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Difference between revisions of "2017 AMC 10A Problems/Problem 24"

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&=\boxed{\bold{(C)\text{ }}\text{-}7007}.\\
 
&=\boxed{\bold{(C)\text{ }}\text{-}7007}.\\
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 +
 +
==Solution 2==
 +
We notice that the constant term of <math>f(x)=c</math> and the constant term in <math>g(x)=10</math>. Because <math>f(x)</math> can be factored as <math>g(x) \cdot (x- r)</math> (where <math>r</math> is the unshared root of <math>f(x)</math>, we see that using the constant term, <math>-10 \cdot r = c</math> and therefore <math>r = -\frac{c}{10}</math>.
 +
Now we once again write <math>f(x)</math> out in factored form:
 +
 +
<cmath>f(x) = g(x)\cdot (x-r) = (x^3+ax^2+x+10)(x+\frac{c}{10})</cmath>.
 +
 +
We can expand the expression on the right-hand side to get:
 +
 +
<cmath>f(x) = x^4+(a+\frac{c}{10})x^3+(1+\frac{ac}{10})x^2+(10+\frac{c}{10})x+c</cmath>
 +
 +
Now we have <math>f(x) = x^4+(a+\frac{c}{10})x^3+(1+\frac{ac}{10})x^2+(10+\frac{c}{10})x+c=x^4+x^3+bx^2+100x+c</math>.
 +
 +
Simply looking at the coefficients for each corresponding term (knowing that they must be equal), we have the equations:
 +
<cmath>10+\frac{c}{10}=100 \Rightarrow c=900</cmath>
 +
<cmath>a+\frac{c}{10} = 1, c=900 \Rightarrow a + 90 =1 \Rightarrow a= -89</cmath>
 +
 +
and finally,
 +
 +
<cmath>1+\frac{ac}{10} = b = 1+\frac{-89 \cdot 900}{10} = b = -8009</cmath>.
 +
 +
We know that <math>f(1)</math> is the sum of its coefficients, hence <math>1+1+b+100+c</math>. We substitute the values we obtained for <math>b</math> and <math>c</math> into this expression to get <math>f(1) = 1 + 1 + (-8009) + 100 + 900 = \boxed{-7007}</math> or <math>\boxed{\text{C}}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=A|num-b=8|num-a=10}}
 
{{AMC10 box|year=2017|ab=A|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:50, 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

$f(x)$ must have four roots, three of which are roots of $g(x)$. Using the fact that every polynomial has a unique factorization into its roots, and since the leading coefficient of $f(x)$ and $g(x)$ are the same, we know that

\[f(x)=g(x)(x-r)\]

where $r\in\mathbb{C}$ is the fourth root of $f(x)$. Substituting $g(x)$ and expanding, we find that

\begin{align*}f(x)&=(x^3+ax^2+x+10)(x-r)\\ &=x^4+(a-r)x^3+(1-ar)x^2+(10-r)x-10r.\end{align*}

Comparing coefficients with $f(x)$, we see that

\begin{align*} a-r&=1\\ 1-ar&=b\\ 10-r&=100\\ -10r&=c.\\ \end{align*}

Let's solve for $a,b,c,$ and $r$. Since $10-r=100$, $r=-90$, so $c=(-10)(-90)=900$. Since $a-r=1$, $a=-89$, and $b=1-ar=-8009$. Thus, we know that

\[f(x)=x^4+x^3-8009x^2+100x+900.\]

Taking $f(1)$, we find that

\begin{align*} f(1)&=1^4+1^3-8009(1)^2+100(1)+900\\ &=1+1-8009+100+900\\ &=\boxed{\bold{(C)\text{ }}\text{-}7007}.\\ \end{align*}

Solution 2

We notice that the constant term of $f(x)=c$ and the constant term in $g(x)=10$. Because $f(x)$ can be factored as $g(x) \cdot (x- r)$ (where $r$ is the unshared root of $f(x)$, we see that using the constant term, $-10 \cdot r = c$ and therefore $r = -\frac{c}{10}$. Now we once again write $f(x)$ out in factored form:

\[f(x) = g(x)\cdot (x-r) = (x^3+ax^2+x+10)(x+\frac{c}{10})\].

We can expand the expression on the right-hand side to get:

\[f(x) = x^4+(a+\frac{c}{10})x^3+(1+\frac{ac}{10})x^2+(10+\frac{c}{10})x+c\]

Now we have $f(x) = x^4+(a+\frac{c}{10})x^3+(1+\frac{ac}{10})x^2+(10+\frac{c}{10})x+c=x^4+x^3+bx^2+100x+c$.

Simply looking at the coefficients for each corresponding term (knowing that they must be equal), we have the equations: \[10+\frac{c}{10}=100 \Rightarrow c=900\] \[a+\frac{c}{10} = 1, c=900 \Rightarrow a + 90 =1 \Rightarrow a= -89\]

and finally,

\[1+\frac{ac}{10} = b = 1+\frac{-89 \cdot 900}{10} = b = -8009\].

We know that $f(1)$ is the sum of its coefficients, hence $1+1+b+100+c$. We substitute the values we obtained for $b$ and $c$ into this expression to get $f(1) = 1 + 1 + (-8009) + 100 + 900 = \boxed{-7007}$ or $\boxed{\text{C}}$.

See Also

2017 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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

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