Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.3 AIME, harder Find the answer

3. Given the polynomial
$\begin{aligned} a_{0}+ & \left(a_{1}+4\right) x+ \\ & \left(a_{2}-10\right) x^{2}+\left(a_{3}+6\right) x^{3}+\left(a_{4}-1\right) x^{4}+ \\ & \left(a_{5}-1\right) x^{5}+a_{6} x^{6}+\cdots+a_{2 \alpha 5} x^{2 \omega 5} \end{aligned}$

can be divided by $x^{2}+3 x-2$ , and $\alpha^{2}+3 \alpha-2=0$ . Then
$a_{0}+a_{1} \alpha+a_{2} \alpha^{2}+\cdots+a_{2 \alpha 6} \alpha^{20 \% 5}$

has the value $\qquad$ .

Your answer

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

Solution

3.0 .

Since $\alpha^{2}+3 \alpha-2=0$ , $\alpha$ is a root of the equation $x^{2}+3 x-2=0$ . Therefore,
$\begin{array}{l} a_{0}+\left(a_{1}+4\right) \alpha+\left(a_{2}-10\right) \alpha^{2}+\left(a_{3}+6\right) \alpha^{3}+ \\ \left(a_{4}-1\right) \alpha^{4}+\left(a_{5}-1\right) \alpha^{5}+a_{6} \alpha^{6}+\cdots+a_{2 \cos } \alpha^{2005} \\ =M\left(\alpha^{2}+3 \alpha-2\right)=0, \end{array}$

where $M$ is the quotient. Therefore,
$\begin{array}{l} a_{0}+a_{1} \alpha+a_{2} \alpha^{2}+\cdots+a_{2} \cos \alpha^{2008} \\ =\alpha^{5}+\alpha^{4}-6 \alpha^{3}+10 \alpha^{2}-4 \alpha \\ =\left(\alpha^{3}-2 \alpha^{2}+2 \alpha\right)\left(\alpha^{2}+3 \alpha-2\right)=0 . \end{array}$

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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.