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Difference between revisions of "2002 AMC 10P Problems/Problem 11"

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== Solution 1==
 
== Solution 1==
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By the factor theorem, <math>x-2</math> is a factor of <math>P(x)</math> if and only if <math>P(2)=0.</math> Therefore, <math>x</math> must equal <math>2.</math> <math>P(2)=0=2^3k+2(2^2)k^2+k^3,</math> which simplifies to <math>k(k^2+8k+8)=0.</math> <math>k=0</math> is a trivial real <math>0.</math> Since <math>8^2 -4(1)(8)=32 > 0,</math> this polynomial does indeed have two real zeros, meaning we can use Vieta’s to conclude that sum of the other two roots are <math>-8.</math>  
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By the [[Factor Theorem | factor theorem]], <math>x-2</math> is a factor of <math>P(x)</math> if and only if <math>P(2)=0.</math> Therefore, <math>x</math> must equal <math>2.</math> <math>P(2)=0=2^3k+2(2^2)k^2+k^3,</math> which simplifies to <math>k(k^2+8k+8)=0.</math> <math>k=0</math> is a trivial real <math>0</math>. Since <math>8^2 -4(1)(8)=32 > 0,</math> this polynomial does indeed have two real zeros, meaning we can use Vieta’s to conclude that sum of the other two roots are <math>-8.</math>  
  
 
Thus, our answer is <math>0-8=\boxed{\textbf{(A)}\ -8}.</math>
 
Thus, our answer is <math>0-8=\boxed{\textbf{(A)}\ -8}.</math>
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{{AMC10 box|year=2002|ab=P|num-b=10|num-a=12}}
 
{{AMC10 box|year=2002|ab=P|num-b=10|num-a=12}}
 
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{{MAA Notice}}
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[[Category: Intermediate Algebra Problems]]

Latest revision as of 21:43, 15 October 2025

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 $0-8=\boxed{\textbf{(A)}\ -8}.$

See also

2002 AMC 10P (Problems • Answer Key • Resources)
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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