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 $f(1)$ ?

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

Video Solution

https://youtu.be/3dfbWzOfJAI?t=4412

~ pi_is_3.14

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

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

Solution 3

Let the roots of $g(x)$ be $p,q,r$ and the roots of $f(x)$ be $p,q,r,s$ . Then by Vietas, $-100=pqr+pqs+prs+qrs = -10+ s(pq+pr+rs) = -10 + s,$ so $s = -90$ . Again by Vietas, $p+q+r+s = -a + s = -1 \implies a = -89$ . Finally, $f(1) = (1-(-90))g(1) = \textbf{(C)}\,-7007$ .

Solution 4

$f(x)$ 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 5

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 6 (Fast)

Let the term $q(x)$ be the linear term that we are solving for in the equation $f(x)=g(x)\cdot q(x)$ . Now, we know that $q(x)=mx-r$ must have $m=1$ , because only $x \cdot x^3=x^4$ . In addition, we know that, by distributing, $10x-rx=100x$ . Therefore, $r=-90$ , and all the other variables are quickly solved for.

Solution 7

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 8

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 9 (Risky)

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 10

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 $c$ . $\dfrac{b-1}{1-a} = a$ , $\therefore b-1=a(1-a)=-89\cdot90=-8010$ and $b=-8009$ . Finally $\dfrac{c}{1-a} = 10$ , $\therefore c=10(1-a)=10\cdot90=900$ Solving the original problem, $f(1)=1 + 1 + b + 100 + c = 102+b+c=102+900-8009=\boxed{\textbf{(C)}\, -7007}$ .

Solution 11

Simple polynomial division is a feasible method. Even though we have variables, we can equate terms at the end of the division so that we can cancel terms. Doing the division of $\frac{f(x)}{g(x)}$ eventually brings us the final step $(1-a)x^3 + (b-1)x^2 + 90x + c$ minus $(1-a)x^3 - (a-a^2)x^2 + (1-a)x + 10(1-a)$ after we multiply $f(x)$ by $(1-a)$ . Now we equate coefficients of same-degree $x$ terms. This gives us $10(1-a) = c, b-1 = a - a^2, 1-a = 90 \Rightarrow a = -89, c = 900, b = -8009$ . We are interested in finding $f(1)$ , which equals $1^4 + 1^3 -8009\cdot1^2 + 100\cdot1 + 900 = \boxed{\textbf{(C)}\,-7007}$ . ~skyscraper

Note

Note that $f(1)$ for any polynomial is simply the sum of the coefficients of the polynomial.

Video Solution 1

https://youtu.be/MBIiz0mroqk

2017 AMC 10A (Problems • Answer Key • Resources)
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2017 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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2017 AMC 12A Problems/Problem 23

Contents

Problem

Video Solution

Solution

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6 (Fast)

Solution 7

Solution 8

Solution 9 (Risky)

Solution 10

Solution 11

Note

Video Solution 1

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