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Algebra Difficulty 4.8 AIME Find the answer

## Task B-1.2.

Factorize into linear factors (x1)(x2)(x3)+(x1)(x2)+1x(x-1)(x-2)(x-3)+(x-1)(x-2)+1-x.

A number or a short expression. Spacing and $ signs are ignored.

Solution

## First solution.

After extracting the common factor (x1)(x-1), we get

(x1)((x2)(x3)+(x2)1)=1 point (x1)(x25x+6+x21)=(x1)(x24x+3)=1 point (x1)(x23xx+3)=1 point  \begin{array}{ll} (x-1)((x-2)(x-3)+(x-2)-1)= & 1 \text { point } \\ (x-1)\left(x^{2}-5 x+6+x-2-1\right)= & \\ (x-1)\left(x^{2}-4 x+3\right)= & 1 \text { point } \\ (x-1)\left(x^{2}-3 x-x+3\right)= & 1 \text { point } \end{array}

Further extraction yields

(x1)(x3)(x1)=(x1)2(x3) (x-1)(x-3)(x-1)=(x-1)^{2}(x-3)

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.