2021 AMC 12B Problems/Problem 16


Let $g(x)$ be a polynomial with leading coefficient $1,$ whose three roots are the reciprocals of the three roots of $f(x)=x^3+ax^2+bx+c,$ where $1<a<b<c.$ What is $g(1)$ in terms of $a,b,$ and $c?$

$\textbf{(A) }\frac{1+a+b+c}c \qquad \textbf{(B) }1+a+b+c \qquad \textbf{(C) }\frac{1+a+b+c}{c^2}\qquad \textbf{(D) }\frac{a+b+c}{c^2} \qquad \textbf{(E) }\frac{1+a+b+c}{a+b+c}$

Solution 1

Note that $f(1/x)$ has the same roots as $g(x)$, if it is multiplied by some monomial so that the leading term is $x^3$ they will be equal. We have \[f(1/x) = \frac{1}{x^3} + \frac{a}{x^2}+\frac{b}{x} + c\] so we can see that \[g(x) = \frac{x^3}{c}f(1/x)\] Therefore \[g(1) = \frac{1}{c}f(1) = \boxed{\textbf{(A) }\frac{1+a+b+c}c}\]

Solution 2 (Vieta's bash)

Let the three roots of $f(x)$ be $d$, $e$, and $f$. (Here e does NOT mean 2.7182818...) We know that $a=-(d+e+f)$, $b=de+ef+df$, and $c=-def$, and that $g(1)=1-\frac{1}{d}-\frac{1}{e}-\frac{1}{f}+\frac{1}{de}+\frac{1}{ef}+\frac{1}{df}-\frac{1}{def}$ (Vieta's). This is equal to $\frac{def-de-df-ef+d+e+f-1}{def}$, which equals $\boxed{(\textbf{A}) \frac{1+a+b+c}{c}}$. -dstanz5

Solution 3 (Fakesolve)

Because the problem doesn't specify what the coefficients of the polynomial actually are, we can just plug in any arbitrary polynomial that satisfies the constraints. Let's take $f(x) = (x+5)^3 = x^3+15x^2+75x+125$. Then $f(x)$ has a triple root of $x = -5$. Then $g(x)$ has a triple root of $-\frac{1}{5}$, and it's monic, so $g(x) = \left(x+\frac{1}{5}\right)^3 = \frac{125x^3+75x^2+15x+1}{125}$. We can see that this is $\frac{1+a+b+c}{c}$, which is answer choice $\boxed{(A)}$.

-Darren Yao

Solution 4

If we let $p, q,$ and $r$ be the roots of $f(x)$, $f(x) = (x-p)(x-q)(x-r)$ and $g(x) = (x-\frac{1}{p})(x-\frac{1}{q})(x-\frac{1}{r})$. The requested value, $g(1)$, is then \[(1-\frac{1}{p})(1-\frac{1}{q})(1-\frac{1}{r}) = \frac{(p-1)(q-1)(r-1)}{pqr}\] The numerator is $-f(1)$ (using the product form of $f(x)$ ) and the denominator is $-c$, so the answer is \[\frac{f(1)}{c} = \boxed{(\textbf{A}) \frac{1+a+b+c}{c}}\]

- gting

Video Solution by Punxsutawney Phil


Video Solution by OmegaLearn (Vieta's Formula)


~ pi_is_3.14

Solution 5 (good at guessing)

g(1) = sum of coefficients. if its (x-r)(x-s)(x-t), then it becomes (x-1/r)(x-1/s)(x-1/t) so -rst becomes -1/rst so c becomes 1/c. Also, there is x^3 so the answer must include 1. the only answer having both of these is A.


Video Solution by Hawk Math


See Also

2021 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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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