AIME 2018 II · 第 6 题

Solution

The polynomial we are given is rather complicated, so let's use Rational Root Theorem to see if it has any retinal roots. By Rational Root Theorem, $x = 1, -1, 2, -2$ are all possible rational roots. Upon plugging these roots into the polynomial, $x = -2$ and $x = 1$ make the polynomial equal 0 and thus, they are roots that we can factor out.

The polynomial becomes:

$(x - 1)(x + 2)(x^2 + (2a - 1)x + 1)$

Since we know $1$ and $-2$ are real numbers, we only need to focus on the quadratic.

Set the discriminant of the quadratic greater than or equal to 0, to ensure the remaining roots are real.

$(2a - 1)^2 - 4 \geq 0$.

This simplifies to:

$a \geq \dfrac{3}{2}$

or

$a \leq -\dfrac{1}{2}$

This means that the interval $\left(-\dfrac{1}{2}, \dfrac{3}{2}\right)$ is the "bad" interval. The length of the interval where $a$ can be chosen from is 38 units long, while the bad interval is 2 units long. Therefore, the "good" interval is 36 units long.

$\dfrac{36}{38} = \dfrac{18}{19}$ $18 + 19 = \boxed{037}$

~First

Solution 2

This solution is essentially the same as solution 1, but it offers a different approach to factoring the polynomial.

We want to isolate $a$, so we rewrite $x^4 + 2ax^3 + (2a - 2)x^2 + (-4a + 3)x - 2$ as $x^4-2x^2+3x-2+2ax^3+2ax^2-4ax$.

We see that $1$ and $-2$ are roots of $x^4-2x^2+3x-2$, so we can factor, yielding $(x-1)(x+2)(x^2-x+1)$.

We can factor out $2ax$ from the other part of the polynomial, yielding $2ax(x^2+x-2)=2ax(x-1)(x+2)$.

Thus,

\begin{align} x^4 + 2ax^3 + (2a - 2)x^2 + (-4a + 3)x - 2 &= x^4-2x^2+3x-2+2ax^3+2ax^2-4ax \\ &= (x-1)(x+2)(x^2-x+1) + 2ax(x-1)(x+2)\\ &= (x-1)(x+2)(x^2+(2a-1)x+1). \\ \end{align}

The rest of the steps are the same as in Solution 1.

Video Solution

https://www.youtube.com/watch?v=q2oc7n-n6aA ~Shreyas S

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