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, I know that the quadratic is bounded between -1 and 1 on the interval [0, 1]. So, for every x in [0, 1], the value of ax² + bx + c is within that range. My goal is to maximize |a| + |b| + |c|.

I remember that sometimes problems like this involve using specific points in the interval to set up equations or inequalities. Since the quadratic is bounded on [0, 1], maybe I can evaluate it at certain key points, like x = 0, x = 1, and maybe the vertex of the parabola? Let me try that.

Let me first note the values of the quadratic at x = 0 and x = 1:

At x = 0: |c| ≤ 1, so c is between -1 and 1.

At x = 1: |a + b + c| ≤ 1.

Also, the vertex of the parabola is at x = -b/(2a). But since the interval is [0, 1], the vertex might lie inside this interval or outside. If it's inside, then the maximum or minimum of the quadratic could be there. However, since the quadratic is bounded between -1 and 1 on the entire interval, the maximum absolute value could occur at the endpoints or at the vertex.

But maybe I need to consider more than just the endpoints. For example, if the quadratic has its maximum or minimum inside the interval, then that point must also satisfy |ax² + bx + c| ≤ 1.

Alternatively, maybe I can use some known inequalities or optimization techniques. Since we need to maximize |a| + |b| + |c|, perhaps I can consider different cases based on the signs of a, b, c? But that might get complicated.

Alternatively, maybe I can use Lagrange multipliers? But since there are infinitely many constraints (for all x in [0,1]), that might not be straightforward. Maybe there's a better way.

Wait, I remember something about Chebyshev polynomials. They are extremal in some sense for polynomial approximation. 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 there's a similar polynomial?

Let me check. If I can find a quadratic polynomial that oscillates between -1 and 1 on [0, 1], then maybe that polynomial would have the maximum coefficients? Because Chebyshev polynomials are known for minimizing the maximum deviation, but here we want to maximize the coefficients given the maximum deviation.

Alternatively, maybe I can perform a change of variables to map [0, 1] to [-1, 1]. Let me try that. Let t = 2x - 1, so when x = 0, t = -1, and when x = 1, t = 1. Then x = (t + 1)/2. Substitute into the quadratic:

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

Let me compute that:

First, expand [(t + 1)/2]^2 = (t² + 2t + 1)/4.

So:

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 I denote this as A t² + B t + C, where:

A = a/4,

B = (a + b)/2,

C = a/4 + b/2 + c.

Now, since the original polynomial is bounded by 1 in absolute value on t ∈ [-1, 1], and Chebyshev polynomials are extremal for such problems. The Chebyshev polynomial of degree 2 is T2(t) = 2t² - 1, which oscillates between -1 and 1 on [-1, 1]. So if I can adjust the coefficients to match this, maybe that would give me the maximum coefficients?

But let me think. If I want the transformed polynomial A t² + B t + C to be equal to T2(t) = 2t² - 1, then:

A = 2,

B = 0,

C = -1.

But then, let's check what that would mean for a, b, c.

From A = a/4 = 2 ⇒ a = 8.

From B = (a + b)/2 = 0 ⇒ (8 + b)/2 = 0 ⇒ 8 + b = 0 ⇒ b = -8.

From C = a/4 + b/2 + c = -1 ⇒ 8/4 + (-8)/2 + c = -1 ⇒ 2 - 4 + c = -1 ⇒ -2 + c = -1 ⇒ c = 1.

So then, the original polynomial would be 8x² - 8x + 1. Let me check if this polynomial is bounded by 1 on [0, 1].

Let me compute its values at x = 0, x = 1, and check the vertex.

At x = 0: 0 - 0 + 1 = 1. Okay, that's within the bound.

At x = 1: 8 - 8 + 1 = 1. Also within the bound.

The vertex is at x = -b/(2a) = 8/(2*8) = 0.5. So at x = 0.5:

8*(0.25) - 8*(0.5) + 1 = 2 - 4 + 1 = -1. So that's -1, which is within the bound. So the polynomial oscillates between 1 and -1 at x=0, x=0.5, and x=1. Therefore, it's exactly the Chebyshev polynomial scaled appropriately.

So this polynomial meets the condition |ax² + bx + c| ≤ 1 on [0, 1], and in this case, |a| + |b| + |c| = |8| + |-8| + |1| = 8 + 8 + 1 = 17.

Is this the maximum? Maybe. But I need to check if there are other polynomials that could give a higher value.

Alternatively, maybe there's a different polynomial that attains a higher |a| + |b| + |c|. Let me think.

Alternatively, maybe I can use the Lagrange interpolation. Suppose that the maximum of |ax² + bx + c| is achieved at three points in [0, 1], which would be the case for the Chebyshev polynomial. Since the Chebyshev polynomials have the minimal maximum norm for a monic polynomial, but here we are trying to maximize the coefficients given the maximum norm.

Alternatively, maybe I can use the fact that the maximum of |a| + |b| + |c| occurs when the polynomial reaches the maximum absolute value at several points. For example, if the polynomial reaches 1 at x=0, -1 at x=0.5, and 1 at x=1, which is exactly the case for the polynomial we found.

But maybe there's another configuration where the polynomial reaches 1 or -1 at different points, leading to a higher |a| + |b| + |c|?

Alternatively, let's think about the maximum of |a| + |b| + |c|. Since c is the value at x=0, which is between -1 and 1. Similarly, a + b + c is the value at x=1, also between -1 and 1. Also, the value at the vertex is between -1 and 1.

But how can we relate these to |a| + |b| + |c|? Maybe we can express a, b, c in terms of the values at certain points. For example, if we know the values at x=0, x=1, and the vertex, we can solve for a, b, c. But since the vertex is dependent on a and b, this might get complicated.

Alternatively, maybe we can use some inequalities. Let me think. Let me consider that for x in [0,1], |ax² + bx + c| ≤ 1. Let me try to express a, b, c in terms of the function values at certain points. For example, if I know the function at x=0, x=1, and x=1/2, then I can set up equations.

Let me denote f(x) = ax² + bx + c. Then:

f(0) = c,

f(1) = a + b + c,

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

If I assume that f(0) = 1, f(1/2) = -1, f(1) = 1, then we can solve for a, b, c. Which is exactly the case of the Chebyshev polynomial we considered earlier. Let me check:

From f(0) = c = 1,

From f(1) = a + b + c = 1,

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

So substituting c = 1 into the other equations:

a + b + 1 = 1 ⇒ a + b = 0 ⇒ b = -a.

Then, substituting into the third equation:

(a/4) + (-a/2) + 1 = -1 ⇒ (-a/4) + 1 = -1 ⇒ -a/4 = -2 ⇒ a = 8. Then b = -8, c = 1. Which is the same as before. So that gives |a| + |b| + |c| = 17.

But maybe if we take different values? For example, suppose that f(0) = 1, f(1) = -1, and f(1/2) = 1. Then solving:

c = 1,

a + b + c = -1 ⇒ a + b = -2,

(a/4) + (b/2) + c = 1 ⇒ (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 second: b = 2. Then from a + 2 = -2 ⇒ a = -4. Then c = 1. So the polynomial is -4x² + 2x + 1. Let me check its maximum absolute value on [0,1].

At x=0: 1, okay.

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

At x=0.5: -4*(0.25) + 2*(0.5) + 1 = -1 + 1 + 1 = 1, okay.

What about the vertex? The vertex is at x = -b/(2a) = -2/(2*(-4)) = -2 / (-8) = 1/4. So x=0.25. Let me compute f(0.25):

-4*(0.0625) + 2*(0.25) + 1 = -0.25 + 0.5 + 1 = 1.25. Wait, that's 1.25, which is greater than 1. That violates the condition. So this polynomial doesn't satisfy |f(x)| ≤ 1 on [0,1]. Therefore, this approach doesn't work. So even though we set the function to reach 1, -1, 1 at three points, the maximum in between is higher. Therefore, this polynomial is invalid.

So that suggests that choosing three points with alternating signs might not be sufficient unless the polynomial is the Chebyshev polynomial. Therefore, maybe the Chebyshev polynomial is the one that attains the maximum.

Alternatively, maybe there's another polynomial where the maximum is achieved at different points. Let me think. For example, suppose that the maximum is achieved at x=0, x=1, and somewhere else. But how?

Alternatively, maybe I can use the concept of the maximum of |a| + |b| + |c|. Let me think about how to express |a| + |b| + |c| in terms of the function values.

Alternatively, maybe I can use the fact that for x in [0,1], the polynomial is bounded by 1. Let me consider evaluating the polynomial at specific points to get inequalities involving a, b, c.

We already have:

At x=0: |c| ≤ 1.

At x=1: |a + b + c| ≤ 1.

Also, maybe consider the derivative? The maximum of the polynomial could be at the endpoints or at critical points. But since we have the bound on the entire interval, maybe we can use some integral or other method?

Alternatively, think of the problem as an optimization problem with constraints. Let me try to set up the problem.

We need to maximize |a| + |b| + |c| subject to |ax² + bx + c| ≤ 1 for all x in [0,1].

This is a constrained optimization problem. However, dealing with infinitely many constraints is difficult. But perhaps we can use the fact that the maximum of |ax² + bx + c| on [0,1] is achieved at certain points, so maybe we can use the maximum at those points as constraints.

In particular, for the maximum to be achieved at three points, as in the Chebyshev case. But how to ensure that?

Alternatively, maybe use the method of Lagrange multipliers with some constraints. But since there are infinitely many constraints, it's not straightforward. However, if we assume that the maximum is achieved at three points, then we can set up equations for those points.

Alternatively, think of the problem as follows: since the polynomial is bounded between -1 and 1 on [0,1], then we can write:

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

To maximize |a| + |b| + |c|, perhaps we need to consider the extremal cases where the polynomial touches the bounds 1 or -1 at several points. For example, the Chebyshev polynomial touches 1 and -1 alternately at three points, which might be the case for maximum coefficients.

Alternatively, maybe there's a way to express |a| + |b| + |c| in terms of the maximum of the polynomial. But I don't recall a direct formula for that.

Alternatively, think of specific values. For example, suppose that c = 1 (maximum possible). Then, we need to maximize |a| + |b| + 1. But then, we have constraints on a and b such that |ax² + bx + 1| ≤ 1 for all x in [0,1]. Let me see.

So, for all x in [0,1], -1 ≤ ax² + bx + 1 ≤ 1. Therefore, ax² + bx + 1 ≤ 1 ⇒ ax² + bx ≤ 0, and ax² + bx + 1 ≥ -1 ⇒ ax² + bx ≥ -2.

So, for all x in [0,1], ax² + bx ≤ 0 and ax² + bx ≥ -2.

Similarly, if c = 1, then maybe we can find a and b such that ax² + bx is between -2 and 0 on [0,1]. But how does that help us maximize |a| + |b|?

Alternatively, maybe consider that the maximum of |a| + |b| + |c| occurs when c is at its maximum or minimum. But I need to check.

Alternatively, think of the problem in terms of linear algebra. Let me consider that the maximum of |a| + |b| + |c| is achieved when the signs of a, b, c are chosen to maximize the sum. However, the constraints are non-linear because of the absolute value.

Alternatively, maybe consider different cases based on the signs of a, b, c. For example, suppose that a, b, c are all positive. Then |a| + |b| + |c| = a + b + c. But since the polynomial is bounded by 1 on [0,1], then at x=1, a + b + c ≤ 1, so the sum a + b + c ≤ 1. But this is probably not the maximum, since if we have negative coefficients, maybe we can get a larger sum.

Alternatively, suppose that c is positive, and a and b are negative. Then |a| + |b| + |c| = -a - b + c. Maybe this can be larger?

Alternatively, maybe the maximum occurs when the polynomial alternates between 1 and -1 at several points, which would be similar to the Chebyshev polynomial. Since the Chebyshev polynomial has the minimal maximum norm for a monic polynomial, but here we want the maximum coefficients given the maximum norm. So maybe the Chebyshev polynomial is the one that gives the maximum coefficients?

In our case, the polynomial 8x² - 8x + 1 has |a| + |b| + |c| = 17. Is there a way to get higher?

Let me try another approach. Let me consider the maximum of |a| + |b| + |c|. Let me think of variables a, b, c. Let me try to express them in terms of the function values at certain points. For example, if I know f(0), f(1), and f(1/2), then I can solve for a, b, c.

Let me denote:

f(0) = c,

f(1) = a + b + c,

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

So, if I have three equations:

1. c = f(0)

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

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

We can solve for a, b, c in terms of f(0), f(1), f(1/2):

From equation 1: c = f(0).

From equation 2: a + b = f(1) - c = f(1) - f(0).

From equation 3: (a/4) + (b/2) = f(1/2) - c = f(1/2) - f(0).

Let me write equation 3 as:

Multiply by 4: a + 2b = 4(f(1/2) - f(0)).

From equation 2: a + b = f(1) - f(0).

Subtract equation 2 from equation 3:

(a + 2b) - (a + b) = [4(f(1/2) - f(0))] - [f(1) - f(0)]

Which simplifies to:

b = 4f(1/2) - 4f(0) - f(1) + f(0) = 4f(1/2) - 3f(0) - f(1)

Then, from equation 2: a = [f(1) - f(0)] - b = [f(1) - f(0)] - [4f(1/2) - 3f(0) - f(1)] = f(1) - f(0) - 4f(1/2) + 3f(0) + f(1) = 2f(1) + 2f(0) - 4f(1/2)

Therefore:

a = 2f(1) + 2f(0) - 4f(1/2)

b = 4f(1/2) - 3f(0) - f(1)

c = f(0)

Therefore, |a| + |b| + |c| can be written in terms of f(0), f(1), f(1/2). But since |f(x)| ≤ 1 for all x in [0,1], then |f(0)| ≤ 1, |f(1)| ≤ 1, |f(1/2)| ≤ 1. However, even if each of these is bounded by 1, the combination might not be straightforward. However, if we can choose f(0), f(1), f(1/2) such that they alternate between 1 and -1, then maybe we can get a higher value.

For example, in the Chebyshev case, f(0) = 1, f(1/2) = -1, f(1) = 1. Let me compute a, b, c:

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

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

c = 1

Which gives |a| + |b| + |c| = 8 + 8 + 1 = 17, as before.

Alternatively, suppose we take f(0) = 1, f(1/2) = 1, f(1) = -1. Then:

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

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

c = 1

So the polynomial is -4x² + 2x + 1. Wait, this is the same polynomial as before which had a maximum of 1.25 at x=0.25. Therefore, this polynomial does not satisfy the condition |f(x)| ≤ 1 on [0,1], so even though we set f(0), f(1/2), f(1) to be within [-1, 1], the polynomial exceeds the bound in between. Therefore, this shows that just setting f(0), f(1/2), f(1) to be within the bounds isn't sufficient, because the polynomial might exceed the bounds in between. Therefore, we need to ensure that the polynomial doesn't exceed the bounds in between.

Therefore, the Chebyshev polynomial case is special because it alternates between the maximum and minimum values at three points, which ensures that the polynomial doesn't exceed the bounds in between. This is due to the equioscillation theorem, which states that the optimal approximating polynomial equioscillates between the maximum error at least n+2 times, where n is the degree. For our case, since we are dealing with a quadratic polynomial (degree 2), the optimal polynomial should equioscillate between the maximum and minimum at least 3 times. Therefore, the Chebyshev polynomial is the one that does this, and hence it's the extremal case for our problem.

Therefore, if we can show that 17 is indeed the maximum, then that would be the answer. But how can I be sure there isn't another polynomial with a higher |a| + |b| + |c|?

Alternatively, let me try to think of another approach. Let me consider that for the polynomial f(x) = ax² + bx + c, we have |f(x)| ≤ 1 on [0,1]. Let me consider evaluating f at three points: 0, 1/2, 1. Then, as above, we can express a, b, c in terms of f(0), f(1/2), f(1). However, even if we set f(0), f(1/2), f(1) to be 1, -1, 1, we get the Chebyshev polynomial. But if we set them to other values, maybe we can get a higher |a| + |b| + |c|?

But how? Let me suppose that f(0) = 1, f(1/2) = -1, f(1) = 1. Then, as we saw, we get a = 8, b = -8, c = 1, which gives |a| + |b| + |c| = 17. Suppose instead that f(0) = 1, f(1/2) = -1, f(1) = something else. Wait, but if we set f(1) to be something else, then maybe we can get a different combination. However, if we set f(1) to be something else, then the polynomial might exceed the bound. For example, if we set f(1) = 1, which is the case for the Chebyshev polynomial, but if we set f(1) to be something else, maybe we can get a different a, b, c. However, since the polynomial must stay within [-1, 1], we can't just set f(1) to be anything.

Alternatively, maybe consider that the maximum of |a| + |b| + |c| occurs when the polynomial reaches the maximum absolute value at three points, which is the case for the Chebyshev polynomial. Therefore, 17 is the maximum.

Alternatively, let me try to compute |a| + |b| + |c| for the Chebyshev polynomial and see if there's a way to get higher. Let me think of another polynomial. For example, suppose that the polynomial reaches 1 at x=0, -1 at x=1, and 1 at x= something else. Wait, but how?

Alternatively, suppose that the polynomial reaches 1 at x=0, -1 at x=1, and maybe another point. Let me try to construct such a polynomial. Let me assume that f(0) = 1, f(1) = -1, and f(t) = 1 for some t in (0,1). Then, solving for a, b, c. Let me try with t = 1/2 again. Wait, that was the case we tried earlier which didn't work. But maybe with another t?

Let me suppose that f(0) = 1, f(1) = -1, and f(t) = 1 for some t in (0,1). Let me set up equations:

c = 1,

a + b + c = -1 ⇒ a + b = -2,

a t² + b t + c = 1 ⇒ a t² + b t = 0.

So from the third equation: a t² + b t = 0 ⇒ a t + b = 0 (since t ≠ 0). Therefore, b = -a t.

From the second equation: a + b = -2 ⇒ a - a t = -2 ⇒ a(1 - t) = -2 ⇒ a = -2 / (1 - t).

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

Therefore, the polynomial is:

f(x) = ax² + bx + c = [-2/(1 - t)]x² + [2t/(1 - t)]x + 1.

Let me check if this polynomial stays within [-1, 1] on [0,1]. For example, let me pick t = 1/3. Then:

a = -2 / (1 - 1/3) = -2 / (2/3) = -3,

b = 2*(1/3) / (2/3) = 1.

So the polynomial is -3x² + x + 1. Let me check its maximum and minimum on [0,1].

At x=0: 1.

At x=1: -3 + 1 + 1 = -1.

At x=1/3: -3*(1/9) + 1/3 + 1 = -1/3 + 1/3 + 1 = 1. Okay, so it meets the conditions at x=0, x=1/3, x=1. Now, check the vertex. The vertex is at x = -b/(2a) = -1/(2*(-3)) = 1/6. Let me compute f(1/6):

-3*(1/36) + (1/6) + 1 = -1/12 + 1/6 + 1 = (-1 + 2)/12 + 1 = 1/12 + 1 = 13/12 ≈ 1.083, which is greater than 1. Therefore, this polynomial exceeds the bound. Therefore, invalid.

So even though we set the polynomial to reach 1 at three points, the maximum in between is higher. Therefore, this approach doesn't work. Therefore, the only way to ensure that the polynomial doesn't exceed the bounds is if it equioscillates between the maximum and minimum at three points, which is the case for the Chebyshev polynomial. Therefore, that must be the extremal case.

Alternatively, let me try to see if there's another polynomial that meets the condition and has a higher |a| + |b| + |c|. For example, suppose that the polynomial reaches 1 at x=0, -1 at x=1, and maybe another point where it reaches -1? Wait, but then it would have to have three points where it reaches the maximum or minimum. However, for a quadratic, it can have at most two extrema, so it can only reach the maximum or minimum at two points. Wait, a quadratic has only one extremum (vertex). Therefore, it can reach the maximum or minimum at two points: one at the vertex and one at an endpoint. Therefore, if it reaches the maximum at two points and the minimum at one, or vice versa. Therefore, the Chebyshev polynomial reaches the maximum at two points and the minimum at one, or alternates.

Wait, in our case, the Chebyshev polynomial reaches 1 at x=0 and x=1, and -1 at x=0.5. So that's three points. But since it's a quadratic, it can only have one extremum. Therefore, the extremum is at x=0.5, which is the minimum. Therefore, the maximum occurs at the endpoints. Therefore, the polynomial reaches the maximum at two points (x=0 and x=1) and the minimum at one point (x=0.5). Therefore, this is the case of equioscillation between three points, which is the minimal maximum deviation.

Therefore, if we want to maximize |a| + |b| + |c|, given that the maximum deviation is 1, then this polynomial is the one that gives the maximum coefficients. Therefore, 17 is the answer.

But to confirm, let me try another approach. Let me consider that for any quadratic polynomial bounded by 1 on [0,1], we can bound |a|, |b|, |c|.

For example, we can use the fact that:

At x=0: |c| ≤ 1.

At x=1: |a + b + c| ≤ 1.

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

So, we have three inequalities:

1. |c| ≤ 1,

2. |a + b + c| ≤ 1,

3. |(a/4) + (b/2) + c| ≤ 1.

Let me try to express a, b, c in terms of these. Let me assume that the maximum is achieved at these three points, so that c = 1, a + b + c = 1, and (a/4) + (b/2) + c = -1. Which gives us the Chebyshev case.

Alternatively, if I consider the maximum of |a| + |b| + |c|, maybe I can use linear algebra. Let me think of variables a, b, c. Let me suppose that c is positive, and a and b are negative. Then |a| + |b| + |c| = -a - b + c. Let me try to maximize this expression under the constraints:

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

But how?

Alternatively, think of the maximum of -a - b + c. Let me consider that c is as large as possible, which is 1, and then try to maximize -a - b. But how?

From the constraints, at x=1: a + b + c ≤ 1. If c = 1, then a + b ≤ 0. Therefore, -a - b ≥ 0. So, if we want to maximize -a - b, given that a + b ≤ 0. But also, we have other constraints. For example, at x=0.5: (a/4) + (b/2) + c ≤ 1. If c = 1, then (a/4) + (b/2) ≤ 0. Multiply by 4: a + 2b ≤ 0. So we have:

a + b ≤ 0,

a + 2b ≤ 0.

Let me denote S = a + b, and T = a + 2b. Then S ≤ 0, T ≤ 0. Also, since T = S + b, so b = T - S. But maybe this is getting too abstract.

Alternatively, let me consider that from a + b ≤ 0 and a + 2b ≤ 0. Let me subtract the first inequality from the second: (a + 2b) - (a + b) = b ≤ 0. Therefore, b ≤ 0. Also, from a + b ≤ 0, since b ≤ 0, then a ≤ -b.

But how does this help? Let me think. If I want to maximize -a - b, given that a + b ≤ 0 and a + 2b ≤ 0. Let me set up the problem:

Maximize -a - b,

subject to:

a + b ≤ 0,

a + 2b ≤ 0,

and also other constraints from the polynomial being bounded by 1 on [0,1]. Wait, but we already used the constraints at x=0, x=1, and x=0.5. However, there are more constraints from other points. For example, the polynomial might have a maximum or minimum inside the interval.

But if we assume that the maximum of |ax² + bx + c| occurs at x=0, x=1, and x=0.5, then we have already considered those constraints. However, if there are other points where the polynomial could exceed the bound, then we need to ensure that. But if we have already set the polynomial to reach the maximum at those three points, then maybe it's safe?

But in the previous example, when we set f(0)=1, f(1)=-1, f(1/2)=1, the polynomial exceeded the bound. Therefore, even if we set the constraints at those three points, we need to ensure that the polynomial doesn't exceed the bound elsewhere. Therefore, maybe the only safe way is to have the polynomial equioscillate between the maximum and minimum at three points, which is the Chebyshev case.

Alternatively, let me try to use the three constraints:

c = 1,

a + b + c = 1,

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

Which gives us a = 8, b = -8, c = 1. Then, check if the polynomial is within bounds. As we saw earlier, it is. Therefore, this is valid.

Alternatively, suppose that instead of c = 1, we have c = -1. Then, similar logic would give us a polynomial that might have |a| + |b| + |c| = 17 as well. For example, if we set f(0) = -1, f(1/2) = 1, f(1) = -1. Then:

c = -1,

a + b + c = -1 ⇒ a + b = 0,

(a/4) + (b/2) + c = 1 ⇒ (a/4) + (b/2) - 1 = 1 ⇒ (a/4) + (b/2) = 2 ⇒ Multiply by 4: a + 2b = 8.

From a + b = 0 ⇒ a = -b. Substitute into a + 2b = 8 ⇒ -b + 2b = 8 ⇒ b = 8 ⇒ a = -8. Therefore, the polynomial is -8x² + 8x - 1. Let me check its values:

At x=0: -1, okay.

At x=1: -8 + 8 -1 = -1, okay.

At x=0.5: -8*(0.25) + 8*(0.5) -1 = -2 + 4 -1 = 1, okay.

Vertex at x = -b/(2a) = -8/(2*(-8)) = 0.5. Wait, so the vertex is at x=0.5, which is the point where it reaches 1. Therefore, the polynomial is symmetric around x=0.5? Let me check. Let me compute f(0.5 + t) and f(0.5 - t):

f(0.5 + t) = -8*(0.5 + t)^2 + 8*(0.5 + t) -1

= -8*(0.25 + t + t²) + 4 + 8t -1

= -2 -8t -8t² + 4 + 8t -1

= (-2 + 4 -1) + (-8t + 8t) + (-8t²)

= 1 - 8t²

Similarly, f(0.5 - t) = -8*(0.5 - t)^2 + 8*(0.5 - t) -1

= -8*(0.25 - t + t²) + 4 - 8t -1

= -2 + 8t -8t² + 4 - 8t -1

= (-2 + 4 -1) + (8t -8t) + (-8t²)

= 1 - 8t²

Therefore, the polynomial is symmetric around x=0.5, and its maximum is 1 at x=0.5, and minimum is -1 at x=0 and x=1. Therefore, this polynomial is also valid, and |a| + |b| + |c| = 8 + 8 + 1 = 17.

Therefore, whether c is 1 or -1, we get the same sum.

Is there a way to get a higher value? Let me think. Suppose that instead of having the polynomial reach 1 and -1 at three points, we have it reach 1 at two points and -1 at one, but with different coefficients? For example, suppose that the polynomial reaches 1 at x=0 and x=1, and reaches some value between -1 and 1 at x=0.5. Then, maybe we can have a larger |a| + |b| + |c|?

Wait, but if the polynomial reaches 1 at x=0 and x=1, then the maximum of the polynomial is 1, but the minimum could be lower. However, since the polynomial is bounded by 1 in absolute value, the minimum must be ≥ -1. So, if we have the polynomial reaching 1 at x=0 and x=1, and reaching some value m ≥ -1 at x=0.5. Then, maybe we can find a polynomial with larger coefficients?

Let me try. Let me assume that f(0) = 1, f(1) = 1, and f(0.5) = m ≥ -1. Then, solving for a, b, c:

c = 1,

a + b + c = 1 ⇒ a + b = 0,

(a/4) + (b/2) + c = m ⇒ (a/4) + (b/2) + 1 = m.

From a + b = 0 ⇒ b = -a. Substitute into the third equation:

(a/4) + (-a)/2 + 1 = m ⇒ (a/4 - 2a/4) + 1 = m ⇒ (-a/4) + 1 = m ⇒ -a/4 = m - 1 ⇒ a = 4(1 - m).

Therefore, a = 4(1 - m), b = -4(1 - m), c = 1.

Therefore, the polynomial is:

f(x) = 4(1 - m)x² - 4(1 - m)x + 1.

Let me compute |a| + |b| + |c| = |4(1 - m)| + | -4(1 - m)| + |1| = 8|1 - m| + 1.

Since m ≥ -1, then 1 - m ≤ 2. Therefore, to maximize 8|1 - m| + 1, we need to maximize |1 - m|. Since m ≥ -1, the maximum of |1 - m| occurs when m is as small as possible, i.e., m = -1. Then |1 - (-1)| = 2, so |a| + |b| + |c| = 8*2 + 1 = 17. Which is the same as before. Therefore, even in this case, the maximum is 17.

Therefore, whether we set m = -1 or not, the maximum is achieved when m = -1, which gives us the Chebyshev polynomial. Therefore, this suggests that 17 is indeed the maximum.

Alternatively, suppose that the polynomial reaches 1 at one point and -1 at two points. For example, f(0) = -1, f(1/2) = 1, f(1) = -1. Which is similar to the previous case, and gives the same result.

Alternatively, suppose that the polynomial reaches 1 at one point and -1 at another, and maybe another point. But since it's a quadratic, it can only have one extremum. Therefore, if it reaches 1 at one endpoint and -1 at the other, then the extremum is somewhere in between. If the extremum is within the interval, then it can reach another value. But to have the maximum absolute value, it would have to reach 1 or -1 at the extremum.

For example, suppose that f(0) = 1, f(1) = -1, and the extremum is at x = t in (0,1) where f(t) = 1 or -1. Let me try this.

Let me assume that f(0) = 1, f(1) = -1, and f(t) = 1 for some t in (0,1). Then, as before, we have:

c = 1,

a + b + c = -1 ⇒ a + b = -2,

a t² + b t + c = 1 ⇒ a t² + b t = 0.

From the third equation: a t² + b t = 0 ⇒ a t + b = 0 (since t ≠ 0) ⇒ b = -a t.

From the second equation: a - a t = -2 ⇒ a(1 - t) = -2 ⇒ a = -2 / (1 - t).

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

Therefore, the polynomial is:

f(x) = [-2/(1 - t)]x² + [2t/(1 - t)]x + 1.

Now, we need to ensure that this polynomial doesn't exceed the bounds of [-1, 1] on [0,1]. The extremum is at x = -b/(2a) = -[2t/(1 - t)] / [2*(-2/(1 - t))] = [2t/(1 - t)] / [4/(1 - t)] = t/2.

So the extremum is at x = t/2. Let me compute f(t/2):

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

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

= [-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 within [-1, 1]. Since we already have f(0) = 1, f(1) = -1, and f(t) = 1, we need to ensure that the extremum at x = t/2 doesn't exceed the bounds.

So, let's compute f(t/2):

[t²/(2(1 - t))] + 1 ≤ 1 ⇒ [t²/(2(1 - t))] ≤ 0. Since t ∈ (0,1), denominator 1 - t > 0, numerator t² ≥ 0. Therefore, [t²/(2(1 - t))] ≥ 0. Therefore, f(t/2) ≥ 1. But we need f(t/2) ≤ 1. Therefore, equality holds only if t = 0, which is not in (0,1). Therefore, this implies that f(t/2) > 1, which violates the condition. Therefore, this polynomial exceeds the bound. Therefore, this approach doesn't work.

Therefore, the only way to have the polynomial stay within the bounds is if the extremum is at the point where it reaches the minimum or maximum. Which is the case for the Chebyshev polynomial. Therefore, the maximum |a| + |b| + |c| is indeed 17.

Alternatively, let me think of another way. Suppose that we want to maximize |a| + |b| + |c|. Let me consider that for the polynomial f(x) = ax² + bx + c, we can write:

At x=0: c = f(0),

At x=1: a + b + c = f(1),

At x=1/2: a/4 + b/2 + c = f(1/2).

Therefore, if we can express |a| + |b| + |c| in terms of f(0), f(1), f(1/2), then maybe we can find the maximum.

Earlier, we found that:

a = 2f(1) + 2f(0) - 4f(1/2),

b = 4f(1/2) - 3f(0) - f(1),

c = f(0).

Therefore, |a| + |b| + |c| = |2f(1) + 2f(0) - 4f(1/2)| + |4f(1/2) - 3f(0) - f(1)| + |f(0)|.

But since |f(x)| ≤ 1 for all x in [0,1], then |f(0)| ≤ 1, |f(1)| ≤ 1, |f(1/2)| ≤ 1. However, the combination of these terms can be complex. However, if we assume that f(0) = 1, f(1) = 1, f(1/2) = -1, then:

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

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

c = 1,

which gives |a| + |b| + |c| = 17.

Alternatively, if we take other values, say f(0) = 1, f(1) = -1, f(1/2) = 1, then:

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

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

c = 1,

which gives |a| + |b| + |c| = 4 + 2 + 1 = 7, which is much less than 17.

Alternatively, if we take f(0) = 1, f(1) = 1, f(1/2) = 1, then:

a = 2*1 + 2*1 - 4*1 = 2 + 2 - 4 = 0,

b = 4*1 - 3*1 - 1 = 4 - 3 - 1 = 0,

c = 1,

so |a| + |b| + |c| = 0 + 0 + 1 = 1, which is worse.

Alternatively, if we take f(0) = 1, f(1) = -1, f(1/2) = -1:

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

b = 4*(-1) - 3*1 - (-1) = -4 - 3 + 1 = -6,

c = 1,

so |a| + |b| + |c| = 4 + 6 + 1 = 11, which is still less than 17.

Alternatively, if we take f(0) = 1, f(1) = 1, f(1/2) = -1, which gives us the Chebyshev case, which gives 17.

Alternatively, suppose that f(0) = 1, f(1) = -1, f(1/2) = -1. Wait, we did that above, which gives 11.

Alternatively, suppose that f(0) = 1, f(1) = 1, f(1/2) = something else. For example, suppose f(1/2) = -1, which gives us the Chebyshev case.

Alternatively, suppose that f(0) = 1, f(1) = 1, f(1/2) = m. Then:

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

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

c = 1.

Then |a| + |b| + |c| = |4 - 4m| + |4m - 4| + 1 = 4|1 - m| + 4|1 - m| + 1 = 8|1 - m| + 1. To maximize this, we need to maximize |1 - m|. Since m is between -1 and 1, the maximum |1 - m| is when m = -1, giving |1 - (-1)| = 2, so total is 8*2 + 1 = 17. Which again gives us the Chebyshev case.

Therefore, regardless of how we approach it, the maximum value of |a| + |b| + |c| is 17.

Therefore, I think the answer is 17.

**Final Answer**
\boxed{17}

To solve the problem, we are given that for all $ x \in [0,1] $, the inequality
$$
|ax^2 + bx + c| \leq 1
$$
holds, and we are to find the **maximum possible value** of $ |a| + |b| + |c| $.

---

### Step 1: Understanding the Constraint

The condition $ |ax^2 + bx + c| \leq 1 $ for all $ x \in [0,1] $ means that the quadratic polynomial $ f(x) = ax^2 + bx + c $ is bounded between $-1$ and $1$ on the interval $[0,1]$. This is a **boundedness constraint** on the polynomial over a closed interval.

Our goal is to **maximize** the sum $ |a| + |b| + |c| $, which is the sum of the absolute values of the coefficients of the polynomial.

---

### Step 2: Strategy and Key Insight

This is a classic optimization problem with a constraint on the maximum of a function over an interval. A powerful tool in such problems is the **Chebyshev polynomial**, which is known to be extremal in the sense of minimizing the maximum deviation from zero over an interval, given a fixed degree.

The **Chebyshev polynomial of degree 2** is:
$$
T_2(x) = 2x^2 - 1
$$
This polynomial oscillates between $-1$ and $1$ on the interval $[-1,1]$, and it is known to be the polynomial of degree 2 with the **smallest maximum absolute value** on that interval among all monic polynomials.

However, our interval is $[0,1]$, not $[-1,1]$. To adapt the Chebyshev polynomial to this interval, we perform a linear transformation:
$$
x = \frac{t + 1}{2}, \quad \text{so that } t \in [-1,1] \text{ corresponds to } x \in [0,1]
$$
Substituting into $ T_2(t) $, we get:
$$
f(x) = 2\left(\frac{t + 1}{2}\right)^2 - 1 = 2\left(\frac{t^2 + 2t + 1}{4}\right) - 1 = \frac{t^2 + 2t + 1}{2} - 1 = \frac{t^2 + 2t - 1}{2}
$$
Now, substituting back $ t = 2x - 1 $, we get:
$$
f(x) = \frac{(2x - 1)^2 + 2(2x - 1) - 1}{2}
$$
Expanding:
$$
(2x - 1)^2 = 4x^2 - 4x + 1, \quad 2(2x - 1) = 4x - 2
$$
So:
$$
f(x) = \frac{4x^2 - 4x + 1 + 4x - 2 - 1}{2} = \frac{4x^2 - 2}{2} = 2x^2 - 1
$$
Thus, the polynomial $ f(x) = 2x^2 - 1 $ is the **Chebyshev polynomial of degree 2** on the interval $[0,1]$, and it oscillates between $-1$ and $1$ at three points: $ x = 0, \frac{1}{2}, 1 $.

---

### Step 3: Compute the Coefficients

From the polynomial $ f(x) = 2x^2 - 1 $, we identify:
- $ a = 2 $
- $ b = 0 $
- $ c = -1 $

So:
$$
|a| + |b| + |c| = |2| + |0| + |-1| = 2 + 0 + 1 = 3
$$

But this is **not** the maximum. Wait — we need to **scale** the polynomial to match the condition $ |f(x)| \leq 1 $ on $[0,1]$, and we want to **maximize** the sum $ |a| + |b| + |c| $.

Let’s consider the polynomial:
$$
f(x) = 8x^2 - 8x + 1
$$
This polynomial is known to oscillate between $-1$ and $1$ at $ x = 0, \frac{1}{2}, 1 $, and it satisfies the condition $ |f(x)| \leq 1 $ on $[0,1]$. Let’s verify:

- $ f(0) = 1 $
- $ f(1) = 8 - 8 + 1 = 1 $
- $ f\left(\frac{1}{2}\right) = 8\cdot\frac{1}{4} - 8\cdot\frac{1}{2} + 1 = 2 - 4 + 1 = -1 $

So it oscillates between $-1$ and $1$ at three points, and the polynomial is bounded in absolute value by 1 on the entire interval. Therefore, it satisfies the condition.

Now compute:
$$
|a| + |b| + |c| = |8| + |-8| + |1| = 8 + 8 + 1 = \boxed{17}
$$

---

### Step 4: Justification of Optimality

This polynomial is the **extremal** one in the class of quadratics bounded by 1 on $[0,1]$, in the sense that it **maximizes** the sum of the absolute values of the coefficients under the constraint $ |f(x)| \leq 1 $ for all $ x \in [0,1] $.

This is a known result from approximation theory, where the **Chebyshev polynomials** are extremal in such problems. The polynomial $ f(x) = 8x^2 - 8x + 1 $ is a scaled version of the Chebyshev polynomial $ T_2(x) $, and it achieves the maximum possible value of $ |a| + |b| + |c| $ under the given constraint.

---

### 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, I know that the quadratic is bounded between -1 and 1 on the interval [0, 1]. So, for every x in [0, 1], the value of ax² + bx + c is within that range. My goal is to maximize |a| + |b| + |c|.

I remember that sometimes problems like this involve using specific points in the interval to set up equations or inequalities. Since the quadratic is bounded on [0, 1], maybe I can evaluate it at certain key points, like x = 0, x = 1, and maybe the vertex of the parabola? Let me try that.

Let me first note the values of the quadratic at x = 0 and x = 1:

At x = 0: |c| ≤ 1, so c is between -1 and 1.

At x = 1: |a + b + c| ≤ 1.

Also, the vertex of the parabola is at x = -b/(2a). But since the interval is [0, 1], the vertex might lie inside this interval or outside. If it's inside, then the maximum or minimum of the quadratic could be there. However, since the quadratic is bounded between -1 and 1 on the entire interval, the maximum absolute value could occur at the endpoints or at the vertex.

But maybe I need to consider more than just the endpoints. For example, if the quadratic has its maximum or minimum inside the interval, then that point must also satisfy |ax² + bx + c| ≤ 1.

Alternatively, maybe I can use some known inequalities or optimization techniques. Since we need to maximize |a| + |b| + |c|, perhaps I can consider different cases based on the signs of a, b, c? But that might get complicated.

Alternatively, maybe I can use Lagrange multipliers? But since there are infinitely many constraints (for all x in [0,1]), that might not be straightforward. Maybe there's a better way.

Wait, I remember something about Chebyshev polynomials. They are extremal in some sense for polynomial approximation. 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 there's a similar polynomial?

Let me check. If I can find a quadratic polynomial that oscillates between -1 and 1 on [0, 1], then maybe that polynomial would have the maximum coefficients? Because Chebyshev polynomials are known for minimizing the maximum deviation, but here we want to maximize the coefficients given the maximum deviation.

Alternatively, maybe I can perform a change of variables to map [0, 1] to [-1, 1]. Let me try that. Let t = 2x - 1, so when x = 0, t = -1, and when x = 1, t = 1. Then x = (t + 1)/2. Substitute into the quadratic:

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

Let me compute that:

First, expand [(t + 1)/2]^2 = (t² + 2t + 1)/4.

So:

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 I denote this as A t² + B t + C, where:

A = a/4,

B = (a + b)/2,

C = a/4 + b/2 + c.

Now, since the original polynomial is bounded by 1 in absolute value on t ∈ [-1, 1], and Chebyshev polynomials are extremal for such problems. The Chebyshev polynomial of degree 2 is T2(t) = 2t² - 1, which oscillates between -1 and 1 on [-1, 1]. So if I can adjust the coefficients to match this, maybe that would give me the maximum coefficients?

But let me think. If I want the transformed polynomial A t² + B t + C to be equal to T2(t) = 2t² - 1, then:

A = 2,

B = 0,

C = -1.

But then, let's check what that would mean for a, b, c.

From A = a/4 = 2 ⇒ a = 8.

From B = (a + b)/2 = 0 ⇒ (8 + b)/2 = 0 ⇒ 8 + b = 0 ⇒ b = -8.

From C = a/4 + b/2 + c = -1 ⇒ 8/4 + (-8)/2 + c = -1 ⇒ 2 - 4 + c = -1 ⇒ -2 + c = -1 ⇒ c = 1.

So then, the original polynomial would be 8x² - 8x + 1. Let me check if this polynomial is bounded by 1 on [0, 1].

Let me compute its values at x = 0, x = 1, and check the vertex.

At x = 0: 0 - 0 + 1 = 1. Okay, that's within the bound.

At x = 1: 8 - 8 + 1 = 1. Also within the bound.

The vertex is at x = -b/(2a) = 8/(2*8) = 0.5. So at x = 0.5:

8*(0.25) - 8*(0.5) + 1 = 2 - 4 + 1 = -1. So that's -1, which is within the bound. So the polynomial oscillates between 1 and -1 at x=0, x=0.5, and x=1. Therefore, it's exactly the Chebyshev polynomial scaled appropriately.

So this polynomial meets the condition |ax² + bx + c| ≤ 1 on [0, 1], and in this case, |a| + |b| + |c| = |8| + |-8| + |1| = 8 + 8 + 1 = 17.

Is this the maximum? Maybe. But I need to check if there are other polynomials that could give a higher value.

Alternatively, maybe there's a different polynomial that attains a higher |a| + |b| + |c|. Let me think.

Alternatively, maybe I can use the Lagrange interpolation. Suppose that the maximum of |ax² + bx + c| is achieved at three points in [0, 1], which would be the case for the Chebyshev polynomial. Since the Chebyshev polynomials have the minimal maximum norm for a monic polynomial, but here we are trying to maximize the coefficients given the maximum norm.

Alternatively, maybe I can use the fact that the maximum of |a| + |b| + |c| occurs when the polynomial reaches the maximum absolute value at several points. For example, if the polynomial reaches 1 at x=0, -1 at x=0.5, and 1 at x=1, which is exactly the case for the polynomial we found.

But maybe there's another configuration where the polynomial reaches 1 or -1 at different points, leading to a higher |a| + |b| + |c|?

Alternatively, let's think about the maximum of |a| + |b| + |c|. Since c is the value at x=0, which is between -1 and 1. Similarly, a + b + c is the value at x=1, also between -1 and 1. Also, the value at the vertex is between -1 and 1.

But how can we relate these to |a| + |b| + |c|? Maybe we can express a, b, c in terms of the values at certain points. For example, if we know the values at x=0, x=1, and the vertex, we can solve for a, b, c. But since the vertex is dependent on a and b, this might get complicated.

Alternatively, maybe we can use some inequalities. Let me think. Let me consider that for x in [0,1], |ax² + bx + c| ≤ 1. Let me try to express a, b, c in terms of the function values at certain points. For example, if I know the function at x=0, x=1, and x=1/2, then I can set up equations.

Let me denote f(x) = ax² + bx + c. Then:

f(0) = c,

f(1) = a + b + c,

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

If I assume that f(0) = 1, f(1/2) = -1, f(1) = 1, then we can solve for a, b, c. Which is exactly the case of the Chebyshev polynomial we considered earlier. Let me check:

From f(0) = c = 1,

From f(1) = a + b + c = 1,

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

So substituting c = 1 into the other equations:

a + b + 1 = 1 ⇒ a + b = 0 ⇒ b = -a.

Then, substituting into the third equation:

(a/4) + (-a/2) + 1 = -1 ⇒ (-a/4) + 1 = -1 ⇒ -a/4 = -2 ⇒ a = 8. Then b = -8, c = 1. Which is the same as before. So that gives |a| + |b| + |c| = 17.

But maybe if we take different values? For example, suppose that f(0) = 1, f(1) = -1, and f(1/2) = 1. Then solving:

c = 1,

a + b + c = -1 ⇒ a + b = -2,

(a/4) + (b/2) + c = 1 ⇒ (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 second: b = 2. Then from a + 2 = -2 ⇒ a = -4. Then c = 1. So the polynomial is -4x² + 2x + 1. Let me check its maximum absolute value on [0,1].

At x=0: 1, okay.

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

At x=0.5: -4*(0.25) + 2*(0.5) + 1 = -1 + 1 + 1 = 1, okay.

What about the vertex? The vertex is at x = -b/(2a) = -2/(2*(-4)) = -2 / (-8) = 1/4. So x=0.25. Let me compute f(0.25):

-4*(0.0625) + 2*(0.25) + 1 = -0.25 + 0.5 + 1 = 1.25. Wait, that's 1.25, which is greater than 1. That violates the condition. So this polynomial doesn't satisfy |f(x)| ≤ 1 on [0,1]. Therefore, this approach doesn't work. So even though we set the function to reach 1, -1, 1 at three points, the maximum in between is higher. Therefore, this polynomial is invalid.

So that suggests that choosing three points with alternating signs might not be sufficient unless the polynomial is the Chebyshev polynomial. Therefore, maybe the Chebyshev polynomial is the one that attains the maximum.

Alternatively, maybe there's another polynomial where the maximum is achieved at different points. Let me think. For example, suppose that the maximum is achieved at x=0, x=1, and somewhere else. But how?

Alternatively, maybe I can use the concept of the maximum of |a| + |b| + |c|. Let me think about how to express |a| + |b| + |c| in terms of the function values.

Alternatively, maybe I can use the fact that for x in [0,1], the polynomial is bounded by 1. Let me consider evaluating the polynomial at specific points to get inequalities involving a, b, c.

We already have:

At x=0: |c| ≤ 1.

At x=1: |a + b + c| ≤ 1.

Also, maybe consider the derivative? The maximum of the polynomial could be at the endpoints or at critical points. But since we have the bound on the entire interval, maybe we can use some integral or other method?

Alternatively, think of the problem as an optimization problem with constraints. Let me try to set up the problem.

We need to maximize |a| + |b| + |c| subject to |ax² + bx + c| ≤ 1 for all x in [0,1].

This is a constrained optimization problem. However, dealing with infinitely many constraints is difficult. But perhaps we can use the fact that the maximum of |ax² + bx + c| on [0,1] is achieved at certain points, so maybe we can use the maximum at those points as constraints.

In particular, for the maximum to be achieved at three points, as in the Chebyshev case. But how to ensure that?

Alternatively, maybe use the method of Lagrange multipliers with some constraints. But since there are infinitely many constraints, it's not straightforward. However, if we assume that the maximum is achieved at three points, then we can set up equations for those points.

Alternatively, think of the problem as follows: since the polynomial is bounded between -1 and 1 on [0,1], then we can write:

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

To maximize |a| + |b| + |c|, perhaps we need to consider the extremal cases where the polynomial touches the bounds 1 or -1 at several points. For example, the Chebyshev polynomial touches 1 and -1 alternately at three points, which might be the case for maximum coefficients.

Alternatively, maybe there's a way to express |a| + |b| + |c| in terms of the maximum of the polynomial. But I don't recall a direct formula for that.

Alternatively, think of specific values. For example, suppose that c = 1 (maximum possible). Then, we need to maximize |a| + |b| + 1. But then, we have constraints on a and b such that |ax² + bx + 1| ≤ 1 for all x in [0,1]. Let me see.

So, for all x in [0,1], -1 ≤ ax² + bx + 1 ≤ 1. Therefore, ax² + bx + 1 ≤ 1 ⇒ ax² + bx ≤ 0, and ax² + bx + 1 ≥ -1 ⇒ ax² + bx ≥ -2.

So, for all x in [0,1], ax² + bx ≤ 0 and ax² + bx ≥ -2.

Similarly, if c = 1, then maybe we can find a and b such that ax² + bx is between -2 and 0 on [0,1]. But how does that help us maximize |a| + |b|?

Alternatively, maybe consider that the maximum of |a| + |b| + |c| occurs when c is at its maximum or minimum. But I need to check.

Alternatively, think of the problem in terms of linear algebra. Let me consider that the maximum of |a| + |b| + |c| is achieved when the signs of a, b, c are chosen to maximize the sum. However, the constraints are non-linear because of the absolute value.

Alternatively, maybe consider different cases based on the signs of a, b, c. For example, suppose that a, b, c are all positive. Then |a| + |b| + |c| = a + b + c. But since the polynomial is bounded by 1 on [0,1], then at x=1, a + b + c ≤ 1, so the sum a + b + c ≤ 1. But this is probably not the maximum, since if we have negative coefficients, maybe we can get a larger sum.

Alternatively, suppose that c is positive, and a and b are negative. Then |a| + |b| + |c| = -a - b + c. Maybe this can be larger?

Alternatively, maybe the maximum occurs when the polynomial alternates between 1 and -1 at several points, which would be similar to the Chebyshev polynomial. Since the Chebyshev polynomial has the minimal maximum norm for a monic polynomial, but here we want the maximum coefficients given the maximum norm. So maybe the Chebyshev polynomial is the one that gives the maximum coefficients?

In our case, the polynomial 8x² - 8x + 1 has |a| + |b| + |c| = 17. Is there a way to get higher?

Let me try another approach. Let me consider the maximum of |a| + |b| + |c|. Let me think of variables a, b, c. Let me try to express them in terms of the function values at certain points. For example, if I know f(0), f(1), and f(1/2), then I can solve for a, b, c.

Let me denote:

f(0) = c,

f(1) = a + b + c,

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

So, if I have three equations:

1. c = f(0)

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

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

We can solve for a, b, c in terms of f(0), f(1), f(1/2):

From equation 1: c = f(0).

From equation 2: a + b = f(1) - c = f(1) - f(0).

From equation 3: (a/4) + (b/2) = f(1/2) - c = f(1/2) - f(0).

Let me write equation 3 as:

Multiply by 4: a + 2b = 4(f(1/2) - f(0)).

From equation 2: a + b = f(1) - f(0).

Subtract equation 2 from equation 3:

(a + 2b) - (a + b) = [4(f(1/2) - f(0))] - [f(1) - f(0)]

Which simplifies to:

b = 4f(1/2) - 4f(0) - f(1) + f(0) = 4f(1/2) - 3f(0) - f(1)

Then, from equation 2: a = [f(1) - f(0)] - b = [f(1) - f(0)] - [4f(1/2) - 3f(0) - f(1)] = f(1) - f(0) - 4f(1/2) + 3f(0) + f(1) = 2f(1) + 2f(0) - 4f(1/2)

Therefore:

a = 2f(1) + 2f(0) - 4f(1/2)

b = 4f(1/2) - 3f(0) - f(1)

c = f(0)

Therefore, |a| + |b| + |c| can be written in terms of f(0), f(1), f(1/2). But since |f(x)| ≤ 1 for all x in [0,1], then |f(0)| ≤ 1, |f(1)| ≤ 1, |f(1/2)| ≤ 1. However, even if each of these is bounded by 1, the combination might not be straightforward. However, if we can choose f(0), f(1), f(1/2) such that they alternate between 1 and -1, then maybe we can get a higher value.

For example, in the Chebyshev case, f(0) = 1, f(1/2) = -1, f(1) = 1. Let me compute a, b, c:

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

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

c = 1

Which gives |a| + |b| + |c| = 8 + 8 + 1 = 17, as before.

Alternatively, suppose we take f(0) = 1, f(1/2) = 1, f(1) = -1. Then:

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

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

c = 1

So the polynomial is -4x² + 2x + 1. Wait, this is the same polynomial as before which had a maximum of 1.25 at x=0.25. Therefore, this polynomial does not satisfy the condition |f(x)| ≤ 1 on [0,1], so even though we set f(0), f(1/2), f(1) to be within [-1, 1], the polynomial exceeds the bound in between. Therefore, this shows that just setting f(0), f(1/2), f(1) to be within the bounds isn't sufficient, because the polynomial might exceed the bounds in between. Therefore, we need to ensure that the polynomial doesn't exceed the bounds in between.

Therefore, the Chebyshev polynomial case is special because it alternates between the maximum and minimum values at three points, which ensures that the polynomial doesn't exceed the bounds in between. This is due to the equioscillation theorem, which states that the optimal approximating polynomial equioscillates between the maximum error at least n+2 times, where n is the degree. For our case, since we are dealing with a quadratic polynomial (degree 2), the optimal polynomial should equioscillate between the maximum and minimum at least 3 times. Therefore, the Chebyshev polynomial is the one that does this, and hence it's the extremal case for our problem.

Therefore, if we can show that 17 is indeed the maximum, then that would be the answer. But how can I be sure there isn't another polynomial with a higher |a| + |b| + |c|?

Alternatively, let me try to think of another approach. Let me consider that for the polynomial f(x) = ax² + bx + c, we have |f(x)| ≤ 1 on [0,1]. Let me consider evaluating f at three points: 0, 1/2, 1. Then, as above, we can express a, b, c in terms of f(0), f(1/2), f(1). However, even if we set f(0), f(1/2), f(1) to be 1, -1, 1, we get the Chebyshev polynomial. But if we set them to other values, maybe we can get a higher |a| + |b| + |c|?

But how? Let me suppose that f(0) = 1, f(1/2) = -1, f(1) = 1. Then, as we saw, we get a = 8, b = -8, c = 1, which gives |a| + |b| + |c| = 17. Suppose instead that f(0) = 1, f(1/2) = -1, f(1) = something else. Wait, but if we set f(1) to be something else, then maybe we can get a different combination. However, if we set f(1) to be something else, then the polynomial might exceed the bound. For example, if we set f(1) = 1, which is the case for the Chebyshev polynomial, but if we set f(1) to be something else, maybe we can get a different a, b, c. However, since the polynomial must stay within [-1, 1], we can't just set f(1) to be anything.

Alternatively, maybe consider that the maximum of |a| + |b| + |c| occurs when the polynomial reaches the maximum absolute value at three points, which is the case for the Chebyshev polynomial. Therefore, 17 is the maximum.

Alternatively, let me try to compute |a| + |b| + |c| for the Chebyshev polynomial and see if there's a way to get higher. Let me think of another polynomial. For example, suppose that the polynomial reaches 1 at x=0, -1 at x=1, and 1 at x= something else. Wait, but how?

Alternatively, suppose that the polynomial reaches 1 at x=0, -1 at x=1, and maybe another point. Let me try to construct such a polynomial. Let me assume that f(0) = 1, f(1) = -1, and f(t) = 1 for some t in (0,1). Then, solving for a, b, c. Let me try with t = 1/2 again. Wait, that was the case we tried earlier which didn't work. But maybe with another t?

Let me suppose that f(0) = 1, f(1) = -1, and f(t) = 1 for some t in (0,1). Let me set up equations:

c = 1,

a + b + c = -1 ⇒ a + b = -2,

a t² + b t + c = 1 ⇒ a t² + b t = 0.

So from the third equation: a t² + b t = 0 ⇒ a t + b = 0 (since t ≠ 0). Therefore, b = -a t.

From the second equation: a + b = -2 ⇒ a - a t = -2 ⇒ a(1 - t) = -2 ⇒ a = -2 / (1 - t).

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

Therefore, the polynomial is:

f(x) = ax² + bx + c = [-2/(1 - t)]x² + [2t/(1 - t)]x + 1.

Let me check if this polynomial stays within [-1, 1] on [0,1]. For example, let me pick t = 1/3. Then:

a = -2 / (1 - 1/3) = -2 / (2/3) = -3,

b = 2*(1/3) / (2/3) = 1.

So the polynomial is -3x² + x + 1. Let me check its maximum and minimum on [0,1].

At x=0: 1.

At x=1: -3 + 1 + 1 = -1.

At x=1/3: -3*(1/9) + 1/3 + 1 = -1/3 + 1/3 + 1 = 1. Okay, so it meets the conditions at x=0, x=1/3, x=1. Now, check the vertex. The vertex is at x = -b/(2a) = -1/(2*(-3)) = 1/6. Let me compute f(1/6):

-3*(1/36) + (1/6) + 1 = -1/12 + 1/6 + 1 = (-1 + 2)/12 + 1 = 1/12 + 1 = 13/12 ≈ 1.083, which is greater than 1. Therefore, this polynomial exceeds the bound. Therefore, invalid.

So even though we set the polynomial to reach 1 at three points, the maximum in between is higher. Therefore, this approach doesn't work. Therefore, the only way to ensure that the polynomial doesn't exceed the bounds is if it equioscillates between the maximum and minimum at three points, which is the case for the Chebyshev polynomial. Therefore, that must be the extremal case.

Alternatively, let me try to see if there's another polynomial that meets the condition and has a higher |a| + |b| + |c|. For example, suppose that the polynomial reaches 1 at x=0, -1 at x=1, and maybe another point where it reaches -1? Wait, but then it would have to have three points where it reaches the maximum or minimum. However, for a quadratic, it can have at most two extrema, so it can only reach the maximum or minimum at two points. Wait, a quadratic has only one extremum (vertex). Therefore, it can reach the maximum or minimum at two points: one at the vertex and one at an endpoint. Therefore, if it reaches the maximum at two points and the minimum at one, or vice versa. Therefore, the Chebyshev polynomial reaches the maximum at two points and the minimum at one, or alternates.

Wait, in our case, the Chebyshev polynomial reaches 1 at x=0 and x=1, and -1 at x=0.5. So that's three points. But since it's a quadratic, it can only have one extremum. Therefore, the extremum is at x=0.5, which is the minimum. Therefore, the maximum occurs at the endpoints. Therefore, the polynomial reaches the maximum at two points (x=0 and x=1) and the minimum at one point (x=0.5). Therefore, this is the case of equioscillation between three points, which is the minimal maximum deviation.

Therefore, if we want to maximize |a| + |b| + |c|, given that the maximum deviation is 1, then this polynomial is the one that gives the maximum coefficients. Therefore, 17 is the answer.

But to confirm, let me try another approach. Let me consider that for any quadratic polynomial bounded by 1 on [0,1], we can bound |a|, |b|, |c|.

For example, we can use the fact that:

At x=0: |c| ≤ 1.

At x=1: |a + b + c| ≤ 1.

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

So, we have three inequalities:

1. |c| ≤ 1,

2. |a + b + c| ≤ 1,

3. |(a/4) + (b/2) + c| ≤ 1.

Let me try to express a, b, c in terms of these. Let me assume that the maximum is achieved at these three points, so that c = 1, a + b + c = 1, and (a/4) + (b/2) + c = -1. Which gives us the Chebyshev case.

Alternatively, if I consider the maximum of |a| + |b| + |c|, maybe I can use linear algebra. Let me think of variables a, b, c. Let me suppose that c is positive, and a and b are negative. Then |a| + |b| + |c| = -a - b + c. Let me try to maximize this expression under the constraints:

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

But how?

Alternatively, think of the maximum of -a - b + c. Let me consider that c is as large as possible, which is 1, and then try to maximize -a - b. But how?

From the constraints, at x=1: a + b + c ≤ 1. If c = 1, then a + b ≤ 0. Therefore, -a - b ≥ 0. So, if we want to maximize -a - b, given that a + b ≤ 0. But also, we have other constraints. For example, at x=0.5: (a/4) + (b/2) + c ≤ 1. If c = 1, then (a/4) + (b/2) ≤ 0. Multiply by 4: a + 2b ≤ 0. So we have:

a + b ≤ 0,

a + 2b ≤ 0.

Let me denote S = a + b, and T = a + 2b. Then S ≤ 0, T ≤ 0. Also, since T = S + b, so b = T - S. But maybe this is getting too abstract.

Alternatively, let me consider that from a + b ≤ 0 and a + 2b ≤ 0. Let me subtract the first inequality from the second: (a + 2b) - (a + b) = b ≤ 0. Therefore, b ≤ 0. Also, from a + b ≤ 0, since b ≤ 0, then a ≤ -b.

But how does this help? Let me think. If I want to maximize -a - b, given that a + b ≤ 0 and a + 2b ≤ 0. Let me set up the problem:

Maximize -a - b,

subject to:

a + b ≤ 0,

a + 2b ≤ 0,

and also other constraints from the polynomial being bounded by 1 on [0,1]. Wait, but we already used the constraints at x=0, x=1, and x=0.5. However, there are more constraints from other points. For example, the polynomial might have a maximum or minimum inside the interval.

But if we assume that the maximum of |ax² + bx + c| occurs at x=0, x=1, and x=0.5, then we have already considered those constraints. However, if there are other points where the polynomial could exceed the bound, then we need to ensure that. But if we have already set the polynomial to reach the maximum at those three points, then maybe it's safe?

But in the previous example, when we set f(0)=1, f(1)=-1, f(1/2)=1, the polynomial exceeded the bound. Therefore, even if we set the constraints at those three points, we need to ensure that the polynomial doesn't exceed the bound elsewhere. Therefore, maybe the only safe way is to have the polynomial equioscillate between the maximum and minimum at three points, which is the Chebyshev case.

Alternatively, let me try to use the three constraints:

c = 1,

a + b + c = 1,

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

Which gives us a = 8, b = -8, c = 1. Then, check if the polynomial is within bounds. As we saw earlier, it is. Therefore, this is valid.

Alternatively, suppose that instead of c = 1, we have c = -1. Then, similar logic would give us a polynomial that might have |a| + |b| + |c| = 17 as well. For example, if we set f(0) = -1, f(1/2) = 1, f(1) = -1. Then:

c = -1,

a + b + c = -1 ⇒ a + b = 0,

(a/4) + (b/2) + c = 1 ⇒ (a/4) + (b/2) - 1 = 1 ⇒ (a/4) + (b/2) = 2 ⇒ Multiply by 4: a + 2b = 8.

From a + b = 0 ⇒ a = -b. Substitute into a + 2b = 8 ⇒ -b + 2b = 8 ⇒ b = 8 ⇒ a = -8. Therefore, the polynomial is -8x² + 8x - 1. Let me check its values:

At x=0: -1, okay.

At x=1: -8 + 8 -1 = -1, okay.

At x=0.5: -8*(0.25) + 8*(0.5) -1 = -2 + 4 -1 = 1, okay.

Vertex at x = -b/(2a) = -8/(2*(-8)) = 0.5. Wait, so the vertex is at x=0.5, which is the point where it reaches 1. Therefore, the polynomial is symmetric around x=0.5? Let me check. Let me compute f(0.5 + t) and f(0.5 - t):

f(0.5 + t) = -8*(0.5 + t)^2 + 8*(0.5 + t) -1

= -8*(0.25 + t + t²) + 4 + 8t -1

= -2 -8t -8t² + 4 + 8t -1

= (-2 + 4 -1) + (-8t + 8t) + (-8t²)

= 1 - 8t²

Similarly, f(0.5 - t) = -8*(0.5 - t)^2 + 8*(0.5 - t) -1

= -8*(0.25 - t + t²) + 4 - 8t -1

= -2 + 8t -8t² + 4 - 8t -1

= (-2 + 4 -1) + (8t -8t) + (-8t²)

= 1 - 8t²

Therefore, the polynomial is symmetric around x=0.5, and its maximum is 1 at x=0.5, and minimum is -1 at x=0 and x=1. Therefore, this polynomial is also valid, and |a| + |b| + |c| = 8 + 8 + 1 = 17.

Therefore, whether c is 1 or -1, we get the same sum.

Is there a way to get a higher value? Let me think. Suppose that instead of having the polynomial reach 1 and -1 at three points, we have it reach 1 at two points and -1 at one, but with different coefficients? For example, suppose that the polynomial reaches 1 at x=0 and x=1, and reaches some value between -1 and 1 at x=0.5. Then, maybe we can have a larger |a| + |b| + |c|?

Wait, but if the polynomial reaches 1 at x=0 and x=1, then the maximum of the polynomial is 1, but the minimum could be lower. However, since the polynomial is bounded by 1 in absolute value, the minimum must be ≥ -1. So, if we have the polynomial reaching 1 at x=0 and x=1, and reaching some value m ≥ -1 at x=0.5. Then, maybe we can find a polynomial with larger coefficients?

Let me try. Let me assume that f(0) = 1, f(1) = 1, and f(0.5) = m ≥ -1. Then, solving for a, b, c:

c = 1,

a + b + c = 1 ⇒ a + b = 0,

(a/4) + (b/2) + c = m ⇒ (a/4) + (b/2) + 1 = m.

From a + b = 0 ⇒ b = -a. Substitute into the third equation:

(a/4) + (-a)/2 + 1 = m ⇒ (a/4 - 2a/4) + 1 = m ⇒ (-a/4) + 1 = m ⇒ -a/4 = m - 1 ⇒ a = 4(1 - m).

Therefore, a = 4(1 - m), b = -4(1 - m), c = 1.

Therefore, the polynomial is:

f(x) = 4(1 - m)x² - 4(1 - m)x + 1.

Let me compute |a| + |b| + |c| = |4(1 - m)| + | -4(1 - m)| + |1| = 8|1 - m| + 1.

Since m ≥ -1, then 1 - m ≤ 2. Therefore, to maximize 8|1 - m| + 1, we need to maximize |1 - m|. Since m ≥ -1, the maximum of |1 - m| occurs when m is as small as possible, i.e., m = -1. Then |1 - (-1)| = 2, so |a| + |b| + |c| = 8*2 + 1 = 17. Which is the same as before. Therefore, even in this case, the maximum is 17.

Therefore, whether we set m = -1 or not, the maximum is achieved when m = -1, which gives us the Chebyshev polynomial. Therefore, this suggests that 17 is indeed the maximum.

Alternatively, suppose that the polynomial reaches 1 at one point and -1 at two points. For example, f(0) = -1, f(1/2) = 1, f(1) = -1. Which is similar to the previous case, and gives the same result.

Alternatively, suppose that the polynomial reaches 1 at one point and -1 at another, and maybe another point. But since it's a quadratic, it can only have one extremum. Therefore, if it reaches 1 at one endpoint and -1 at the other, then the extremum is somewhere in between. If the extremum is within the interval, then it can reach another value. But to have the maximum absolute value, it would have to reach 1 or -1 at the extremum.

For example, suppose that f(0) = 1, f(1) = -1, and the extremum is at x = t in (0,1) where f(t) = 1 or -1. Let me try this.

Let me assume that f(0) = 1, f(1) = -1, and f(t) = 1 for some t in (0,1). Then, as before, we have:

c = 1,

a + b + c = -1 ⇒ a + b = -2,

a t² + b t + c = 1 ⇒ a t² + b t = 0.

From the third equation: a t² + b t = 0 ⇒ a t + b = 0 (since t ≠ 0) ⇒ b = -a t.

From the second equation: a - a t = -2 ⇒ a(1 - t) = -2 ⇒ a = -2 / (1 - t).

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

Therefore, the polynomial is:

f(x) = [-2/(1 - t)]x² + [2t/(1 - t)]x + 1.

Now, we need to ensure that this polynomial doesn't exceed the bounds of [-1, 1] on [0,1]. The extremum is at x = -b/(2a) = -[2t/(1 - t)] / [2*(-2/(1 - t))] = [2t/(1 - t)] / [4/(1 - t)] = t/2.

So the extremum is at x = t/2. Let me compute f(t/2):

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

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

= [-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 within [-1, 1]. Since we already have f(0) = 1, f(1) = -1, and f(t) = 1, we need to ensure that the extremum at x = t/2 doesn't exceed the bounds.

So, let's compute f(t/2):

[t²/(2(1 - t))] + 1 ≤ 1 ⇒ [t²/(2(1 - t))] ≤ 0. Since t ∈ (0,1), denominator 1 - t > 0, numerator t² ≥ 0. Therefore, [t²/(2(1 - t))] ≥ 0. Therefore, f(t/2) ≥ 1. But we need f(t/2) ≤ 1. Therefore, equality holds only if t = 0, which is not in (0,1). Therefore, this implies that f(t/2) > 1, which violates the condition. Therefore, this polynomial exceeds the bound. Therefore, this approach doesn't work.

Therefore, the only way to have the polynomial stay within the bounds is if the extremum is at the point where it reaches the minimum or maximum. Which is the case for the Chebyshev polynomial. Therefore, the maximum |a| + |b| + |c| is indeed 17.

Alternatively, let me think of another way. Suppose that we want to maximize |a| + |b| + |c|. Let me consider that for the polynomial f(x) = ax² + bx + c, we can write:

At x=0: c = f(0),

At x=1: a + b + c = f(1),

At x=1/2: a/4 + b/2 + c = f(1/2).

Therefore, if we can express |a| + |b| + |c| in terms of f(0), f(1), f(1/2), then maybe we can find the maximum.

Earlier, we found that:

a = 2f(1) + 2f(0) - 4f(1/2),

b = 4f(1/2) - 3f(0) - f(1),

c = f(0).

Therefore, |a| + |b| + |c| = |2f(1) + 2f(0) - 4f(1/2)| + |4f(1/2) - 3f(0) - f(1)| + |f(0)|.

But since |f(x)| ≤ 1 for all x in [0,1], then |f(0)| ≤ 1, |f(1)| ≤ 1, |f(1/2)| ≤ 1. However, the combination of these terms can be complex. However, if we assume that f(0) = 1, f(1) = 1, f(1/2) = -1, then:

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

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

c = 1,

which gives |a| + |b| + |c| = 17.

Alternatively, if we take other values, say f(0) = 1, f(1) = -1, f(1/2) = 1, then:

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

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

c = 1,

which gives |a| + |b| + |c| = 4 + 2 + 1 = 7, which is much less than 17.

Alternatively, if we take f(0) = 1, f(1) = 1, f(1/2) = 1, then:

a = 2*1 + 2*1 - 4*1 = 2 + 2 - 4 = 0,

b = 4*1 - 3*1 - 1 = 4 - 3 - 1 = 0,

c = 1,

so |a| + |b| + |c| = 0 + 0 + 1 = 1, which is worse.

Alternatively, if we take f(0) = 1, f(1) = -1, f(1/2) = -1:

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

b = 4*(-1) - 3*1 - (-1) = -4 - 3 + 1 = -6,

c = 1,

so |a| + |b| + |c| = 4 + 6 + 1 = 11, which is still less than 17.

Alternatively, if we take f(0) = 1, f(1) = 1, f(1/2) = -1, which gives us the Chebyshev case, which gives 17.

Alternatively, suppose that f(0) = 1, f(1) = -1, f(1/2) = -1. Wait, we did that above, which gives 11.

Alternatively, suppose that f(0) = 1, f(1) = 1, f(1/2) = something else. For example, suppose f(1/2) = -1, which gives us the Chebyshev case.

Alternatively, suppose that f(0) = 1, f(1) = 1, f(1/2) = m. Then:

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

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

c = 1.

Then |a| + |b| + |c| = |4 - 4m| + |4m - 4| + 1 = 4|1 - m| + 4|1 - m| + 1 = 8|1 - m| + 1. To maximize this, we need to maximize |1 - m|. Since m is between -1 and 1, the maximum |1 - m| is when m = -1, giving |1 - (-1)| = 2, so total is 8*2 + 1 = 17. Which again gives us the Chebyshev case.

Therefore, regardless of how we approach it, the maximum value of |a| + |b| + |c| is 17.

Therefore, I think the answer is 17.

**Final Answer**
\boxed{17}

To solve the problem, we are given that for all $ x \in [0,1] $, the inequality
$$
|ax^2 + bx + c| \leq 1
$$
holds, and we are to find the **maximum possible value** of $ |a| + |b| + |c| $.

---

### Step 1: Understanding the Constraint

The condition $ |ax^2 + bx + c| \leq 1 $ for all $ x \in [0,1] $ means that the quadratic polynomial $ f(x) = ax^2 + bx + c $ is bounded between $-1$ and $1$ on the interval $[0,1]$. This is a **boundedness constraint** on the polynomial over a closed interval.

Our goal is to **maximize** the sum $ |a| + |b| + |c| $, which is the sum of the absolute values of the coefficients of the polynomial.

---

### Step 2: Strategy and Key Insight

This is a classic optimization problem with a constraint on the maximum of a function over an interval. A powerful tool in such problems is the **Chebyshev polynomial**, which is known to be extremal in the sense of minimizing the maximum deviation from zero over an interval, given a fixed degree.

The **Chebyshev polynomial of degree 2** is:
$$
T_2(x) = 2x^2 - 1
$$
This polynomial oscillates between $-1$ and $1$ on the interval $[-1,1]$, and it is known to be the polynomial of degree 2 with the **smallest maximum absolute value** on that interval among all monic polynomials.

However, our interval is $[0,1]$, not $[-1,1]$. To adapt the Chebyshev polynomial to this interval, we perform a linear transformation:
$$
x = \frac{t + 1}{2}, \quad \text{so that } t \in [-1,1] \text{ corresponds to } x \in [0,1]
$$
Substituting into $ T_2(t) $, we get:
$$
f(x) = 2\left(\frac{t + 1}{2}\right)^2 - 1 = 2\left(\frac{t^2 + 2t + 1}{4}\right) - 1 = \frac{t^2 + 2t + 1}{2} - 1 = \frac{t^2 + 2t - 1}{2}
$$
Now, substituting back $ t = 2x - 1 $, we get:
$$
f(x) = \frac{(2x - 1)^2 + 2(2x - 1) - 1}{2}
$$
Expanding:
$$
(2x - 1)^2 = 4x^2 - 4x + 1, \quad 2(2x - 1) = 4x - 2
$$
So:
$$
f(x) = \frac{4x^2 - 4x + 1 + 4x - 2 - 1}{2} = \frac{4x^2 - 2}{2} = 2x^2 - 1
$$
Thus, the polynomial $ f(x) = 2x^2 - 1 $ is the **Chebyshev polynomial of degree 2** on the interval $[0,1]$, and it oscillates between $-1$ and $1$ at three points: $ x = 0, \frac{1}{2}, 1 $.

---

### Step 3: Compute the Coefficients

From the polynomial $ f(x) = 2x^2 - 1 $, we identify:
- $ a = 2 $
- $ b = 0 $
- $ c = -1 $

So:
$$
|a| + |b| + |c| = |2| + |0| + |-1| = 2 + 0 + 1 = 3
$$

But this is **not** the maximum. Wait — we need to **scale** the polynomial to match the condition $ |f(x)| \leq 1 $ on $[0,1]$, and we want to **maximize** the sum $ |a| + |b| + |c| $.

Let’s consider the polynomial:
$$
f(x) = 8x^2 - 8x + 1
$$
This polynomial is known to oscillate between $-1$ and $1$ at $ x = 0, \frac{1}{2}, 1 $, and it satisfies the condition $ |f(x)| \leq 1 $ on $[0,1]$. Let’s verify:

- $ f(0) = 1 $
- $ f(1) = 8 - 8 + 1 = 1 $
- $ f\left(\frac{1}{2}\right) = 8\cdot\frac{1}{4} - 8\cdot\frac{1}{2} + 1 = 2 - 4 + 1 = -1 $

So it oscillates between $-1$ and $1$ at three points, and the polynomial is bounded in absolute value by 1 on the entire interval. Therefore, it satisfies the condition.

Now compute:
$$
|a| + |b| + |c| = |8| + |-8| + |1| = 8 + 8 + 1 = \boxed{17}
$$

---

### Step 4: Justification of Optimality

This polynomial is the **extremal** one in the class of quadratics bounded by 1 on $[0,1]$, in the sense that it **maximizes** the sum of the absolute values of the coefficients under the constraint $ |f(x)| \leq 1 $ for all $ x \in [0,1] $.

This is a known result from approximation theory, where the **Chebyshev polynomials** are extremal in such problems. The polynomial $ f(x) = 8x^2 - 8x + 1 $ is a scaled version of the Chebyshev polynomial $ T_2(x) $, and it achieves the maximum possible value of $ |a| + |b| + |c| $ under the given constraint.

---

### 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 did (3 + 4i) divided by (1 + 2i) instead, and he got 11/5 - 2/5 i. Now, the question is asking what answer he should have obtained, meaning the correct answer for (-3 + 4i)/(1 + 2i).

First, I need to recall how to divide complex numbers. I think the standard method is to multiply the numerator and the denominator by the complex conjugate of the denominator to rationalize it. That way, the denominator becomes a real number, and then you can separate the real and imaginary parts.

So, let me start by writing down the original problem: (-3 + 4i) divided by (1 + 2i). To divide these, 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)]

First, let me compute the denominator because that might be simpler.

Denominator: (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.

Okay, so the denominator simplifies to 5. That's straightforward.

Now, the numerator: (-3 + 4i)(1 - 2i). Let me expand this using the 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.

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

So, combining all these terms: -3 + 6i + 4i - 8i².

Combine like terms: the real parts are -3 and the imaginary parts are 6i + 4i = 10i. Then, the -8i² term.

But remember that i² is -1, so -8i² becomes -8*(-1) = +8.

So, putting it all together: -3 + 8 + 10i = 5 + 10i.

Therefore, the numerator simplifies to 5 + 10i, and the denominator is 5.

So, dividing numerator by denominator: (5 + 10i)/5 = 5/5 + 10i/5 = 1 + 2i.

Wait, so the correct answer should be 1 + 2i? Let me check that again because I might have made a mistake in the calculation.

Let me recompute the numerator step by step.

Original numerator: (-3 + 4i)(1 - 2i).

First term: -3*1 = -3.

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

Third term: 4i*1 = +4i.

Fourth term: 4i*(-2i) = -8i².

So, combining: -3 + 6i + 4i -8i².

Combine real parts: -3 -8i². Wait, but -8i² is +8, so -3 + 8 = 5.

Imaginary parts: 6i + 4i = 10i.

So total is 5 + 10i. Divided by 5 gives 1 + 2i.

Hmm, that seems correct. But let me check with another method. Maybe plugging into the original problem.

Alternatively, maybe I can check by multiplying the result by the denominator to see if I get the original numerator.

If the result is 1 + 2i, then multiplying by (1 + 2i) should give (-3 + 4i).

Let me compute (1 + 2i)(1 + 2i) = 1^2 + 2*1*2i + (2i)^2 = 1 + 4i + 4i² = 1 + 4i - 4 = -3 + 4i.

Yes! That's exactly the original numerator. So that checks out.

Therefore, the correct answer is 1 + 2i.

But wait, the problem says that William missed the minus sign and did (3 + 4i)/(1 + 2i) and got 11/5 - 2/5 i. Let me verify that as well to make sure.

So, let's compute (3 + 4i)/(1 + 2i). Multiply numerator and denominator by (1 - 2i):

Numerator: (3 + 4i)(1 - 2i). Let's compute that.

First term: 3*1 = 3.

Second term: 3*(-2i) = -6i.

Third term: 4i*1 = 4i.

Fourth term: 4i*(-2i) = -8i².

Combine terms: 3 -6i +4i -8i².

Simplify: 3 - 2i + 8 (since -8i² is +8).

So total is 3 + 8 = 11, and -2i. Therefore, numerator is 11 - 2i.

Denominator is still 5.

Therefore, (11 - 2i)/5 = 11/5 - (2/5)i. Which matches what William got.

So that's correct. Therefore, the correct answer for the original problem is 1 + 2i.

But let me just make sure that I didn't make a mistake in the first calculation.

Original problem: (-3 + 4i)/(1 + 2i).

Multiply numerator and denominator by (1 - 2i):

Numerator: (-3 + 4i)(1 - 2i).

Let me compute this again.

First, distribute -3 over (1 - 2i): -3*1 + (-3)*(-2i) = -3 + 6i.

Then distribute 4i over (1 - 2i): 4i*1 + 4i*(-2i) = 4i - 8i².

So combining all terms: -3 + 6i + 4i -8i².

Which is -3 + 10i + 8 (since -8i² is +8).

So -3 + 8 = 5, so 5 + 10i.

Divide by 5: 1 + 2i.

Yes, that's correct.

Alternatively, maybe I can use another approach. Let me write the division as a complex number.

Let me denote z = (-3 + 4i)/(1 + 2i). Let me write z = a + bi, and solve for a and b.

So, z*(1 + 2i) = -3 + 4i.

Expanding the left side: (a + bi)(1 + 2i) = a*1 + a*2i + bi*1 + bi*2i = a + 2ai + bi + 2bi².

Simplify: a + (2a + b)i + 2b(-1) = a - 2b + (2a + b)i.

Set equal to -3 + 4i.

Therefore, equating real and imaginary parts:

Real: a - 2b = -3

Imaginary: 2a + b = 4

So, we have a system of equations:

1) a - 2b = -3

2) 2a + b = 4

Let me solve this system.

From equation 1: a = -3 + 2b.

Substitute into equation 2:

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

=> -6 + 4b + b = 4

=> -6 + 5b = 4

=> 5b = 10

=> b = 2

Then, substitute back into a = -3 + 2b = -3 + 2*2 = -3 + 4 = 1.

Therefore, a = 1, b = 2. So z = 1 + 2i.

Same result as before. So that's correct.

Therefore, the answer William should have obtained is 1 + 2i.

But let me check once more with another method. Maybe converting to polar form? Although that might be more complicated, but let's try.

First, for the original numerator: -3 + 4i.

The modulus is sqrt((-3)^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5.

The argument θ is arctangent(4 / -3). Since the real part is negative and the imaginary part is positive, it's in the second quadrant.

So θ = π - arctan(4/3).

For the denominator: 1 + 2i.

Modulus is sqrt(1^2 + 2^2) = sqrt(1 + 4) = sqrt(5).

Argument φ is arctangent(2/1) = arctan(2).

Therefore, dividing the two complex numbers: modulus is 5 / sqrt(5) = sqrt(5).

Argument is θ - φ = [π - arctan(4/3)] - arctan(2).

Hmm, not sure if that helps directly, but maybe we can compute it.

Alternatively, maybe convert the result back to rectangular form.

But perhaps this is more complicated. Alternatively, since we already have two methods giving 1 + 2i, and verifying by multiplying back, I think that's solid.

Therefore, the correct answer is 1 + 2i.

But let me check once again with decimal approximations.

Let me compute (-3 + 4i)/(1 + 2i).

If I approximate the values, maybe I can check.

First, compute denominator 1 + 2i. Its magnitude is sqrt(1 + 4) = sqrt(5) ≈ 2.236.

The numerator is -3 + 4i, magnitude 5, as before.

So the magnitude of the result is 5 / sqrt(5) = sqrt(5) ≈ 2.236.

Now, the result we found is 1 + 2i, whose magnitude is sqrt(1^2 + 2^2) = sqrt(5) ≈ 2.236. So that matches.

Also, the argument of the result 1 + 2i is arctan(2/1) = arctan(2) ≈ 63.43 degrees.

The argument of the numerator is arctan(4 / -3) which is in the second quadrant, so π - arctan(4/3). Let me compute arctan(4/3).

arctan(4/3) is approximately 53.13 degrees. Therefore, the argument of the numerator is 180 - 53.13 = 126.87 degrees.

The argument of the denominator is arctan(2/1) ≈ 63.43 degrees.

Therefore, the argument of the result is 126.87 - 63.43 ≈ 63.44 degrees, which matches the argument of 1 + 2i (which is arctan(2) ≈ 63.43 degrees).

So that also checks out.

Therefore, I can be confident that the correct answer is 1 + 2i.

So, William should have obtained 1 + 2i, but he mistakenly did 11/5 - 2/5 i.

Therefore, the answer is 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 method of rationalizing the denominator by multiplying both the numerator and denominator by the **complex conjugate** of the denominator. The complex conjugate of $1 + 2i$ is $1 - 2i$.

---

### Step 1: Multiply numerator and denominator by the conjugate

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

---

### Step 2: Compute the denominator

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

---

### Step 3: Compute the numerator

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

Use the distributive property (FOIL):

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

Now combine all terms:

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

---

### Step 4: Divide numerator by denominator

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

---

### Final Answer

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

This is the correct result that William Sydney Porter should have obtained.

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

First, I need to recall how to divide complex numbers. I think the standard method is to multiply the numerator and the denominator by the complex conjugate of the denominator to rationalize it. That way, the denominator becomes a real number, and then you can separate the real and imaginary parts.

So, let me start by writing down the original problem: (-3 + 4i) divided by (1 + 2i). To divide these, 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)]

First, let me compute the denominator because that might be simpler.

Denominator: (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.

Okay, so the denominator simplifies to 5. That's straightforward.

Now, the numerator: (-3 + 4i)(1 - 2i). Let me expand this using the 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.

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

So, combining all these terms: -3 + 6i + 4i - 8i².

Combine like terms: the real parts are -3 and the imaginary parts are 6i + 4i = 10i. Then, the -8i² term.

But remember that i² is -1, so -8i² becomes -8*(-1) = +8.

So, putting it all together: -3 + 8 + 10i = 5 + 10i.

Therefore, the numerator simplifies to 5 + 10i, and the denominator is 5.

So, dividing numerator by denominator: (5 + 10i)/5 = 5/5 + 10i/5 = 1 + 2i.

Wait, so the correct answer should be 1 + 2i? Let me check that again because I might have made a mistake in the calculation.

Let me recompute the numerator step by step.

Original numerator: (-3 + 4i)(1 - 2i).

First term: -3*1 = -3.

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

Third term: 4i*1 = +4i.

Fourth term: 4i*(-2i) = -8i².

So, combining: -3 + 6i + 4i -8i².

Combine real parts: -3 -8i². Wait, but -8i² is +8, so -3 + 8 = 5.

Imaginary parts: 6i + 4i = 10i.

So total is 5 + 10i. Divided by 5 gives 1 + 2i.

Hmm, that seems correct. But let me check with another method. Maybe plugging into the original problem.

Alternatively, maybe I can check by multiplying the result by the denominator to see if I get the original numerator.

If the result is 1 + 2i, then multiplying by (1 + 2i) should give (-3 + 4i).

Let me compute (1 + 2i)(1 + 2i) = 1^2 + 2*1*2i + (2i)^2 = 1 + 4i + 4i² = 1 + 4i - 4 = -3 + 4i.

Yes! That's exactly the original numerator. So that checks out.

Therefore, the correct answer is 1 + 2i.

But wait, the problem says that William missed the minus sign and did (3 + 4i)/(1 + 2i) and got 11/5 - 2/5 i. Let me verify that as well to make sure.

So, let's compute (3 + 4i)/(1 + 2i). Multiply numerator and denominator by (1 - 2i):

Numerator: (3 + 4i)(1 - 2i). Let's compute that.

First term: 3*1 = 3.

Second term: 3*(-2i) = -6i.

Third term: 4i*1 = 4i.

Fourth term: 4i*(-2i) = -8i².

Combine terms: 3 -6i +4i -8i².

Simplify: 3 - 2i + 8 (since -8i² is +8).

So total is 3 + 8 = 11, and -2i. Therefore, numerator is 11 - 2i.

Denominator is still 5.

Therefore, (11 - 2i)/5 = 11/5 - (2/5)i. Which matches what William got.

So that's correct. Therefore, the correct answer for the original problem is 1 + 2i.

But let me just make sure that I didn't make a mistake in the first calculation.

Original problem: (-3 + 4i)/(1 + 2i).

Multiply numerator and denominator by (1 - 2i):

Numerator: (-3 + 4i)(1 - 2i).

Let me compute this again.

First, distribute -3 over (1 - 2i): -3*1 + (-3)*(-2i) = -3 + 6i.

Then distribute 4i over (1 - 2i): 4i*1 + 4i*(-2i) = 4i - 8i².

So combining all terms: -3 + 6i + 4i -8i².

Which is -3 + 10i + 8 (since -8i² is +8).

So -3 + 8 = 5, so 5 + 10i.

Divide by 5: 1 + 2i.

Yes, that's correct.

Alternatively, maybe I can use another approach. Let me write the division as a complex number.

Let me denote z = (-3 + 4i)/(1 + 2i). Let me write z = a + bi, and solve for a and b.

So, z*(1 + 2i) = -3 + 4i.

Expanding the left side: (a + bi)(1 + 2i) = a*1 + a*2i + bi*1 + bi*2i = a + 2ai + bi + 2bi².

Simplify: a + (2a + b)i + 2b(-1) = a - 2b + (2a + b)i.

Set equal to -3 + 4i.

Therefore, equating real and imaginary parts:

Real: a - 2b = -3

Imaginary: 2a + b = 4

So, we have a system of equations:

1) a - 2b = -3

2) 2a + b = 4

Let me solve this system.

From equation 1: a = -3 + 2b.

Substitute into equation 2:

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

=> -6 + 4b + b = 4

=> -6 + 5b = 4

=> 5b = 10

=> b = 2

Then, substitute back into a = -3 + 2b = -3 + 2*2 = -3 + 4 = 1.

Therefore, a = 1, b = 2. So z = 1 + 2i.

Same result as before. So that's correct.

Therefore, the answer William should have obtained is 1 + 2i.

But let me check once more with another method. Maybe converting to polar form? Although that might be more complicated, but let's try.

First, for the original numerator: -3 + 4i.

The modulus is sqrt((-3)^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5.

The argument θ is arctangent(4 / -3). Since the real part is negative and the imaginary part is positive, it's in the second quadrant.

So θ = π - arctan(4/3).

For the denominator: 1 + 2i.

Modulus is sqrt(1^2 + 2^2) = sqrt(1 + 4) = sqrt(5).

Argument φ is arctangent(2/1) = arctan(2).

Therefore, dividing the two complex numbers: modulus is 5 / sqrt(5) = sqrt(5).

Argument is θ - φ = [π - arctan(4/3)] - arctan(2).

Hmm, not sure if that helps directly, but maybe we can compute it.

Alternatively, maybe convert the result back to rectangular form.

But perhaps this is more complicated. Alternatively, since we already have two methods giving 1 + 2i, and verifying by multiplying back, I think that's solid.

Therefore, the correct answer is 1 + 2i.

But let me check once again with decimal approximations.

Let me compute (-3 + 4i)/(1 + 2i).

If I approximate the values, maybe I can check.

First, compute denominator 1 + 2i. Its magnitude is sqrt(1 + 4) = sqrt(5) ≈ 2.236.

The numerator is -3 + 4i, magnitude 5, as before.

So the magnitude of the result is 5 / sqrt(5) = sqrt(5) ≈ 2.236.

Now, the result we found is 1 + 2i, whose magnitude is sqrt(1^2 + 2^2) = sqrt(5) ≈ 2.236. So that matches.

Also, the argument of the result 1 + 2i is arctan(2/1) = arctan(2) ≈ 63.43 degrees.

The argument of the numerator is arctan(4 / -3) which is in the second quadrant, so π - arctan(4/3). Let me compute arctan(4/3).

arctan(4/3) is approximately 53.13 degrees. Therefore, the argument of the numerator is 180 - 53.13 = 126.87 degrees.

The argument of the denominator is arctan(2/1) ≈ 63.43 degrees.

Therefore, the argument of the result is 126.87 - 63.43 ≈ 63.44 degrees, which matches the argument of 1 + 2i (which is arctan(2) ≈ 63.43 degrees).

So that also checks out.

Therefore, I can be confident that the correct answer is 1 + 2i.

So, William should have obtained 1 + 2i, but he mistakenly did 11/5 - 2/5 i.

Therefore, the answer is 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 method of rationalizing the denominator by multiplying both the numerator and denominator by the **complex conjugate** of the denominator. The complex conjugate of $1 + 2i$ is $1 - 2i$.

---

### Step 1: Multiply numerator and denominator by the conjugate

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

---

### Step 2: Compute the denominator

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

---

### Step 3: Compute the numerator

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

Use the distributive property (FOIL):

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

Now combine all terms:

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

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### Step 4: Divide numerator by denominator

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

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### Final Answer

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

This is the correct result that William Sydney Porter should have obtained.

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*}