# Difference between revisions of "2017 AMC 12A Problems/Problem 23"

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==Solution== | ==Solution== | ||

− | Let <math>r_1,r_2,</math> and <math>r_3</math> be the roots of <math>g(x)</math>. Let <math>r_4</math> be the additional root of <math>f(x)</math>. Then | + | Let <math>r_1,r_2,</math> and <math>r_3</math> be the roots of <math>g(x)</math>. Let <math>r_4</math> be the additional root of <math>f(x)</math>. Then from Vieta's formulas on the quadratic term of <math>g(x)</math> and the cubic term of <math>f(x)</math>, we obtain the following: |

− | + | <cmath>\begin{align*} | |

+ | r_1+r_2+r_3&=-a \\ | ||

+ | r_1+r_2+r_3+r_4&=-1 | ||

+ | \end{align*}</cmath> | ||

− | + | so <math>r_4=a-1</math>. | |

− | Now we can factor <math>f(x)</math> in terms of <math>g(x)</math> as < | + | Now applying Vieta's formulas on the constant term of <math>g(x)</math>, the linear term of <math>g(x)</math>, and the linear term of <math>f(x)</math>, we obtain: |

+ | |||

+ | <cmath>\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*}</cmath> | ||

+ | |||

+ | Substituting for <math>r_1r_2r_3</math> and factoring the remainder of the expression, we obtain: | ||

+ | |||

+ | <cmath>-10+(r_1r_2+r_2r_3+r_3r_1)r_4=-10+r_4=-100</cmath> | ||

+ | |||

+ | It follows that <math>r_4=-90</math>. But <math>r_4=a-1</math> so <math>a=-89</math> | ||

+ | |||

+ | Now we can factor <math>f(x)</math> in terms of <math>g(x)</math> as | ||

+ | |||

+ | <cmath>f(x)=(x-r_4)g(x)=(x+90)g(x)</cmath> | ||

+ | |||

+ | Then <math>f(1)=91g(1)</math> and | ||

+ | |||

+ | <cmath>g(1)=1^3-89\cdot 1^2+1+10=-77</cmath> | ||

Hence <math>f(1)=91\cdot(-77)=\boxed{\textbf{(D)}\,-7007}</math>. | Hence <math>f(1)=91\cdot(-77)=\boxed{\textbf{(D)}\,-7007}</math>. |

## Revision as of 15:57, 8 February 2017

## Problem

For certain real numbers , , and , the polynomial has three distinct roots, and each root of is also a root of the polynomial What is ?

## Solution

Let and be the roots of . Let be the additional root of . Then from Vieta's formulas on the quadratic term of and the cubic term of , we obtain the following:

so .

Now applying Vieta's formulas on the constant term of , the linear term of , and the linear term of , we obtain:

Substituting for and factoring the remainder of the expression, we obtain:

It follows that . But so

Now we can factor in terms of as

Then and

Hence .