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HMMT February

Algebra Difficulty 4.8 AIME Find the answer

Find all the roots of (x2+3x+2)(x2−7x+12)(x2−2x−1)+24=0\left(x^{2}+3 x+2\right)\left(x^{2}-7 x+12\right)\left(x^{2}-2 x-1\right)+24=0.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

We re-factor as (x+1)(x−3)(x+2)(x−4)(x2−2x−1)+24(x+1)(x-3)(x+2)(x-4)\left(x^{2}-2 x-1\right)+24, or (x2−2x−3)(x2−2x−8)(x2−2x−1)+24\left(x^{2}-2 x-3\right)\left(x^{2}-2 x-8\right)\left(x^{2}-2 x-1\right)+24, and this becomes (y−4)(y−9)(y−2)+24(y-4)(y-9)(y-2)+24 where y=(x−1)2y=(x-1)^{2}. Now, (y−4)(y−9)(y−2)+24=(y−8)(y−6)(y−1)(y-4)(y-9)(y-2)+24=(y-8)(y-6)(y-1), so yy is 1, 6, or 8. Thus the roots of the original polynomial are 0,2,1±6,1±22\mathbf{0}, \mathbf{2}, \mathbf{1} \pm \sqrt{6}, 1 \pm 2 \sqrt{2}.

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