Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.1 AIME, harder Find the answer

Let $b$ and $c$ be real numbers, and define the polynomial $P(x)=x^{2}+b x+c$ . Suppose that $P(P(1))=P(P(2))=0$ , and that $P(1) \neq P(2)$ . Find $P(0)$ .

Your answer

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

Solution

Since $P(P(1))=P(P(2))=0$ , but $P(1) \neq P(2)$ , it follows that $P(1)=1+b+c$ and $P(2)=4+2 b+c$ are the distinct roots of the polynomial $P(x)$ . Thus, $P(x)$ factors: $P(x) =x^{2}+b x+c =(x-(1+b+c))(x-(4+2 b+c)) =x^{2}-(5+3 b+2 c) x+(1+b+c)(4+2 b+c)$ It follows that $-(5+3 b+2 c)=b$ , and that $c=(1+b+c)(4+2 b+c)$ . From the first equation, we find $2 b+c=-5 / 2$ . Plugging in $c=-5 / 2-2 b$ into the second equation yields $-5 / 2-2 b=(1+(-5 / 2)-b)(4+(-5 / 2))$ Solving this equation yields $b=-\frac{1}{2}$ , so $c=-5 / 2-2 b=-\frac{3}{2}$ .

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