Autumn Tournament

Algebra Difficulty 5.0 AIME Prove it Bulgaria

Let $a$ be the largest value of the expression $24y-9y^2$ , where $y$ is a rational number and $b$ is the smallest integer satisfying the inequality
$(t+3)^3 - (6t-7)^2 - (t-9)^3 < 3.$
Factor into irreducible factors with integer coefficients the expression
$a(x-1)x^3 + bx - 2x - 1.$

Solution

We have $24y - 9y^2 = 16 - (3y - 4)^2$ whose largest value $a = 16$ is reached for $y = \frac{4}{3}$ . The given inequality is equivalent to
$\begin{aligned} & t^3 + 9t^2 + 27t + 27 - 36t^2 + 84t - 49 - t^3 + 27t^2 - 243t + 729 < 3 \\ & -132t + 704 < 0, \end{aligned}$
i.e. $t > \frac{16}{3}$ and $b = 6$ . Substituting $a = 16$ , $b = 6$ in the given expression, we get
$\begin{aligned} & 16(x-1)x^3 + 6x - 2x - 1 = 16x^4 - 16x^3 + 4x - 1 \\ & = (16x^4 - 1) - 4x(4x^2 - 1) = (4x^2 - 1)(4x^2 + 1) - 4x(4x^2 - 1) \\ & = (4x^2 - 4x + 1)(2x - 1)(2x + 1) = (2x - 1)^3(2x + 1). \end{aligned}$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.