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Difference between revisions of "2017 AMC 12A Problems/Problem 23"

(Problem statement and solution to 2017 AMC 12A #23)
 
m (→‎Solution)
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Now we can factor <math>f(x)</math> in terms of <math>g(x)</math> as <math>f(x)=(x-r_4)g(x)=(x+90)g(x)</math>. Then <math>f(1)=91g(1)</math> and <math>g(1)=1^3-89\cdot 1^2+1+10=-77</math>.
 
Now we can factor <math>f(x)</math> in terms of <math>g(x)</math> as <math>f(x)=(x-r_4)g(x)=(x+90)g(x)</math>. Then <math>f(1)=91g(1)</math> and <math>g(1)=1^3-89\cdot 1^2+1+10=-77</math>.
  
−
Hence <math>f(1)=91\cdot(-77)=\boxed{-7007 \qquad\textbf{(D)}}</math>.
+
Hence <math>f(1)=91\cdot(-77)=\boxed{\textbf{(D)}\,-7007}</math>.

Revision as of 15:39, 8 February 2017

Problem

For certain real numbers $a$, $b$, and $c$, the polynomial \[g(x) = x^3 + ax^2 + x + 10\]has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\]What is $f(1)$?

$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005$

Solution

Let $r_1,r_2,$ and $r_3$ be the roots of $g(x)$. Let $r_4$ be the additional root of $f(x)$. Then by Vieta's formulas, $r_1+r_2+r_3=-a$ and $r_1+r_2+r_3+r_4=-1$ so $r_4=a-1$.

Also, Vieta's formulas tell us that $r_1r_2+r_2r_3+r_3r_1=1,$ $r_1r_2r_3=-10.$ and $r_1r_2r_3+r_2r_3r_4+r_3r_4r_1+r_4r_1r_2=-100$.

Hence $-10+(r_1r_2+r_2r_3+r_3r_1)r_4=-100$ so that $r_4=-90$. But $r_4=a-1$ so $a=-89$.

Now we can factor $f(x)$ in terms of $g(x)$ as $f(x)=(x-r_4)g(x)=(x+90)g(x)$. Then $f(1)=91g(1)$ and $g(1)=1^3-89\cdot 1^2+1+10=-77$.

Hence $f(1)=91\cdot(-77)=\boxed{\textbf{(D)}\,-7007}$.

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