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

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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 1

$f(x)$ 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

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

(Solution 1.1 picks up here.)

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

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

Solution 1.1

A faster ending to Solution 1 is as follows. We shall solve for only $a$ and $r$ . Since $10-r=100$ , $r=-90$ , and since $a-r=1$ , $a=-89$ . Then, $\begin{align*} f(1)&=(1-r)(1^3+a\cdot1^2+1+10)\\ &=(91)(-77)\\ &=\boxed{\bold{(C)}\, -7007}.\\ \end{align*}$

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 $f(1) = 1 + 1 + (-8009) + 100 + 900 = \boxed{\textbf{(C)}\,-7007}$ .

Solution 3

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 4 (Slight guessing)

Let the roots of $g(x)$ be $r_1$ , $r_2$ , and $r_3$ . Let the roots of $f(x)$ be $r_1$ , $r_2$ , $r_3$ , and $r_4$ . From Vieta's, we have: $\begin{align*} r_1+r_2+r_3=-a \\ r_1+r_2+r_3+r_4=-1 \\ r_4=a-1 \end{align*}$ The fourth root is $a-1$ . Since $r_1$ , $r_2$ , and $r_3$ are common roots, we have: $\begin{align*} f(x)=g(x)(x-(a-1)) \\ f(1)=g(1)(1-(a-1)) \\ f(1)=(a+12)(2-a) \\ f(1)=-(a+12)(a-2) \\ \end{align*}$ Let $a-2=k$ : $\begin{align*} f(1)=-k(k+14) \end{align*}$ Note that $-7007=-1001\cdot(7)=-(7\cdot(11)\cdot(13))\cdot(7)=-91\cdot(77)$ This gives us a pretty good guess of $\boxed{\textbf{(C)}\, -7007}$ .

Solution 5

First off, let's get rid of the $x^4$ term by finding $h(x)=f(x)-xg(x)$ . This polynomial consists of the difference of two polynomials with $3$ common factors, so it must also have these factors. The polynomial is $h(x)=(1-a)x^3 + (b-1)x^2 + 90x + c$ , and must be equal to $(1-a)g(x)$ . Equating the coefficients, we get $3$ equations. We will tackle the situation one equation at a time, starting the $x$ terms. Looking at the coefficients, we get $\dfrac{90}{1-a} = 1$ . $\therefore 90=1-a.$ The solution to the previous is obviously $a=-89$ . We can now find $b$ and $c$ . $\dfrac{b-1}{1-a} = a$ , $\therefore b-1=a(1-a)=-89*90=-8010$ and $b=-8009$ . Finally $\dfrac{c}{1-a} = 10$ , $\therefore c=10(1-a)=10*90=900$ Solving the original problem, $f(1)=1 + 1 + b + 100 + 1 = 102+b+c=102+900-8009=\boxed{\textbf{(C)}\, -7007}$ .

@@ Line 41: / Line 41: @@
 We shall solve for only <math>a</math> and <math>r</math>. Since <math>10-r=100</math>, <math>r=-90</math>, and since <math>a-r=1</math>, <math>a=-89</math>. Then,
 <cmath>\begin{align*}
-f(1)&=(1-r)(x^3+ax^2+x+10)\\
+f(1)&=(1-r)(1^3+a\cdot1^2+1+10)\\
 &=(91)(-77)\\
 &=\boxed{\bold{(C)}\, -7007}.\\
 \end{align*}</cmath>
-Thanks for reading, Rowechen Zhong.
 ==Solution 2==

2017 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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