2002 AMC 10P Problems/Problem 11

Problem

Let $P(x)=kx^3 + 2k^2x^2+k^3.$ Find the sum of all real numbers $k$ for which $x-2$ is a factor of $P(x).$

$\text{(A) }-8 \qquad \text{(B) }-4 \qquad \text{(C) }0 \qquad \text{(D) }5 \qquad \text{(E) }8$

Solution 1

By the factor theorem, $x-2$ is a factor of $P(x)$ if and only if $P(2)=0.$ Therefore, $x$ must equal $2.$ $P(2)=0=2^3k+2(2^2)k^2+k^3,$ which simplifies to $k(k^2+8k+8)=0.$ $k=0$ is a trivial real $0.$ Since $8^2 -4(1)(8)=32 > 0,$ this polynomial does indeed have two real zeros, meaning we can use Vieta’s to conclude that sum of the other two roots are $-8.$

Thus, our answer is $0-8=\boxed{\textbf{(A)}\ -8}.$

2002 AMC 10P (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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All AMC 10 Problems and Solutions

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2002 AMC 10P Problems/Problem 11

Problem

Solution 1

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