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2017 AMC 12A Problems/Problem 23

Revision as of 17:51, 8 February 2017 by Pieater314159 (talk | contribs) (Solution 2)


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 $f(1)$?

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


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:


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


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,


Expanding and equating coefficients we get that


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


See Also

2017 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 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

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