or minimum at $ x = -\frac{b}{2a} $, assuming $ a \ne 0 $).
- The **maximum and minimum** of $ f $ on the interval $ [0,1] $ must therefore occur **either at the endpoints** $ x = 0 $, $ x = 1 $, or at the critical point.
- The given condition $ |f(x)| \leq 1 $ on $ [0,1] $ implies that:
  $$
  -1 \leq f(x) \leq 1 \quad \text{for all } x \in [0, 1].
  $$

---

### Strategy: Extremal Polynomials

A powerful method to tackle such extremal problems is to consider **Chebyshev polynomials**, which are known to provide extremal values under maximum deviation constraints. Specifically, the **Chebyshev polynomial** of degree 2 is:

$$
T_2(t) = 2t^2 - 1,
$$
which oscillates between $ -1 $ and $ 1 $ on the interval $ [-1, 1] $, and is **extremal** in the sense that any other degree-2 polynomial with the same maximum deviation will have **smaller coefficients**.

Now, we shift and scale the interval to match $ [0,1] $:

- Let $ t = 2x - 1 $, so that $ x = 0 \mapsto t = -1 $, $ x = 1 \mapsto t = 1 $, and $ x = \frac{1}{2} \mapsto t = 0 $.

- Then, substitute back:
$$
f(x) = T_2(2x - 1) = 2(2x - 1)^2 - 1 = 8x^2 - 8x + 1.
$$

This is a quadratic function that satisfies:

- $ f(0) = 1 $,
- $ f(1) = 1 $,
- $ f\left( \frac{1}{2} \right) = -1 $,
- and $ |f(x)| \leq 1 $ for all $ x \in [0, 1] $ (as ensured by the construction).

---

### Computing $ |a| + |b| + |c| $

For $ f(x) = 8x^2 - 8x + 1 $, we have:

$$
a = 8, \quad b = -8, \quad c = 1.
$$

Then:

$$
|a| + |b| + |c| = |8| + |-8| + |1| = 8 + 8 + 1 = \boxed{17}.
$$

---

### Verifying Optimality

We now confirm that this is the **maximum possible** value. Any other quadratic polynomial satisfying the given condition would have a maximum deviation of 1 over $ [0,1] $, and the Chebyshev polynomial is known to be the **unique extremal polynomial** that achieves this bound with maximal coefficients.

Several alternate configurations (e.g., using different points or signs for the values of $ f(x) $) were tried, but they either:

- Violated the constraint $ |f(x)| \leq 1 $, or
- Resulted in smaller values of $ |a| + |b| + |c| $.

Thus, none yield a larger value than 17.

---

### Final Answer

$$
\boxed{17}
$$

[Assistant]
Explanation:
Okay, so I need to find the largest possible value of |a| + |b| + |c| given that the absolute value of the quadratic function ax² + bx + c is less than or equal to 1 for all x between 0 and 1. Hmm, let me think about how to approach this.

First, since the quadratic is bounded between -1 and 1 on the interval [0, 1], maybe I can use some properties of polynomials or optimization techniques. Maybe Lagrange multipliers? Or perhaps considering specific points in the interval where the maximum absolute value occurs?

Let me recall that for a quadratic function, the maximum or minimum occurs either at the vertex or at the endpoints of the interval. Since the interval is [0, 1], the critical points are at x = 0, x = 1, and the vertex x = -b/(2a). But I need to ensure that the function doesn't exceed 1 or go below -1 anywhere in between.

But the problem is about maximizing |a| + |b| + |c| under the constraint that |ax² + bx + c| ≤ 1 for x in [0, 1]. So maybe I can find expressions for a, b, c that satisfy the constraints and then compute |a| + |b| + |c|?

Alternatively, perhaps I can use some known inequalities or transformations. Let me think about some specific values.

First, let me note that when x = 0, the function evaluates to c, so |c| ≤ 1. Similarly, when x = 1, the function is a + b + c, so |a + b + c| ≤ 1. Also, the vertex of the quadratic is at x = -b/(2a), but if that lies within [0, 1], then the maximum or minimum could be there. However, if the vertex is outside [0, 1], then the maximum and minimum would occur at the endpoints.

But since the quadratic is bounded between -1 and 1 on [0, 1], we need to ensure that the function doesn't exceed those bounds anywhere in the interval.

But how do I relate this to |a| + |b| + |c|? Maybe I can parametrize the quadratic in terms of its values at certain points?

Another idea: perhaps use Chebyshev polynomials? Because Chebyshev polynomials are known for minimizing the maximum deviation, but I need the opposite—maximizing |a| + |b| + |c| under the maximum deviation constraint. Maybe not directly, but perhaps similar techniques can be applied.

Wait, Chebyshev polynomials have extrema at certain points. Let me recall that the Chebyshev polynomial of degree 2 is T₂(x) = 2x² - 1, which oscillates between -1 and 1 on the interval [-1, 1]. But our interval is [0, 1], so maybe we can adjust it?

If I make a substitution, let me map the interval [0, 1] to [-1, 1]. Let t = 2x - 1, so when x = 0, t = -1, and x = 1, t = 1. Then, substituting x = (t + 1)/2 into the quadratic, we get:

ax² + bx + c = a[(t + 1)/2]^2 + b[(t + 1)/2] + c.

Expanding this:

a[(t² + 2t + 1)/4] + b[(t + 1)/2] + c

= (a/4)t² + (a/2)t + a/4 + (b/2)t + b/2 + c

Combine like terms:

= (a/4)t² + [(a/2) + (b/2)]t + [a/4 + b/2 + c]

So if we want this to be similar to the Chebyshev polynomial T₂(t) = 2t² - 1, which has maximum absolute value 1 on [-1, 1], maybe we can set coefficients accordingly?

But our quadratic is bounded by 1 in absolute value on [0, 1], which corresponds to t in [-1, 1]. So maybe if we scale the Chebyshev polynomial appropriately?

But let me think. Suppose we want our quadratic to attain the maximum absolute value 1 at several points in [0, 1]. For Chebyshev polynomials, they alternate between 1 and -1 at n+1 points, which is maximally oscillating. Maybe if our quadratic is constructed similarly, it could achieve the maximal |a| + |b| + |c|?

Alternatively, maybe consider that to maximize |a| + |b| + |c|, we need the quadratic to have maximum coefficients. However, since the quadratic is constrained to be within [-1, 1] on [0,1], maybe we need to find the quadratic that touches the bounds 1 or -1 at several points, such that the coefficients are as large as possible.

For instance, if we can find a quadratic that reaches 1 at x=0, -1 at x=1, and 1 at some point in between, maybe that would create the maximum coefficients? Or some combination like that.

Alternatively, maybe consider specific cases where the maximum of |ax² + bx + c| is achieved at three points, which would be necessary for the maximum coefficients?

Wait, for a quadratic, the maximum on an interval can be at most two points, unless it's constant. Wait, but since it's a parabola, it can have only one extremum. So the maximum absolute value can be achieved at the endpoints and the critical point. Hence, maybe three points? For example, the maximum is achieved at x=0, x=1, and x=-b/(2a) if that's within [0,1].

So perhaps if we have the quadratic reaching 1 at x=0, -1 at x=1, and 1 at the critical point, or some combination like that. Let me try to set up equations.

Suppose that the quadratic reaches maximum absolute value at three points. For example, let me assume that the maximum is achieved at x=0, x=1, and the critical point. Let me assume that:

At x=0: |c| = 1, so c = 1 or -1.

At x=1: |a + b + c| = 1.

At the critical point x = -b/(2a): |ax² + bx + c| = 1.

But this seems a bit vague. Let me think of a specific example. Suppose we have the quadratic that attains 1 at x=0, -1 at x=1, and 1 at the midpoint x=1/2. Then, we can set up equations.

Let me try that.

Let me assume that:

At x=0: ax² + bx + c = 1 => c = 1.

At x=1: ax² + bx + c = -1 => a + b + c = -1.

At x=1/2: a*(1/2)^2 + b*(1/2) + c = 1 => (a/4) + (b/2) + c = 1.

Since c = 1, substitute into the other equations.

From x=1: a + b + 1 = -1 => a + b = -2.

From x=1/2: (a/4) + (b/2) + 1 = 1 => (a/4) + (b/2) = 0 => multiply by 4: a + 2b = 0.

So we have:

a + b = -2,

a + 2b = 0.

Subtract first equation from the second: b = 2. Then from first equation: a = -2 - b = -2 - 2 = -4.

Thus, the quadratic is -4x² + 2x + 1.

Let me check if this quadratic stays within [-1, 1] on [0, 1].

First, at x=0: 1, which is okay.

At x=1: -4 + 2 + 1 = -1, okay.

At x=1/2: -4*(1/4) + 2*(1/2) + 1 = -1 + 1 + 1 = 1, okay.

Now check the vertex. The critical point is at x = -b/(2a) = -2/(2*(-4)) = -2/(-8) = 1/4.

So at x=1/4, the value is:

-4*(1/4)^2 + 2*(1/4) + 1 = -4*(1/16) + 1/2 + 1 = -1/4 + 1/2 + 1 = (-1/4 + 1/2) + 1 = 1/4 + 1 = 5/4 = 1.25.

Wait, that's 1.25, which is greater than 1! That violates the condition. So this quadratic does not satisfy the constraint. So my assumption that it reaches 1 at x=0, -1 at x=1, and 1 at x=1/2 is wrong because it exceeds 1 in between. Hence, this approach is flawed.

Hmm, so maybe I need to ensure that the maximum is not exceeded. So perhaps the maximum is achieved at the endpoints and the critical point, but with alternating signs? Let me think again.

Alternatively, maybe set up the quadratic so that it reaches 1 at two points and -1 at another, such that the maximum is achieved at those points. However, for a quadratic, it can only have one extremum. So if it reaches 1 at two points and -1 at one, or similar. But how?

Alternatively, maybe use the Lagrange interpolation. Suppose the quadratic has maximum absolute value 1 at three points: x=0, x=1/2, x=1. Let me suppose that it alternates between 1 and -1 at these points. For instance, suppose at x=0: 1, x=1/2: -1, x=1: 1. Then we can find the quadratic passing through these points.

Let me try that. Let me denote the quadratic as f(x) = ax² + bx + c.

We have:

f(0) = c = 1.

f(1/2) = a*(1/4) + b*(1/2) + c = -1.

f(1) = a + b + c = 1.

So substituting c = 1 into the other equations:

From f(1/2): (a/4) + (b/2) + 1 = -1 => (a/4) + (b/2) = -2.

Multiply by 4: a + 2b = -8.

From f(1): a + b + 1 = 1 => a + b = 0.

So from a + b = 0, we have a = -b. Substitute into a + 2b = -8: -b + 2b = -8 => b = -8. Then a = 8.

Thus, the quadratic is 8x² - 8x + 1.

Let me check its value at the critical point. The critical point is at x = -b/(2a) = 8/(2*8) = 8/16 = 1/2. Wait, that's the same as the point where we have f(1/2) = -1. So the vertex is at x=1/2, which is already one of the points where the function reaches -1. Therefore, the maximum and minimum are at the endpoints and the vertex. But let's check the value at x=1/2, which is -1, and the other points are 1 at x=0 and x=1. Let's check other points in between to see if the function stays within [-1, 1].

Take x=1/4:

f(1/4) = 8*(1/16) - 8*(1/4) + 1 = 0.5 - 2 + 1 = -0.5, which is between -1 and 1.

Take x=3/4:

f(3/4) = 8*(9/16) -8*(3/4) + 1 = 4.5 - 6 + 1 = -0.5, same as above.

So it seems that this quadratic oscillates between 1 and -1 at x=0, 1/2, 1, and stays within the bounds in between. Therefore, it satisfies the condition |f(x)| ≤ 1 on [0, 1].

Now, compute |a| + |b| + |c| for this quadratic: |8| + |-8| + |1| = 8 + 8 + 1 = 17. Hmm, that's quite large. Is this the maximum?

But wait, let me check if there are other quadratics with higher |a| + |b| + |c|.

Alternatively, maybe considering another configuration where the quadratic reaches 1 at three points or alternates signs differently? But since a quadratic is determined by three points, if we set it up to pass through three points with alternating signs, maybe that's the maximum?

Wait, but maybe there's a quadratic that reaches 1 at x=0, -1 at x=1, and 1 at some other point, but arranged such that the maximum is not exceeded. However, earlier when I tried such a case, the vertex exceeded the bound. Let me try again with different points.

Alternatively, maybe take another approach. Let me use the method of Lagrange multipliers. Let me try to maximize |a| + |b| + |c| subject to |ax² + bx + c| ≤ 1 for all x in [0,1].

But dealing with absolute values and inequalities over an interval is complicated. Maybe instead, consider that the maximum of |ax² + bx + c| over [0, 1] is ≤ 1. Therefore, the maximum of ax² + bx + c is ≤ 1, and the minimum is ≥ -1.

Alternatively, since the quadratic is continuous on [0,1], we can write that:

-1 ≤ ax² + bx + c ≤ 1 for all x in [0,1].

To maximize |a| + |b| + |c|, perhaps we need to find the extremal quadratic which touches the bounds at several points. The previous example gave |a| + |b| + |c| = 17. Let me check if that's indeed the maximum.

Alternatively, consider that when we have the quadratic 8x² - 8x + 1, which we saw satisfies the condition. Also, let's check the value of |a| + |b| + |c| for other quadratics.

Alternatively, suppose that we set up the quadratic to reach maximum at x=0, x=1, and the critical point. Wait, but if we do that, we might have similar equations.

But maybe another approach: express the quadratic in terms of its values at three points. Since a quadratic is uniquely determined by its values at three points. Let me use the three points x=0, x=1/2, x=1, and set up equations such that the quadratic is extremal.

Alternatively, maybe use some linear algebra. Let me think.

Let me consider that the quadratic function is bounded by 1 in absolute value on [0, 1]. Let me denote f(x) = ax² + bx + c.

We have the constraints:

For all x in [0, 1], -1 ≤ f(x) ≤ 1.

We need to maximize |a| + |b| + |c|.

This seems like an optimization problem with infinitely many constraints. However, perhaps the maximum is attained when the function f(x) touches the bounds ±1 at several points, which is similar to the Chebyshev polynomial idea.

Earlier, with the quadratic 8x² - 8x + 1, we saw that it reaches 1 at x=0 and x=1, and -1 at x=1/2, and stays within the bounds. So this is similar to the Chebyshev polynomial oscillating between extremes.

Alternatively, maybe there's a quadratic that oscillates between 1 and -1 three times? But a quadratic can only have one extremum. So maximum two alternations? Wait, on the interval [0, 1], if the function has one extremum, then it can alternate between max and min at most two times. For example, going from 1 at x=0, down to -1 at x=1/2, then back up to 1 at x=1. Which is exactly what the quadratic 8x² - 8x + 1 does. So this is the maximal oscillation for a quadratic on [0, 1], which might lead to the maximal coefficients.

So if this is the case, then |a| + |b| + |c| = 8 + 8 + 1 = 17. But let me check if there are other quadratics with higher values.

Alternatively, consider if we set the quadratic to reach 1 at x=0, -1 at x=1, and then have another extremum? Wait, but the extremum is already at x=1/2.

Alternatively, what if we have maximum at x=0, minimum at x=1, and maximum at some other point? But the quadratic can't have two maxima or minima. So only one extremum.

Alternatively, suppose the quadratic has maximum at x=0 and x=1, and minimum somewhere in between? But then the quadratic would be concave down. For example, like -8x² + 8x - 1? Wait, let me check.

Wait, let me try this quadratic: Let me assume that f(0) = 1, f(1) = 1, and f(1/2) = -1. Wait, that's the same as before but with a sign change. Wait, actually, the previous quadratic was 8x² -8x +1, which gives f(1/2) = -1. If we take the negative of that, -8x² +8x -1, then f(0) = -1, f(1) = -1, and f(1/2)=1. But then |a| + |b| + |c| is same: |-8| + |8| + |-1| = 8 + 8 +1 = 17. So same value.

Alternatively, maybe there's a way to have larger coefficients?

Wait, suppose I set up the quadratic so that it reaches 1 at x=0, -1 at x=1, and also reaches 1 at some other point within the interval. But earlier when I tried that, the vertex exceeded the bound. However, maybe if we adjust the points?

Wait, let me try again. Suppose f(0) = 1, f(1) = -1, and f(1/2) = 1. Then set up equations:

f(0) = c = 1.

f(1) = a + b + c = -1 => a + b = -2.

f(1/2) = a*(1/4) + b*(1/2) + c = 1 => (a/4) + (b/2) + 1 = 1 => (a/4) + (b/2) = 0 => a + 2b = 0.

So solving:

From a + b = -2 and a + 2b = 0, subtract first equation from second:

b = 2. Then a = -2 - b = -4. So quadratic is -4x² + 2x + 1. Earlier, this had a vertex at x=1/4 with value f(1/4) = -4*(1/16) + 2*(1/4) + 1 = -0.25 + 0.5 + 1 = 1.25. Which is over the bound. Therefore, this quadratic violates the condition. Therefore, this approach doesn't work.

Therefore, maybe the previous quadratic that attains 1 at x=0 and x=1, and -1 at x=1/2 is the maximal in some sense. But how to confirm?

Alternatively, maybe consider other extremal points. Let me think about how to maximize |a| + |b| + |c|. Since c is evaluated at x=0, which is bounded by 1 in absolute value, so |c| ≤ 1. Similarly, a + b + c is evaluated at x=1, so |a + b + c| ≤ 1. Also, if we can find other constraints.

Alternatively, let me think in terms of linear algebra. Suppose I want to maximize |a| + |b| + |c|, given that for x in [0, 1], |ax² + bx + c| ≤ 1.

But this is a complex constraint. Maybe we can use some specific points.

For example, the maximum of |a| + |b| + |c| could be achieved when c is maximized or minimized, along with a and b. However, since c is bounded by 1, and similarly a + b + c is bounded by 1, perhaps there's a relationship between these terms.

Alternatively, perhaps consider that the maximum of |a| + |b| + |c| occurs when the quadratic is extremal in some direction.

Alternatively, think of the problem as an optimization problem with variables a, b, c, and constraints that for all x in [0, 1], -1 ≤ ax² + bx + c ≤ 1. To maximize |a| + |b| + |c|.

But solving this requires knowledge of the maximum of the quadratic on [0, 1]. However, since the quadratic is continuous, its maximum and minimum on [0,1] occur either at critical points or endpoints.

Therefore, maybe the constraints can be written as:

-1 ≤ c ≤ 1,

-1 ≤ a + b + c ≤ 1,

and if the critical point x_v = -b/(2a) is in [0, 1], then -1 ≤ a(x_v)^2 + b(x_v) + c ≤ 1.

But this seems complicated due to the dependence on a and b.

Alternatively, maybe consider specific cases where the quadratic takes the maximum absolute value at the endpoints and the critical point. For example, suppose that the maximum is achieved at three points: x=0, x=1, and x= t in (0,1). Then, we can set up equations such that f(0) = 1, f(1) = 1, f(t) = -1 or something like that.

Alternatively, maybe use the previous example where the quadratic reaches 1 at x=0, -1 at x=1/2, and 1 at x=1. Which gives us |a| + |b| + |c| = 17. Let me check if there is a quadratic with larger |a| + |b| + |c|.

Alternatively, maybe consider another quadratic which attains maximum absolute value at different points. For example, suppose it attains 1 at x=0 and x=1, and another value at the critical point. Wait, but we saw that if we set it to attain 1 at x=0, x=1, and -1 at x=1/2, then it gives a quadratic with |a| + |b| + |c| = 17. What if we do something else?

Alternatively, suppose we take the quadratic that attains 1 at x=0, and -1 at x=1, and then another extrema. Wait, but the quadratic can only have one extremum. So maybe the extremum is somewhere else.

Alternatively, think of a quadratic that has maximum slope or something? Not sure.

Alternatively, consider that the maximum of |a| + |b| + |c| could be achieved when a, b, c have the same sign? Or different signs?

Wait, in the previous example, a=8, b=-8, c=1. So |a| + |b| + |c| = 17. If I can find a quadratic with larger coefficients, that would be better.

Alternatively, consider the case where the quadratic reaches 1 at x=0, and -1 at x=1, and also reaches another extremum. But how?

Alternatively, think of the following: if I can write the quadratic in terms of its values at x=0, x=1, and x=1/2, then express a, b, c in terms of those values.

Let me denote:

Let’s say f(0) = c = y₀,

f(1) = a + b + c = y₁,

f(1/2) = a(1/4) + b(1/2) + c = y₂.

Then, solving for a, b, c:

From f(0): c = y₀.

From f(1): a + b = y₁ - y₀.

From f(1/2): (a/4) + (b/2) = y₂ - y₀.

Multiply the third equation by 4: a + 2b = 4(y₂ - y₀).

So subtract the second equation (a + b = y₁ - y₀) from this: (a + 2b) - (a + b) = b = 4(y₂ - y₀) - (y₁ - y₀) = 4y₂ - 4y₀ - y₁ + y₀ = 4y₂ - y₁ - 3y₀.

Then, from a + b = y₁ - y₀, we get a = y₁ - y₀ - b = y₁ - y₀ - [4y₂ - y₁ - 3y₀] = y₁ - y₀ - 4y₂ + y₁ + 3y₀ = 2y₁ + 2y₀ - 4y₂.

Therefore,

a = 2y₁ + 2y₀ - 4y₂,

b = 4y₂ - y₁ - 3y₀,

c = y₀.

Now, if we want |a| + |b| + |c| to be maximized, given that |y₀| ≤ 1, |y₁| ≤ 1, |y₂| ≤ 1, and also the maximum of |ax² + bx + c| on [0,1] is ≤ 1.

But this seems complicated, but maybe we can consider the case where the maximum is achieved at the three points x=0, x=1/2, x=1. So if we set y₀, y₁, y₂ to be ±1, such that the quadratic oscillates between 1 and -1.

For example, if we set y₀ = 1, y₁ = 1, y₂ = -1, which gives us the quadratic we discussed earlier:

a = 2*1 + 2*1 - 4*(-1) = 2 + 2 + 4 = 8,

b = 4*(-1) - 1 - 3*1 = -4 -1 -3 = -8,

c = 1.

So that gives |a| + |b| + |c| = 8 + 8 + 1 = 17.

Alternatively, if we set y₀ = 1, y₁ = -1, y₂ = 1, then:

a = 2*(-1) + 2*1 - 4*1 = -2 + 2 -4 = -4,

b = 4*1 - (-1) - 3*1 = 4 +1 -3 = 2,

c = 1.

Which gives quadratic -4x² + 2x +1, which as we saw earlier has a maximum of 1.25 at x=1/4, which violates the constraint. Therefore, this case is invalid.

Therefore, the previous case with y₀=1, y₁=1, y₂=-1 is valid, giving the maximum of 17.

Alternatively, what if we set y₀ = -1, y₁ = -1, y₂ = 1? Then similarly, a = 2*(-1) + 2*(-1) - 4*1 = -2 -2 -4 = -8, b = 4*1 - (-1) - 3*(-1) = 4 +1 +3 = 8, c = -1. So the quadratic is -8x² +8x -1, which gives |a| + |b| + |c| = 8 +8 +1=17 again.

So that also gives 17. So that seems consistent.

Alternatively, if we set y₀ = -1, y₁ = 1, y₂ = -1? Let's check:

a = 2*1 + 2*(-1) - 4*(-1) = 2 -2 +4 = 4,

b = 4*(-1) -1 -3*(-1) = -4 -1 +3 = -2,

c = -1.

Thus, quadratic is 4x² -2x -1.

Check its maximum. Let's compute the vertex at x = -b/(2a) = 2/(2*4) = 1/4. At x=1/4:

f(1/4) = 4*(1/16) -2*(1/4) -1 = 0.25 -0.5 -1 = -1.25, which is less than -1. So it violates the constraint. Therefore, this is invalid.

So only when we set y₀ and y₁ to the same sign and y₂ to the opposite, we get a valid quadratic. Therefore, 17 might be the maximum.

But let me check if there are other configurations with different y₀, y₁, y₂ that could result in a larger |a| + |b| + |c|.

For example, suppose y₀ = 1, y₁ = 1, y₂ = -1 gives 17. What if we take y₀ = 1, y₁ = something else, and y₂ = something else?

Wait, but if we want to maximize |a| + |b| + |c|, then we need to maximize |2y₁ + 2y₀ - 4y₂| + |4y₂ - y₁ -3y₀| + |y₀|.

This seems complex, but maybe we can consider that y₀, y₁, y₂ are constrained by the fact that the quadratic must not exceed 1 or -1 in between. Therefore, perhaps the maximal |a| + |b| + |c| occurs when y₀, y₁, y₂ are set to ±1, with the appropriate alternations.

Alternatively, let's think of the following: since we already have a quadratic with |a| + |b| + |c| = 17, can we find a quadratic with larger sum?

Suppose we consider another case where f(0) = 1, f(1) = -1, and f(1/2) = something else. Wait, but earlier this led to a violation. However, maybe if we adjust the values?

Wait, perhaps not. Let me think. Suppose we want to maximize |a| + |b| + |c|. If we can have some of a, b, c to be larger in magnitude, maybe by adjusting the points where the function attains the maximum absolute value.

Wait, another approach: Let me consider the general quadratic and try to express |a| + |b| + |c| in terms of some other variables.

Alternatively, think about the maximum of |a| + |b| + |c| given the constraints on f(x). Let me note that at x=0, c is bounded by 1, so |c| ≤ 1. Similarly, at x=1, |a + b + c| ≤ 1. Also, the derivative at x=0 is b, and at x=1 is 2a + b. Maybe not helpful.

Alternatively, let me think about the coefficients in terms of the function's values. For example, if I can relate a, b, c to the function's values at certain points.

But maybe using the concept of the maximum norm. The problem is similar to finding the maximum of the coefficients of a quadratic polynomial bounded by 1 in the maximum norm on [0, 1].

There might be known results for this. For example, in approximation theory, the Chebyshev polynomials are extremal for certain problems.

But since we are dealing with maximum norm on [0, 1], perhaps scaling the Chebyshev polynomial.

The standard Chebyshev polynomial of degree 2 is T₂(t) = 2t² - 1, which has extrema of ±1 on [-1, 1]. If we map the interval [0, 1] to [-1, 1], then t = 2x - 1 as before.

Then, substituting into T₂(t):

T₂(2x - 1) = 2(2x - 1)^2 - 1 = 2(4x² -4x +1) -1 = 8x² -8x +2 -1 = 8x² -8x +1.

Ah! Which is exactly the quadratic we found earlier: 8x² -8x +1. Which has maximum absolute value 1 on [0,1], and |a| + |b| + |c| = 8 + 8 +1 = 17. Therefore, this is the Chebyshev polynomial on [0,1], scaled appropriately. Hence, this might be the extremal polynomial for our problem.

Therefore, it's likely that 17 is the maximum value of |a| + |b| + |c|.

But to confirm, let me see if there's a way to get a larger sum. Suppose someone claims that there is a quadratic with |a| + |b| + |c| > 17, but still satisfying |ax² + bx + c| ≤ 1 on [0,1].

Assume such a quadratic exists. Let's suppose |a| + |b| + |c| = S > 17.

But since |c| ≤ 1, and |a + b + c| ≤ 1, then |a + b| ≤ |a + b + c| + |c| ≤ 1 + 1 = 2. Also, from the Chebyshev polynomial case, we have a relation between a, b, c.

Alternatively, maybe use some inequalities. Let me think.

Let me consider that for any quadratic f(x) = ax² + bx + c with |f(x)| ≤ 1 on [0, 1], then we can bound |a| + |b| + |c|.

Alternatively, express a, b, c in terms of the values of f at three points. For instance, take x=0, x=1, and x=1/2.

As we did earlier, with:

c = f(0),

a + b + c = f(1),

and (a/4) + (b/2) + c = f(1/2).

Which gives us expressions for a, b, c in terms of f(0), f(1), f(1/2). If we assume that f(0)=1, f(1)=1, f(1/2)=-1, we get a=8, b=-8, c=1, which gives S=17. Suppose we try other values. For example, if f(0)=1, f(1)=1, f(1/2)=k, then:

a = 2*1 + 2*1 - 4k = 4 - 4k,

b = 4k - 1 - 3*1 = 4k -4,

c =1.

Then |a| + |b| + |c| = |4 - 4k| + |4k - 4| + 1.

Let me compute this expression. Note that |4 - 4k| + |4k - 4| = 2*|4 -4k| if we consider that 4k -4 = - (4 -4k). Therefore, this is 2*|4 -4k| + 1.

To maximize this expression, we need to maximize |4 -4k|. Since f(1/2)=k must satisfy |k| ≤ 1 (since |f(x)| ≤1 on [0,1]). Therefore, k ∈ [-1, 1].

Therefore, 4 - 4k ranges from 4 -4*(1)=0 to 4 -4*(-1)=8. Therefore, |4 -4k| ranges from 0 to 8. Therefore, the maximum of 2*|4 -4k| +1 is 2*8 +1=17, which occurs when k=-1. Which corresponds to our previous quadratic. Hence, this case indeed gives the maximum of 17.

Similarly, if we take other combinations of f(0), f(1), f(1/2), we might not get higher values. For example, if we take f(0)=1, f(1)=-1, f(1/2)=k, but then the quadratic may not satisfy the constraints, as seen earlier, leading to violations.

Alternatively, if we take other points, such as x=0, x=1, and some other x in between?

Alternatively, suppose we take three points not equally spaced, but maybe that complicates things. However, since the Chebyshev polynomial gives the maximal alternation, it is likely optimal.

Alternatively, think of the problem in terms of linear algebra. Suppose we want to maximize |a| + |b| + |c| given that -1 ≤ ax² + bx + c ≤1 for all x in [0,1]. This is equivalent to maximizing a linear function over a convex set defined by infinitely many constraints. The maximum should occur at an extreme point of this convex set. The extreme points are likely the polynomials which achieve the maximum absolute value at several points, similar to the Chebyshev polynomials.

Therefore, it's reasonable to conjecture that the maximum is 17.

But to be thorough, let me check if there are other quadratics where |a| + |b| + |c| could be larger. For instance, suppose we consider quadratics that are not passing through the three points x=0, x=1, x=1/2, but maybe have different extrema.

Alternatively, consider the case where the quadratic reaches 1 at x=0, and -1 at some point t in (0,1), and also reaches -1 at x=1. Wait, but that would require the quadratic to have two points where it reaches -1, which would mean it has a minimum at two points, but a quadratic can only have one extremum. Hence, impossible unless it's a constant function. Therefore, that's not possible.

Alternatively, suppose the quadratic reaches 1 at x=0 and x=1, and has a minimum somewhere inside the interval. Which is what we already considered, leading to the quadratic 8x² -8x +1. Similarly, if we have a maximum at x=0 and x=1, and minimum somewhere else.

Alternatively, suppose the quadratic reaches 1 at x=0 and at another point, and -1 at x=1. For example, suppose f(0) = 1, f(1) = -1, and f(t) = 1 for some t in (0,1). Then, solving this would give us a quadratic, but we need to ensure that it doesn't exceed 1 or -1 in between. However, when I tried this earlier with t=1/2, the quadratic exceeded the bound. Perhaps with a different t, it could be adjusted?

Let me try with t= something else. Let me suppose that f(0) = 1, f(1) = -1, and f(t) = 1. Let me find a quadratic that satisfies these three conditions.

So, f(0) = c = 1.

f(1) = a + b + c = -1 ⇒ a + b = -2.

f(t) = a t² + b t + c = 1 ⇒ a t² + b t + 1 = 1 ⇒ a t² + b t = 0 ⇒ a t² = -b t ⇒ a t = -b (assuming t ≠ 0).

So from a t = -b, we have b = -a t. Substitute into a + b = -2:

a - a t = -2 ⇒ a(1 - t) = -2 ⇒ a = -2 / (1 - t).

Therefore, b = -a t = 2 t / (1 - t).

Therefore, the quadratic is f(x) = ax² + bx + c = [-2/(1 - t)]x² + [2t/(1 - t)]x + 1.

Now, we need to ensure that |f(x)| ≤ 1 for all x in [0,1]. Let's analyze this quadratic.

First, note that the critical point is at x = -b/(2a) = [-2t/(1 - t)] / [2*(-2/(1 - t))] = [ -2t/(1 - t) ] / [ -4/(1 - t) ] = ( -2t ) / ( -4 ) = t/2.

Therefore, the critical point is at x = t/2. So, depending on the value of t, this critical point is in [0, 1] as long as t ∈ [0, 2], but since t ∈ (0,1), then x = t/2 ∈ (0, 0.5).

Therefore, we need to check the value of the quadratic at x = t/2 and ensure that it lies within [-1, 1].

So compute f(t/2):

f(t/2) = a(t/2)^2 + b(t/2) + c.

Substitute a, b:

= [ -2/(1 - t) ]*(t² / 4) + [ 2t/(1 - t) ]*(t / 2 ) + 1

= [ -2t² / (4(1 - t)) ] + [ 2t² / (2(1 - t)) ] + 1

Simplify:

= [ -t² / (2(1 - t)) ] + [ t² / (1 - t) ] + 1

= [ (-t² + 2t² ) / (2(1 - t)) ] + 1

= [ t² / (2(1 - t)) ] + 1

We need this value to be between -1 and 1. Let's compute:

t² / [ 2(1 - t) ] + 1 ≤ 1 ⇒ t² / [ 2(1 - t) ] ≤ 0 ⇒ since t ∈ (0,1), denominator is positive, numerator t² is positive ⇒ t² / [ 2(1 - t) ] > 0 ⇒ 0 + 1 > 1 ⇒ which would mean f(t/2) > 1, violating the constraint. Therefore, regardless of t ∈ (0,1), this quadratic will have a value greater than 1 at x = t/2. Therefore, this quadratic does not satisfy the constraint. Therefore, this approach doesn't work.

Therefore, the only valid case where the quadratic touches the bounds at three points is when it alternates between 1 and -1 at three points, which is the Chebyshev polynomial case, leading to |a| + |b| + |c| = 17.

Alternatively, is there a quadratic that attains 1 at x=0 and x=1, and has its maximum at the critical point, but has the maximum absolute value at the same time? Wait, in the case of the Chebyshev polynomial, the critical point is where it reaches -1. So this seems to be the maximal case.

Alternatively, what if we consider quadratics with different signs? For example, suppose c = -1, a + b + c = 1, and the critical point is -1. But that would be similar to the previous case.

Alternatively, if I set c = -1, f(1) = 1, and f(1/2) = 1. Then solving:

c = -1,

a + b + c = 1 ⇒ a + b = 2,

at x=1/2: (a/4) + (b/2) + (-1) = 1 ⇒ (a/4) + (b/2) = 2 ⇒ multiply by 4: a + 2b = 8.

Therefore:

a + b = 2,

a + 2b = 8.

Subtracting gives b = 6, then a = 2 - 6 = -4. So quadratic is -4x² + 6x -1.

Check its value at critical point x = -b/(2a) = -6/(2*(-4)) = 6/8 = 3/4.

At x = 3/4:

f(3/4) = -4*(9/16) + 6*(3/4) -1 = -9/4 + 9/2 -1 = (-9/4 + 18/4 - 4/4) = 5/4 = 1.25 > 1. Again violating the constraint. Therefore invalid.

Hence, this approach also doesn't work.

Therefore, after trying various configurations, it seems that the only valid quadratic that achieves the maximal |a| + |b| + |c| without violating the constraints is the Chebyshev polynomial scaled to [0, 1], which gives 17. Hence, I think that 17 is indeed the maximum value.

Therefore, the answer should be \boxed{17}.

**Final Answer**
\boxed{17}

To find the largest possible value of $ |a| + |b| + |c| $ given the constraint that the quadratic expression $ f(x) = ax^2 + bx + c $ satisfies
$$
|f(x)| \leq 1 \quad \text{for all } x \in [0, 1],
$$
we proceed by analyzing the extremal properties of such quadratics and leveraging the structure of the function over the interval.

---

### Key Observations

- The function $ f(x) = ax^2 + bx + c $ is continuous and differentiable on $ [0, 1] $.
- Since it is a quadratic function, it has at most **one critical point** (i.e., a maximum or minimum at $ x = -\frac{b}{2a} $, assuming $ a \ne 0 $).
- The **maximum and minimum** of $ f $ on the interval $ [0,1] $ must therefore occur **either at the endpoints** $ x = 0 $, $ x = 1 $, or at the critical point.
- The given condition $ |f(x)| \leq 1 $ on $ [0,1] $ implies that:
  $$
  -1 \leq f(x) \leq 1 \quad \text{for all } x \in [0, 1].
  $$

---

### Strategy: Extremal Polynomials

A powerful method to tackle such extremal problems is to consider **Chebyshev polynomials**, which are known to provide extremal values under maximum deviation constraints. Specifically, the **Chebyshev polynomial** of degree 2 is:

$$
T_2(t) = 2t^2 - 1,
$$
which oscillates between $ -1 $ and $ 1 $ on the interval $ [-1, 1] $, and is **extremal** in the sense that any other degree-2 polynomial with the same maximum deviation will have **smaller coefficients**.

Now, we shift and scale the interval to match $ [0,1] $:

- Let $ t = 2x - 1 $, so that $ x = 0 \mapsto t = -1 $, $ x = 1 \mapsto t = 1 $, and $ x = \frac{1}{2} \mapsto t = 0 $.

- Then, substitute back:
$$
f(x) = T_2(2x - 1) = 2(2x - 1)^2 - 1 = 8x^2 - 8x + 1.
$$

This is a quadratic function that satisfies:

- $ f(0) = 1 $,
- $ f(1) = 1 $,
- $ f\left( \frac{1}{2} \right) = -1 $,
- and $ |f(x)| \leq 1 $ for all $ x \in [0, 1] $ (as ensured by the construction).

---

### Computing $ |a| + |b| + |c| $

For $ f(x) = 8x^2 - 8x + 1 $, we have:

$$
a = 8, \quad b = -8, \quad c = 1.
$$

Then:

$$
|a| + |b| + |c| = |8| + |-8| + |1| = 8 + 8 + 1 = \boxed{17}.
$$

---

### Verifying Optimality

We now confirm that this is the **maximum possible** value. Any other quadratic polynomial satisfying the given condition would have a maximum deviation of 1 over $ [0,1] $, and the Chebyshev polynomial is known to be the **unique extremal polynomial** that achieves this bound with maximal coefficients.

Several alternate configurations (e.g., using different points or signs for the values of $ f(x) $) were tried, but they either:

- Violated the constraint $ |f(x)| \leq 1 $, or
- Resulted in smaller values of $ |a| + |b| + |c| $.

Thus, none yield a larger value than 17.

---

### Final Answer

$$
\boxed{17}
$$

Solution:
Setting $x = 0,$ we get $|c| \le 1.$  Setting $x = 1,$ we get
\[|a + b + c| \le 1.\]Setting $x = \frac{1}{2},$ we get
\[\left| \frac{a}{4} + \frac{b}{2} + c \right| \le 1.\]Let
\begin{align*}
p &= c, \\
q &= \frac{a}{4} + \frac{b}{2} + c, \\
r &= a + b + c,
\end{align*}so $-1 \le p,$ $q,$ $r \le 1.$  Solving for $a,$ $b,$ and $c,$ we find
\begin{align*}
a &= 2p - 4q + 2r, \\
b &= -3p + 4q - r, \\
c &= p.
\end{align*}Hence, by Triangle Inequality,
\begin{align*}
|a| &= |2p - 4q + 2r| \le |2p| + |4q| + |2r| = 8, \\
|b| &= |-3p + 4q - r| \le |3p| + |4q| + |r| = 8, \\
|c| &= |p| \le 1.
\end{align*}Therefore, $|a| + |b| + |c| = 8 + 8 + 1 = 17.$

Consider the quadratic $f(x) = 8x^2 - 8x + 1.$  We can write
\[f(x) = 8 \left( x - \frac{1}{2} \right)^2 - 1.\]For $0 \le x \le 1,$ $0 \le \left( x - \frac{1}{2} \right)^2 \le \frac{1}{4},$ so $-1 \le f(x) \le 1.$

Therefore, the largest possible value of $|a| + |b| + |c|$ is $\boxed{17}.$


Explanation:
Okay, so William Sydney Porter tried to calculate (-3 + 4i) divided by (1 + 2i), but he missed the minus sign and ended up with (3 + 4i)/(1 + 2i) = 11/5 - 2/5 i. Now, the question is asking what answer he should have obtained, meaning the correct result of (-3 + 4i)/(1 + 2i).

First, I need to recall how to divide complex numbers. The standard method is to multiply the numerator and the denominator by the complex conjugate of the denominator. This process rationalizes the denominator, similar to how we rationalize denominators with radicals.

So, let me write down the original problem again: (-3 + 4i) divided by (1 + 2i). To perform this division, I need to multiply both the numerator and the denominator by the complex conjugate of the denominator. The complex conjugate of (1 + 2i) is (1 - 2i).

So, let me set that up:

[(-3 + 4i)/(1 + 2i)] * [(1 - 2i)/(1 - 2i)]

Multiplying the numerators together and the denominators together. Let me compute the denominator first because that might be simpler. The denominator will be (1 + 2i)(1 - 2i). Using the formula (a + b)(a - b) = a² - b². Here, a is 1 and b is 2i.

So, (1)^2 - (2i)^2 = 1 - 4i². Since i² is -1, this becomes 1 - 4(-1) = 1 + 4 = 5. So the denominator simplifies to 5. That seems straightforward.

Now, let me compute the numerator: (-3 + 4i)(1 - 2i). Let me expand this using distributive property (FOIL method).

First, multiply -3 by 1: -3*1 = -3

Then, -3 multiplied by -2i: -3*(-2i) = 6i

Next, 4i multiplied by 1: 4i*1 = 4i

Lastly, 4i multiplied by -2i: 4i*(-2i) = -8i²

Now, combine all those terms: -3 + 6i + 4i - 8i²

Combine like terms: The real parts are -3 and the -8i². The imaginary parts are 6i and 4i.

First, let me handle the imaginary parts: 6i + 4i = 10i

Now the real parts: -3 - 8i². Remember that i² = -1, so -8i² becomes -8*(-1) = +8. Therefore, the real part is -3 + 8 = 5.

Putting it all together, the numerator simplifies to 5 + 10i.

Therefore, the entire expression becomes (5 + 10i)/5. Let me split this into two fractions: 5/5 + 10i/5, which simplifies to 1 + 2i.

Wait, that seems too simple. Let me check my calculations again because if the numerator was 5 + 10i divided by 5, then yes, that would be 1 + 2i. Let me verify the multiplication step again because sometimes signs can be tricky.

Original numerator multiplication:

(-3 + 4i)(1 - 2i)

First term: -3*1 = -3

Outer term: -3*(-2i) = +6i

Inner term: 4i*1 = +4i

Last term: 4i*(-2i) = -8i²

So altogether:

-3 + 6i + 4i -8i²

Combine like terms:

-3 + (6i + 4i) + (-8i²) = -3 + 10i + (-8)(-1) = -3 + 10i + 8 = (-3 + 8) + 10i = 5 + 10i. Yes, that's correct. Then divide by 5, getting 1 + 2i.

Wait, but according to the problem statement, William missed the minus sign and computed (3 + 4i)/(1 + 2i) and got 11/5 - 2/5 i. Let me check that as well to see if my answer is consistent. Maybe I can cross-verify.

Let me compute (3 + 4i)/(1 + 2i) using the same method. Multiply numerator and denominator by (1 - 2i):

[(3 + 4i)(1 - 2i)] / [(1 + 2i)(1 - 2i)]

Denominator is again 1^2 - (2i)^2 = 1 - 4i² = 1 + 4 = 5. So denominator is 5.

Numerator: 3*1 + 3*(-2i) + 4i*1 + 4i*(-2i) = 3 - 6i + 4i - 8i². Combine like terms:

Real parts: 3 - 8i². Imaginary parts: -6i + 4i = -2i.

Again, -8i² is +8. So real part is 3 + 8 = 11, and imaginary part is -2i. Thus numerator is 11 - 2i. Therefore, (11 - 2i)/5 = 11/5 - (2/5)i. Which matches the problem statement.

Therefore, William's incorrect calculation was (3 + 4i)/(1 + 2i) = 11/5 - 2/5 i. So that checks out.

Therefore, the correct calculation should be (-3 + 4i)/(1 + 2i) = 1 + 2i? Wait, but let me check once again.

Alternatively, maybe I made a mistake in the sign when computing the original numerator. Let me check again:

Original numerator: (-3 + 4i)(1 - 2i). Let me compute this again step by step.

First, expand:

-3 * 1 = -3

-3 * (-2i) = +6i

4i * 1 = +4i

4i * (-2i) = -8i²

Adding them up: -3 + 6i + 4i -8i². Combine terms:

-3 + (6i + 4i) + (-8i²) = -3 + 10i + 8 = 5 + 10i. So yes, that's correct. Divided by 5, gives 1 + 2i.

But maybe I should check with another method. Let me convert both complex numbers to polar form and perform division? Although that might be more complicated, but just to verify.

First, let me recall that division of complex numbers in polar form is done by dividing their magnitudes and subtracting their angles.

But let me try for the original problem: (-3 + 4i) divided by (1 + 2i).

First, compute magnitude and angle for numerator and denominator.

Numerator: z1 = -3 + 4i.

Magnitude: |z1| = sqrt((-3)^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5.

Angle θ1: arctangent of (4/(-3)), but since the real part is negative and imaginary part is positive, it's in the second quadrant. So θ1 = π - arctan(4/3).

Denominator: z2 = 1 + 2i.

Magnitude: |z2| = sqrt(1^2 + 2^2) = sqrt(1 + 4) = sqrt(5).

Angle θ2: arctangent(2/1) = arctan(2), which is in the first quadrant.

Therefore, the division z1/z2 has magnitude |z1| / |z2| = 5 / sqrt(5) = sqrt(5).

Angle: θ1 - θ2 = [π - arctan(4/3)] - arctan(2).

Hmm, this seems more complicated. Let me compute the result in rectangular form. If we know that the result should be 1 + 2i, let's check its magnitude and angle.

Result: 1 + 2i.

Magnitude: sqrt(1^2 + 2^2) = sqrt(1 + 4) = sqrt(5). Which matches the magnitude from before.

Angle: arctangent(2/1) = arctan(2). But according to the previous angle calculation, the angle of z1/z2 is θ1 - θ2 = [π - arctan(4/3)] - arctan(2). Let me see if this equals arctan(2).

Wait, this seems conflicting. Let me compute θ1 - θ2.

Let me compute θ1 = π - arctan(4/3). θ2 = arctan(2). Then θ1 - θ2 = π - arctan(4/3) - arctan(2).

But if the result is 1 + 2i, which has angle θ = arctan(2). So π - arctan(4/3) - arctan(2) = arctan(2)?

Wait, that would imply π - arctan(4/3) = 2 arctan(2). Hmm, not sure. Maybe this approach is not the best. Let me check numerically.

Compute θ1 = π - arctan(4/3). Let me compute arctan(4/3). 4/3 is approximately 1.333, arctan(1.333) is approximately 53.13 degrees (since tan(53.13°) ≈ 1.333). Therefore θ1 is π - 53.13°, which is 180° - 53.13° = 126.87°.

θ2 is arctan(2) ≈ 63.43°.

Therefore, θ1 - θ2 ≈ 126.87° - 63.43° ≈ 63.44°, which is arctan(2). Therefore, the angle of the result is arctan(2), which matches the angle of 1 + 2i. Therefore, this confirms that the result is indeed 1 + 2i.

Therefore, my previous calculation seems correct.

Alternatively, perhaps I can use another approach by expressing the original fraction as a + bi and solving for a and b.

Let me denote (-3 + 4i)/(1 + 2i) = a + bi, where a and b are real numbers. Then, multiplying both sides by (1 + 2i):

-3 + 4i = (a + bi)(1 + 2i)

Expanding the right-hand side:

a*(1) + a*(2i) + bi*(1) + bi*(2i) = a + 2ai + bi + 2bi²

Simplify:

a + (2a + b)i + 2b*(-1) = a - 2b + (2a + b)i

Therefore, equating real and imaginary parts:

Real: a - 2b = -3

Imaginary: 2a + b = 4

So, we have the system of equations:

1) a - 2b = -3

2) 2a + b = 4

Let me solve this system. Let me use substitution or elimination. Let's do elimination.

Multiply equation 1 by 2: 2a - 4b = -6

Subtract equation 2 from this:

(2a - 4b) - (2a + b) = -6 - 4

Which is 2a - 4b - 2a - b = -10

Simplify: -5b = -10 => b = 2

Then substitute back into equation 2: 2a + 2 = 4 => 2a = 2 => a = 1

Therefore, a = 1, b = 2, so the result is 1 + 2i.

Therefore, this method also gives the same answer. Therefore, I can be confident that the correct answer is 1 + 2i.

But let me check once again with another approach. Suppose I use decimal approximations. Let me compute (-3 + 4i)/(1 + 2i).

First, compute 1 + 2i ≈ 1 + 2i. Let me compute its magnitude sqrt(1 + 4) = sqrt(5) ≈ 2.236. The angle is arctan(2/1) ≈ 63.43 degrees.

The numerator -3 + 4i has magnitude 5, as before, angle in second quadrant. Let me compute its angle: arctan(4/-3). Since it's in second quadrant, arctan(4/3) ≈ 53.13 degrees, so total angle is 180 - 53.13 ≈ 126.87 degrees.

Dividing the magnitudes: 5 / 2.236 ≈ 2.236. Which is sqrt(5) ≈ 2.236. The angle is 126.87 - 63.43 ≈ 63.44 degrees, which is arctan(2) as before. So converting back to rectangular form: magnitude sqrt(5) at angle arctan(2).

Which is sqrt(5)*(cos(arctan(2)) + i sin(arctan(2))). Let me compute cos(arctan(2)) and sin(arctan(2)).

If θ = arctan(2), then tanθ = 2 = opposite/adjacent = 2/1. Therefore, imagine a right triangle with opposite side 2, adjacent side 1, hypotenuse sqrt(1 + 4) = sqrt(5). Therefore:

cosθ = adjacent/hypotenuse = 1/sqrt(5)

sinθ = opposite/hypotenuse = 2/sqrt(5)

Therefore, the rectangular form is sqrt(5)*(1/sqrt(5) + i*(2/sqrt(5))) = 1 + 2i.

Yes, so this confirms again that the result is 1 + 2i. Therefore, I think this is definitely the correct answer.

But just to make sure I didn't make a mistake in the original problem. The problem says William missed the minus sign, so he calculated (3 + 4i)/(1 + 2i) and got 11/5 - 2/5 i, which we confirmed earlier. Then the correct answer is (-3 + 4i)/(1 + 2i) = 1 + 2i.

Alternatively, if I wanted to check with decimal approximations, let me compute both results numerically.

First, compute (3 + 4i)/(1 + 2i). Let me do it numerically.

First, multiply numerator and denominator by (1 - 2i):

Numerator: (3 + 4i)(1 - 2i) = 3*1 + 3*(-2i) + 4i*1 + 4i*(-2i) = 3 - 6i + 4i - 8i² = 3 - 2i + 8 = 11 - 2i, as before. Divided by 5, (11 - 2i)/5 = 11/5 - 2/5 i ≈ 2.2 - 0.4i.

Now, if we compute (-3 + 4i)/(1 + 2i), which we found to be 1 + 2i, which is 1 + 2i ≈ 1 + 2i.

Alternatively, let me compute this division with decimals. Let me write (-3 + 4i) divided by (1 + 2i). Let me approximate it numerically.

Alternatively, think of it as a complex number division. Let me use the formula:

If z = a + bi and w = c + di, then z/w = [(ac + bd) + (bc - ad)i]/(c² + d²)

Wait, let me recall the formula.

The formula for division is:

(a + bi)/(c + di) = [(ac + bd) + (bc - ad)i]/(c² + d²)

Wait, let me check:

Multiply numerator and denominator by (c - di):

[(a + bi)(c - di)] / (c² + d²)

Expanding numerator: a*c - a*d i + b i c - b i d i = (ac + bd) + (bc - ad)i

Yes, so the formula is [(ac + bd) + (bc - ad)i]/(c² + d²)

So for our original problem, numerator is (-3 + 4i), denominator is (1 + 2i). Therefore:

a = -3, b = 4, c = 1, d = 2.

Therefore, compute:

ac + bd = (-3)(1) + (4)(2) = -3 + 8 = 5

bc - ad = (4)(1) - (-3)(2) = 4 + 6 = 10

Denominator: c² + d² = 1² + 2² = 1 + 4 = 5

Therefore, the result is (5 + 10i)/5 = 1 + 2i, same as before.

Therefore, this confirms again the answer.

Therefore, after multiple methods and checks, I can conclude that the correct answer is 1 + 2i.

Therefore, William should have obtained 1 + 2i.

**Final Answer**
The correct answer William should have obtained is \boxed{1 + 2i}.

To solve the complex division problem
$$
\frac{-3 + 4i}{1 + 2i}
$$
we follow the standard process for dividing complex numbers: multiply both the numerator and denominator by the **complex conjugate of the denominator**. This ensures the denominator becomes a real number, simplifying the expression.

---

### Step 1: Multiply by the Conjugate
The complex conjugate of $1 + 2i$ is $1 - 2i$. Multiply both numerator and denominator by this conjugate:

$$
\frac{-3 + 4i}{1 + 2i} \cdot \frac{1 - 2i}{1 - 2i}
$$

---

### Step 2: Simplify the Denominator
Using the identity $(a + b)(a - b) = a^2 - b^2$:

$$
(1 + 2i)(1 - 2i) = 1^2 - (2i)^2 = 1 - 4i^2 = 1 - 4(-1) = 5
$$

So the denominator is simply 5.

---

### Step 3: Expand the Numerator
Now compute the product of the numerator:

$$
(-3 + 4i)(1 - 2i)
$$

Using the distributive property (FOIL method):

- $ -3 \cdot 1 = -3 $
- $ -3 \cdot (-2i) = +6i $
- $ 4i \cdot 1 = +4i $
- $ 4i \cdot (-2i) = -8i^2 = +8 $ (since $i^2 = -1$)

So, combining all parts:

$$
-3 + 6i + 4i + 8 = ( -3 + 8 ) + ( 6i + 4i ) = 5 + 10i
$$

---

### Step 4: Divide by the Denominator

$$
\frac{5 + 10i}{5} = \frac{5}{5} + \frac{10i}{5} = 1 + 2i
$$

---

### Final Answer

Thus, the correct result of the calculation is:

$$
\boxed{1 + 2i}
$$

[Assistant]
Explanation:
Okay, so William Sydney Porter tried to calculate (-3 + 4i) divided by (1 + 2i), but he missed the minus sign and ended up with (3 + 4i)/(1 + 2i) = 11/5 - 2/5 i. Now, the question is asking what answer he should have obtained, meaning the correct result of (-3 + 4i)/(1 + 2i).

First, I need to recall how to divide complex numbers. The standard method is to multiply the numerator and the denominator by the complex conjugate of the denominator. This process rationalizes the denominator, similar to how we rationalize denominators with radicals.

So, let me write down the original problem again: (-3 + 4i) divided by (1 + 2i). To perform this division, I need to multiply both the numerator and the denominator by the complex conjugate of the denominator. The complex conjugate of (1 + 2i) is (1 - 2i).

So, let me set that up:

[(-3 + 4i)/(1 + 2i)] * [(1 - 2i)/(1 - 2i)]

Multiplying the numerators together and the denominators together. Let me compute the denominator first because that might be simpler. The denominator will be (1 + 2i)(1 - 2i). Using the formula (a + b)(a - b) = a² - b². Here, a is 1 and b is 2i.

So, (1)^2 - (2i)^2 = 1 - 4i². Since i² is -1, this becomes 1 - 4(-1) = 1 + 4 = 5. So the denominator simplifies to 5. That seems straightforward.

Now, let me compute the numerator: (-3 + 4i)(1 - 2i). Let me expand this using distributive property (FOIL method).

First, multiply -3 by 1: -3*1 = -3

Then, -3 multiplied by -2i: -3*(-2i) = 6i

Next, 4i multiplied by 1: 4i*1 = 4i

Lastly, 4i multiplied by -2i: 4i*(-2i) = -8i²

Now, combine all those terms: -3 + 6i + 4i - 8i²

Combine like terms: The real parts are -3 and the -8i². The imaginary parts are 6i and 4i.

First, let me handle the imaginary parts: 6i + 4i = 10i

Now the real parts: -3 - 8i². Remember that i² = -1, so -8i² becomes -8*(-1) = +8. Therefore, the real part is -3 + 8 = 5.

Putting it all together, the numerator simplifies to 5 + 10i.

Therefore, the entire expression becomes (5 + 10i)/5. Let me split this into two fractions: 5/5 + 10i/5, which simplifies to 1 + 2i.

Wait, that seems too simple. Let me check my calculations again because if the numerator was 5 + 10i divided by 5, then yes, that would be 1 + 2i. Let me verify the multiplication step again because sometimes signs can be tricky.

Original numerator multiplication:

(-3 + 4i)(1 - 2i)

First term: -3*1 = -3

Outer term: -3*(-2i) = +6i

Inner term: 4i*1 = +4i

Last term: 4i*(-2i) = -8i²

So altogether:

-3 + 6i + 4i -8i²

Combine like terms:

-3 + (6i + 4i) + (-8i²) = -3 + 10i + (-8)(-1) = -3 + 10i + 8 = (-3 + 8) + 10i = 5 + 10i. Yes, that's correct. Then divide by 5, getting 1 + 2i.

Wait, but according to the problem statement, William missed the minus sign and computed (3 + 4i)/(1 + 2i) and got 11/5 - 2/5 i. Let me check that as well to see if my answer is consistent. Maybe I can cross-verify.

Let me compute (3 + 4i)/(1 + 2i) using the same method. Multiply numerator and denominator by (1 - 2i):

[(3 + 4i)(1 - 2i)] / [(1 + 2i)(1 - 2i)]

Denominator is again 1^2 - (2i)^2 = 1 - 4i² = 1 + 4 = 5. So denominator is 5.

Numerator: 3*1 + 3*(-2i) + 4i*1 + 4i*(-2i) = 3 - 6i + 4i - 8i². Combine like terms:

Real parts: 3 - 8i². Imaginary parts: -6i + 4i = -2i.

Again, -8i² is +8. So real part is 3 + 8 = 11, and imaginary part is -2i. Thus numerator is 11 - 2i. Therefore, (11 - 2i)/5 = 11/5 - (2/5)i. Which matches the problem statement.

Therefore, William's incorrect calculation was (3 + 4i)/(1 + 2i) = 11/5 - 2/5 i. So that checks out.

Therefore, the correct calculation should be (-3 + 4i)/(1 + 2i) = 1 + 2i? Wait, but let me check once again.

Alternatively, maybe I made a mistake in the sign when computing the original numerator. Let me check again:

Original numerator: (-3 + 4i)(1 - 2i). Let me compute this again step by step.

First, expand:

-3 * 1 = -3

-3 * (-2i) = +6i

4i * 1 = +4i

4i * (-2i) = -8i²

Adding them up: -3 + 6i + 4i -8i². Combine terms:

-3 + (6i + 4i) + (-8i²) = -3 + 10i + 8 = 5 + 10i. So yes, that's correct. Divided by 5, gives 1 + 2i.

But maybe I should check with another method. Let me convert both complex numbers to polar form and perform division? Although that might be more complicated, but just to verify.

First, let me recall that division of complex numbers in polar form is done by dividing their magnitudes and subtracting their angles.

But let me try for the original problem: (-3 + 4i) divided by (1 + 2i).

First, compute magnitude and angle for numerator and denominator.

Numerator: z1 = -3 + 4i.

Magnitude: |z1| = sqrt((-3)^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5.

Angle θ1: arctangent of (4/(-3)), but since the real part is negative and imaginary part is positive, it's in the second quadrant. So θ1 = π - arctan(4/3).

Denominator: z2 = 1 + 2i.

Magnitude: |z2| = sqrt(1^2 + 2^2) = sqrt(1 + 4) = sqrt(5).

Angle θ2: arctangent(2/1) = arctan(2), which is in the first quadrant.

Therefore, the division z1/z2 has magnitude |z1| / |z2| = 5 / sqrt(5) = sqrt(5).

Angle: θ1 - θ2 = [π - arctan(4/3)] - arctan(2).

Hmm, this seems more complicated. Let me compute the result in rectangular form. If we know that the result should be 1 + 2i, let's check its magnitude and angle.

Result: 1 + 2i.

Magnitude: sqrt(1^2 + 2^2) = sqrt(1 + 4) = sqrt(5). Which matches the magnitude from before.

Angle: arctangent(2/1) = arctan(2). But according to the previous angle calculation, the angle of z1/z2 is θ1 - θ2 = [π - arctan(4/3)] - arctan(2). Let me see if this equals arctan(2).

Wait, this seems conflicting. Let me compute θ1 - θ2.

Let me compute θ1 = π - arctan(4/3). θ2 = arctan(2). Then θ1 - θ2 = π - arctan(4/3) - arctan(2).

But if the result is 1 + 2i, which has angle θ = arctan(2). So π - arctan(4/3) - arctan(2) = arctan(2)?

Wait, that would imply π - arctan(4/3) = 2 arctan(2). Hmm, not sure. Maybe this approach is not the best. Let me check numerically.

Compute θ1 = π - arctan(4/3). Let me compute arctan(4/3). 4/3 is approximately 1.333, arctan(1.333) is approximately 53.13 degrees (since tan(53.13°) ≈ 1.333). Therefore θ1 is π - 53.13°, which is 180° - 53.13° = 126.87°.

θ2 is arctan(2) ≈ 63.43°.

Therefore, θ1 - θ2 ≈ 126.87° - 63.43° ≈ 63.44°, which is arctan(2). Therefore, the angle of the result is arctan(2), which matches the angle of 1 + 2i. Therefore, this confirms that the result is indeed 1 + 2i.

Therefore, my previous calculation seems correct.

Alternatively, perhaps I can use another approach by expressing the original fraction as a + bi and solving for a and b.

Let me denote (-3 + 4i)/(1 + 2i) = a + bi, where a and b are real numbers. Then, multiplying both sides by (1 + 2i):

-3 + 4i = (a + bi)(1 + 2i)

Expanding the right-hand side:

a*(1) + a*(2i) + bi*(1) + bi*(2i) = a + 2ai + bi + 2bi²

Simplify:

a + (2a + b)i + 2b*(-1) = a - 2b + (2a + b)i

Therefore, equating real and imaginary parts:

Real: a - 2b = -3

Imaginary: 2a + b = 4

So, we have the system of equations:

1) a - 2b = -3

2) 2a + b = 4

Let me solve this system. Let me use substitution or elimination. Let's do elimination.

Multiply equation 1 by 2: 2a - 4b = -6

Subtract equation 2 from this:

(2a - 4b) - (2a + b) = -6 - 4

Which is 2a - 4b - 2a - b = -10

Simplify: -5b = -10 => b = 2

Then substitute back into equation 2: 2a + 2 = 4 => 2a = 2 => a = 1

Therefore, a = 1, b = 2, so the result is 1 + 2i.

Therefore, this method also gives the same answer. Therefore, I can be confident that the correct answer is 1 + 2i.

But let me check once again with another approach. Suppose I use decimal approximations. Let me compute (-3 + 4i)/(1 + 2i).

First, compute 1 + 2i ≈ 1 + 2i. Let me compute its magnitude sqrt(1 + 4) = sqrt(5) ≈ 2.236. The angle is arctan(2/1) ≈ 63.43 degrees.

The numerator -3 + 4i has magnitude 5, as before, angle in second quadrant. Let me compute its angle: arctan(4/-3). Since it's in second quadrant, arctan(4/3) ≈ 53.13 degrees, so total angle is 180 - 53.13 ≈ 126.87 degrees.

Dividing the magnitudes: 5 / 2.236 ≈ 2.236. Which is sqrt(5) ≈ 2.236. The angle is 126.87 - 63.43 ≈ 63.44 degrees, which is arctan(2) as before. So converting back to rectangular form: magnitude sqrt(5) at angle arctan(2).

Which is sqrt(5)*(cos(arctan(2)) + i sin(arctan(2))). Let me compute cos(arctan(2)) and sin(arctan(2)).

If θ = arctan(2), then tanθ = 2 = opposite/adjacent = 2/1. Therefore, imagine a right triangle with opposite side 2, adjacent side 1, hypotenuse sqrt(1 + 4) = sqrt(5). Therefore:

cosθ = adjacent/hypotenuse = 1/sqrt(5)

sinθ = opposite/hypotenuse = 2/sqrt(5)

Therefore, the rectangular form is sqrt(5)*(1/sqrt(5) + i*(2/sqrt(5))) = 1 + 2i.

Yes, so this confirms again that the result is 1 + 2i. Therefore, I think this is definitely the correct answer.

But just to make sure I didn't make a mistake in the original problem. The problem says William missed the minus sign, so he calculated (3 + 4i)/(1 + 2i) and got 11/5 - 2/5 i, which we confirmed earlier. Then the correct answer is (-3 + 4i)/(1 + 2i) = 1 + 2i.

Alternatively, if I wanted to check with decimal approximations, let me compute both results numerically.

First, compute (3 + 4i)/(1 + 2i). Let me do it numerically.

First, multiply numerator and denominator by (1 - 2i):

Numerator: (3 + 4i)(1 - 2i) = 3*1 + 3*(-2i) + 4i*1 + 4i*(-2i) = 3 - 6i + 4i - 8i² = 3 - 2i + 8 = 11 - 2i, as before. Divided by 5, (11 - 2i)/5 = 11/5 - 2/5 i ≈ 2.2 - 0.4i.

Now, if we compute (-3 + 4i)/(1 + 2i), which we found to be 1 + 2i, which is 1 + 2i ≈ 1 + 2i.

Alternatively, let me compute this division with decimals. Let me write (-3 + 4i) divided by (1 + 2i). Let me approximate it numerically.

Alternatively, think of it as a complex number division. Let me use the formula:

If z = a + bi and w = c + di, then z/w = [(ac + bd) + (bc - ad)i]/(c² + d²)

Wait, let me recall the formula.

The formula for division is:

(a + bi)/(c + di) = [(ac + bd) + (bc - ad)i]/(c² + d²)

Wait, let me check:

Multiply numerator and denominator by (c - di):

[(a + bi)(c - di)] / (c² + d²)

Expanding numerator: a*c - a*d i + b i c - b i d i = (ac + bd) + (bc - ad)i

Yes, so the formula is [(ac + bd) + (bc - ad)i]/(c² + d²)

So for our original problem, numerator is (-3 + 4i), denominator is (1 + 2i). Therefore:

a = -3, b = 4, c = 1, d = 2.

Therefore, compute:

ac + bd = (-3)(1) + (4)(2) = -3 + 8 = 5

bc - ad = (4)(1) - (-3)(2) = 4 + 6 = 10

Denominator: c² + d² = 1² + 2² = 1 + 4 = 5

Therefore, the result is (5 + 10i)/5 = 1 + 2i, same as before.

Therefore, this confirms again the answer.

Therefore, after multiple methods and checks, I can conclude that the correct answer is 1 + 2i.

Therefore, William should have obtained 1 + 2i.

**Final Answer**
The correct answer William should have obtained is \boxed{1 + 2i}.

To solve the complex division problem
$$
\frac{-3 + 4i}{1 + 2i}
$$
we follow the standard process for dividing complex numbers: multiply both the numerator and denominator by the **complex conjugate of the denominator**. This ensures the denominator becomes a real number, simplifying the expression.

---

### Step 1: Multiply by the Conjugate
The complex conjugate of $1 + 2i$ is $1 - 2i$. Multiply both numerator and denominator by this conjugate:

$$
\frac{-3 + 4i}{1 + 2i} \cdot \frac{1 - 2i}{1 - 2i}
$$

---

### Step 2: Simplify the Denominator
Using the identity $(a + b)(a - b) = a^2 - b^2$:

$$
(1 + 2i)(1 - 2i) = 1^2 - (2i)^2 = 1 - 4i^2 = 1 - 4(-1) = 5
$$

So the denominator is simply 5.

---

### Step 3: Expand the Numerator
Now compute the product of the numerator:

$$
(-3 + 4i)(1 - 2i)
$$

Using the distributive property (FOIL method):

- $ -3 \cdot 1 = -3 $
- $ -3 \cdot (-2i) = +6i $
- $ 4i \cdot 1 = +4i $
- $ 4i \cdot (-2i) = -8i^2 = +8 $ (since $i^2 = -1$)

So, combining all parts:

$$
-3 + 6i + 4i + 8 = ( -3 + 8 ) + ( 6i + 4i ) = 5 + 10i
$$

---

### Step 4: Divide by the Denominator

$$
\frac{5 + 10i}{5} = \frac{5}{5} + \frac{10i}{5} = 1 + 2i
$$

---

### Final Answer

Thus, the correct result of the calculation is:

$$
\boxed{1 + 2i}
$$

Solution:
To perform division in complex numbers, we multiply both the numerator and the denominator by the conjugate of the denominator. In this case, the conjugate of $1+2i$ is $1-2i$. Multiplying: \begin{align*}
\frac{-3+4i}{1+2i}&=\frac{(-3+4i)(1-2i)}{(1+2i)(1-2i)}\\
&=\frac{-3+4i+6i-8i^2}{1+2i-2i-4i^2}\\
&=\frac{5+10i}{5}\\
&=\boxed{1+2i}
\end{align*}