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

## 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 $r_1r_2r_3$ in the bottom equation 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 $f(1)=91\cdot(-77)=\boxed{\textbf{(C)}\,-7007}$.

## Solution 2

Since all of the roots of $g(x)$ are distinct and are roots of $f(x)$, and the degree of $f$ is one more than the degree of $g$, we have that

$$f(x) = C(x-k)g(x)$$

for some number $k$. By comparing $x^4$ coefficients, we see that $C=1$. Thus,

$$x^4+x^3+bx^2+100x+c=(x-k)(x^3+ax^2+x+10)$$

Expanding and equating coefficients we get that

$$a-k=1,1-ak=b,10-k=100,-10k=c$$

The third equation yields $k=-90$, and the first equation yields $a=-89$. So we have that

$f(1)=(1+90)g(1)=91(1-89+1+10)=(91)(-77)=\boxed{\textbf{(C)}\,-7007}$

## See Also

 2017 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 22 Followed byProblem 24 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 12 Problems and Solutions

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

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