Difference between revisions of "2017 AMC 10A Problems/Problem 24"
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&=\boxed{\bold{(C)\text{ }}\text{-}7007}.\ | &=\boxed{\bold{(C)\text{ }}\text{-}7007}.\ | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
+ | |||
+ | ==Solution 2== | ||
+ | We notice that the constant term of <math>f(x)=c</math> and the constant term in <math>g(x)=10</math>. Because <math>f(x)</math> can be factored as <math>g(x) \cdot (x- r)</math> (where <math>r</math> is the unshared root of <math>f(x)</math>, we see that using the constant term, <math>-10 \cdot r = c</math> and therefore <math>r = -\frac{c}{10}</math>. | ||
+ | Now we once again write <math>f(x)</math> out in factored form: | ||
+ | |||
+ | <cmath>f(x) = g(x)\cdot (x-r) = (x^3+ax^2+x+10)(x+\frac{c}{10})</cmath>. | ||
+ | |||
+ | We can expand the expression on the right-hand side to get: | ||
+ | |||
+ | <cmath>f(x) = x^4+(a+\frac{c}{10})x^3+(1+\frac{ac}{10})x^2+(10+\frac{c}{10})x+c</cmath> | ||
+ | |||
+ | Now we have <math>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</math>. | ||
+ | |||
+ | Simply looking at the coefficients for each corresponding term (knowing that they must be equal), we have the equations: | ||
+ | <cmath>10+\frac{c}{10}=100 \Rightarrow c=900</cmath> | ||
+ | <cmath>a+\frac{c}{10} = 1, c=900 \Rightarrow a + 90 =1 \Rightarrow a= -89</cmath> | ||
+ | |||
+ | and finally, | ||
+ | |||
+ | <cmath>1+\frac{ac}{10} = b = 1+\frac{-89 \cdot 900}{10} = b = -8009</cmath>. | ||
+ | |||
+ | We know that <math>f(1)</math> is the sum of its coefficients, hence <math>1+1+b+100+c</math>. We substitute the values we obtained for <math>b</math> and <math>c</math> into this expression to get <math>f(1) = 1 + 1 + (-8009) + 100 + 900 = \boxed{-7007}</math> or <math>\boxed{\text{C}}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=A|num-b=8|num-a=10}} | {{AMC10 box|year=2017|ab=A|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:50, 8 February 2017
Contents
[hide]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
must have four roots, three of which are roots of . Using the fact that every polynomial has a unique factorization into its roots, and since the leading coefficient of and are the same, we know that
where is the fourth root of . Substituting and expanding, we find that
Comparing coefficients with , we see that
Let's solve for and . Since , , so . Since , , and . Thus, we know that
Taking , we find that
Solution 2
We notice that the constant term of and the constant term in . Because can be factored as (where is the unshared root of , we see that using the constant term, and therefore . Now we once again write out in factored form:
.
We can expand the expression on the right-hand side to get:
Now we have .
Simply looking at the coefficients for each corresponding term (knowing that they must be equal), we have the equations:
and finally,
.
We know that is the sum of its coefficients, hence . We substitute the values we obtained for and into this expression to get or .
See Also
2017 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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 10 Problems and Solutions |
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