Difference between revisions of "2017 AMC 12A Problems/Problem 23"
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==Solution 3== | ==Solution 3== | ||
Let the roots of <math>g(x)</math> be <math>p,q,r</math> and the roots of <math>f(x)</math> be <math>p,q,r,s</math>. Then by Vietas, <cmath>-100=pqr+pqs+prs+qrs = -10+ s(pq+pr+rs) = -10 + s,</cmath>so <math>s = -90</math>. Again by Vietas, <math>p+q+r+s = -a + s = -1 \implies a = -89</math>. Finally, <math>f(1) = (1-(-90))g(1) = \textbf{(C)}\,-7007</math>. | Let the roots of <math>g(x)</math> be <math>p,q,r</math> and the roots of <math>f(x)</math> be <math>p,q,r,s</math>. Then by Vietas, <cmath>-100=pqr+pqs+prs+qrs = -10+ s(pq+pr+rs) = -10 + s,</cmath>so <math>s = -90</math>. Again by Vietas, <math>p+q+r+s = -a + s = -1 \implies a = -89</math>. Finally, <math>f(1) = (1-(-90))g(1) = \textbf{(C)}\,-7007</math>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | <math>f(x)</math> must have four roots, three of which are roots of <math>g(x)</math>. Using the fact that every polynomial has a unique factorization into its roots, and since the leading coefficient of <math>f(x)</math> and <math>g(x)</math> are the same, we know that | ||
+ | |||
+ | <cmath>f(x)=g(x)(x-r)</cmath> | ||
+ | |||
+ | where <math>r\in\mathbb{C}</math> is the fourth root of <math>f(x)</math>. Substituting <math>g(x)</math> and expanding, we find that | ||
+ | |||
+ | <cmath>\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*}</cmath> | ||
+ | |||
+ | Comparing coefficients with <math>f(x)</math>, we see that | ||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | a-r&=1\\ | ||
+ | 1-ar&=b\\ | ||
+ | 10-r&=100\\ | ||
+ | -10r&=c.\\ | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | (Solution 1.1 picks up here.) | ||
+ | |||
+ | Let's solve for <math>a,b,c,</math> and <math>r</math>. Since <math>10-r=100</math>, <math>r=-90</math>, so <math>c=(-10)(-90)=900</math>. Since <math>a-r=1</math>, <math>a=-89</math>, and <math>b=1-ar=-8009</math>. Thus, we know that | ||
+ | |||
+ | <cmath>f(x)=x^4+x^3-8009x^2+100x+900.</cmath> | ||
+ | |||
+ | Taking <math>f(1)</math>, we find that | ||
+ | |||
+ | <cmath>\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*}</cmath> | ||
+ | |||
+ | ==Solution 5== | ||
+ | A faster ending to Solution 1 is as follows. | ||
+ | We shall solve for only <math>a</math> and <math>r</math>. Since <math>10-r=100</math>, <math>r=-90</math>, and since <math>a-r=1</math>, <math>a=-89</math>. Then, | ||
+ | <cmath>\begin{align*} | ||
+ | f(1)&=(1-r)(1^3+a\cdot1^2+1+10)\\ | ||
+ | &=(91)(-77)\\ | ||
+ | &=\boxed{\bold{(C)}\, -7007}.\\ | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | ==Solution 6 (Fast)== | ||
+ | Let the term <math>q(x)</math> be the linear term that we are solving for in the equation <math>f(x)=g(x)\cdot q(x)</math>. Now, we know that <math>q(x)=mx-r</math> must have <math>m=1</math>, because only <math>x \cdot x^3=x^4</math>. In addition, we know that, by distributing, <math>10x-rx=100x</math>. Therefore, <math>r=-90</math>, and all the other variables are quickly solved for. | ||
+ | |||
+ | ==Solution 7== | ||
+ | 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{\textbf{(C)}\,-7007}</math>. | ||
+ | |||
+ | ==Solution 8== | ||
+ | |||
+ | 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> | ||
+ | |||
+ | Thus <math>r_4=a-1</math>. | ||
+ | |||
+ | 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> in the bottom equation 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{(C)}\,-7007}</math>. | ||
+ | |||
+ | ==Solution 9 (risky)== | ||
+ | Let the roots of <math>g(x)</math> be <math>r_1</math>, <math>r_2</math>, and <math>r_3</math>. Let the roots of <math>f(x)</math> be <math>r_1</math>, <math>r_2</math>, <math>r_3</math>, and <math>r_4</math>. From Vieta's, we have: | ||
+ | <cmath>\begin{align*} | ||
+ | r_1+r_2+r_3=-a \\ | ||
+ | r_1+r_2+r_3+r_4=-1 \\ | ||
+ | r_4=a-1 | ||
+ | \end{align*}</cmath> | ||
+ | The fourth root is <math>a-1</math>. Since <math>r_1</math>, <math>r_2</math>, and <math>r_3</math> are common roots, we have: | ||
+ | <cmath>\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*}</cmath> | ||
+ | Let <math>a-2=k</math>: | ||
+ | <cmath>\begin{align*} | ||
+ | f(1)=-k(k+14) | ||
+ | \end{align*}</cmath> | ||
+ | Note that <math>-7007=-1001\cdot(7)=-(7\cdot(11)\cdot(13))\cdot(7)=-91\cdot(77)</math> | ||
+ | This gives us a pretty good guess of <math>\boxed{\textbf{(C)}\, -7007}</math>. | ||
+ | |||
+ | ==Solution 10== | ||
+ | First off, let's get rid of the <math>x^4</math> term by finding <math>h(x)=f(x)-xg(x)</math>. This polynomial consists of the difference of two polynomials with <math>3</math> common factors, so it must also have these factors. The polynomial is <math>h(x)=(1-a)x^3 + (b-1)x^2 + 90x + c</math>, and must be equal to <math>(1-a)g(x)</math>. Equating the coefficients, we get <math>3</math> equations. We will tackle the situation one equation at a time, starting the <math>x</math> terms. Looking at the coefficients, we get <math>\dfrac{90}{1-a} = 1</math>. <cmath>\therefore 90=1-a.</cmath> The solution to the previous is obviously <math>a=-89</math>. We can now find <math>b</math> and <math>c</math>. <math>\dfrac{b-1}{1-a} = a</math>, <cmath>\therefore b-1=a(1-a)=-89*90=-8010</cmath> and <math>b=-8009</math>. Finally <math>\dfrac{c}{1-a} = 10</math>, <cmath>\therefore c=10(1-a)=10*90=900</cmath> Solving the original problem, <math>f(1)=1 + 1 + b + 100 + c = 102+b+c=102+900-8009=\boxed{\textbf{(C)}\, -7007}</math>. | ||
+ | |||
+ | ==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 <math>\frac{f(x)}{g(x)}</math> eventually brings us the final step <math>(1-a)x^3 + (b-1)x^2 + 90x + c</math> minus <math>(1-a)x^3 - (a-a^2)x^2 + (1-a)x + 10(1-a)</math> after we multiply <math>f(x)</math> by <math>(1-a)</math>. Now we equate coefficients of same-degree <math>x</math> terms. This gives us <math> 10(1-a) = c, b-1 = a - a^2, 1-a = 90 \Rightarrow a = -89, c = 900, b = -8009</math>. We are interested in finding <math>f(1)</math>, which equals <math>1^4 + 1^3 -8009\cdot1^2 + 100\cdot1 + 900 = \boxed{\textbf{(C)}\,-7007}</math>. ~skyscraper | ||
+ | |||
+ | ==Note== | ||
+ | |||
+ | Note that <math>f(1)</math> for any polynomial is simply the sum of the coefficients of the polynomial. | ||
+ | |||
==Video Solution== | ==Video Solution== |
Revision as of 23:45, 14 September 2020
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
Solution 3
Let the roots of be and the roots of be . Then by Vietas, so . Again by Vietas, . Finally, .
Solution 4
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
(Solution 1.1 picks up here.)
Let's solve for and . Since , , so . Since , , and . Thus, we know that
Taking , we find that
Solution 5
A faster ending to Solution 1 is as follows. We shall solve for only and . Since , , and since , . Then,
Solution 6 (Fast)
Let the term be the linear term that we are solving for in the equation . Now, we know that must have , because only . In addition, we know that, by distributing, . Therefore, , and all the other variables are quickly solved for.
Solution 7
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 .
Solution 8
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 9 (risky)
Let the roots of be , , and . Let the roots of be , , , and . From Vieta's, we have: The fourth root is . Since , , and are common roots, we have: Let : Note that This gives us a pretty good guess of .
Solution 10
First off, let's get rid of the term by finding . This polynomial consists of the difference of two polynomials with common factors, so it must also have these factors. The polynomial is , and must be equal to . Equating the coefficients, we get equations. We will tackle the situation one equation at a time, starting the terms. Looking at the coefficients, we get . The solution to the previous is obviously . We can now find and . , and . Finally , Solving the original problem, .
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 eventually brings us the final step minus after we multiply by . Now we equate coefficients of same-degree terms. This gives us . We are interested in finding , which equals . ~skyscraper
Note
Note that for any polynomial is simply the sum of the coefficients of the polynomial.
Video Solution
See Also
2017 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 |
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.