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

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 1: / Line 1: @@
-== Problem 12 ==
+== Problem ==
-For <math>f_n(x)=x^n</math> and <math>a \neq 1</math> consider
+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>
-<math>\text{I. } (f_{11}(a)f_{13}(a))^{14}</math>
-<math>\text{II. } f_{11}(a)f_{13}(a)f_{14}(a)</math>
-<math>\text{III. } (f_{11}(f_{13}(a)))^{14}</math>
-<math>\text{IV. } f_{11}(f_{13}(f_{14}(a)))</math>
-Which of these equal <math>f_{2002}(a)?</math>
 <math>
-\text{(A) I and II only}
+\text{(A) }-8
 \qquad
-\text{(B) II and III only}
+\text{(B) }-4
 \qquad
-\text{(C) III and IV only}
+\text{(C) }0
 \qquad
-\text{(D) II, III, and IV only}
+\text{(D) }5
 \qquad
-\text{(E) all of them}
+\text{(E) }8
 </math>
 == Solution 1==
+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>
 == See also ==
 {{AMC10 box|year=2002|ab=P|num-b=10|num-a=12}}
 {{MAA Notice}}

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

Latest revision as of 06:42, 15 July 2024

Problem

Solution 1

See also