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2017 AMC 12A Problems/Problem 23

Revision as of 15:59, 8 February 2017 by Acegikmoqsuwy2000 (talk | contribs) (→‎Solution)

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 from Vieta's formulas on the quadratic term of $g(x)$ and the cubic term of $f(x)$, we obtain the following:

\begin{align*} r_1+r_2+r_3&=-a \\ r_1+r_2+r_3+r_4&=-1 \end{align*}

Thus $r_4=a-1$.

Now applying Vieta's formulas on the constant term of $g(x)$, the linear term of $g(x)$, and the linear term of $f(x)$, we obtain:

\begin{align*} r_1r_2r_3 & = -10\\ r_1r_2+r_2r_3+r_3r_1 &= 1\\ r_1r_2r_3+r_2r_3r_4+r_3r_4r_1+r_4r_1r_2 & = -100\\ \end{align*}

Substituting for $r_1r_2r_3$ and factoring the remainder of the expression, we obtain:

\[-10+(r_1r_2+r_2r_3+r_3r_1)r_4=-10+r_4=-100\]

It follows 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 $f(1)=91\cdot(-77)=\boxed{\textbf{(C)}\,-7007}$.

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