Difference between revisions of "2017 AMC 12A Problems/Problem 23"
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Hence <math>f(1)=91\cdot(-77)=\boxed{\textbf{(C)}\,-7007}</math>. | Hence <math>f(1)=91\cdot(-77)=\boxed{\textbf{(C)}\,-7007}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Since all of the roots of <math>g(x)</math> are distinct and are roots of <math>f(x)</math>, and the degree of <math>f</math> is one more than the degree of <math>g</math>, we have that | ||
+ | |||
+ | <cmath>f(x) = C(x-k)g(x)</cmath> | ||
+ | |||
+ | for some number <math>k</math>. By comparing <math>x^4</math> coefficients, we see that <math>C=1</math>. Thus, | ||
+ | |||
+ | <cmath>x^4+x^3+bx^2+100x+c=(x-k)(x^3+ax^2+x+10)</cmath> | ||
+ | |||
+ | Expanding and equating coefficients we get that | ||
+ | |||
+ | <cmath>a-k=1,1-ak=b,10-k=100,-10k=c</cmath> | ||
+ | |||
+ | The third equation yields <math>k=-90</math>, and the first equation yields <math>a=-89</math>. So we have that | ||
+ | |||
+ | <math>f(1)=(1+90)g(1)=91(1-89+1+10)=(91)(-77)=boxed{\textbf{(C)}\,-7007}</math> | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2017|ab=A|num-b=22|num-a=24}} | {{AMC12 box|year=2017|ab=A|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 17:50, 8 February 2017
Contents
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:
Thus .
Now applying Vieta's formulas on the constant term of , the linear term of , and the linear term of , we obtain:
Substituting for in the bottom equation 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 .
Solution 2
Since all of the roots of are distinct and are roots of , and the degree of is one more than the degree of , we have that
for some number . By comparing coefficients, we see that . Thus,
Expanding and equating coefficients we get that
The third equation yields , and the first equation yields . So we have that
See Also
2017 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.