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

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== Problem ==
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Let <math>P(x)=kx^3 + 2k^2x^2+k^3.</math> Find the sum of all real numbers <math>k</math> for which <math>x-2</math> is a factor of <math>P(x).</math>
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<math>
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\text{(A) }-8
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\qquad
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\text{(B) }-4
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\qquad
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\text{(C) }0
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\qquad
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\text{(D) }5
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\qquad
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\text{(E) }8
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</math>
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== Solution 1==
 
== Solution 1==
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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>
  
 
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Thus, our answer is <math>0-8=\boxed{\textbf{(A)}\ -8}.</math>
  
 
== See also ==
 
== See also ==
 
{{AMC10 box|year=2002|ab=P|num-b=10|num-a=12}}
 
{{AMC10 box|year=2002|ab=P|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 06:42, 15 July 2024

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 (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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