2002 AMC 10P Problems/Problem 11

Revision as of 21:09, 14 July 2024 by Wes (talk | contribs) (Solution 1)

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. 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$\boxed{\textbf{(A)}\ -8}.

See also

2002 AMC 10P (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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