Difference between revisions of "1980 AHSME Problems/Problem 2"

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The degree of <math>(x^2+1)^4 (x^3+1)^3</math> as a polynomial in <math>x</math> is
 
The degree of <math>(x^2+1)^4 (x^3+1)^3</math> as a polynomial in <math>x</math> is
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<math>\text{(A)} \ 5 \qquad \text{(B)} \ 7 \qquad \text{(C)} \ 12 \qquad \text{(D)} \ 17 \qquad \text{(E)} \ 72</math>
 
<math>\text{(A)} \ 5 \qquad \text{(B)} \ 7 \qquad \text{(C)} \ 12 \qquad \text{(D)} \ 17 \qquad \text{(E)} \ 72</math>
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== Solution 1==
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It becomes <math> (x^{8}+...)(x^{9}+...) </math> with 8 being the degree of the first factor and 9 being the degree of the second factor, making the degree of the whole thing 17, or <math>\boxed{(D)}</math>
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==Solution 2==
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First note that given a polynomial <math>P(x)</math> and a polynomial <math>Q(x)</math>:
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<math>deg(P(x))^n = ndeg(P(x))</math> and <math>deg(P(x)Q(x)) = deg(P(x))+deg(Q(x))</math>.
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We let <math>x^2+1=P(x)</math> and <math>x^3+1 = Q(x)</math>.
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Hence <math>deg(P(x)) = 2</math> and <math>deg(Q(x)) = 3</math>
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So <math>deg(P(x))^4) = 4\cdot2 = 8</math> and <math>deg(Q(x)^3) = 3\cdot3 = 9</math>
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Now we let <math>P(x)^4 = R(x)</math> and <math>Q(x)^3 = S(x)</math>
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We want to find <math>deg(R(x)S(x)) = deg(R(x))+deg(S(x)) = 9+8 = 17</math>.
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So the answer is '''(D)''' 17.
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*Solution 2 by mihirb
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== See also ==
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{{AHSME box|year=1980|num-b=1|num-a=3}}
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{{MAA Notice}}

Latest revision as of 15:39, 20 March 2018

Problem

The degree of $(x^2+1)^4 (x^3+1)^3$ as a polynomial in $x$ is

$\text{(A)} \ 5 \qquad \text{(B)} \ 7 \qquad \text{(C)} \ 12 \qquad \text{(D)} \ 17 \qquad \text{(E)} \ 72$

Solution 1

It becomes $(x^{8}+...)(x^{9}+...)$ with 8 being the degree of the first factor and 9 being the degree of the second factor, making the degree of the whole thing 17, or $\boxed{(D)}$

Solution 2

First note that given a polynomial $P(x)$ and a polynomial $Q(x)$:


$deg(P(x))^n = ndeg(P(x))$ and $deg(P(x)Q(x)) = deg(P(x))+deg(Q(x))$.


We let $x^2+1=P(x)$ and $x^3+1 = Q(x)$.

Hence $deg(P(x)) = 2$ and $deg(Q(x)) = 3$

So $deg(P(x))^4) = 4\cdot2 = 8$ and $deg(Q(x)^3) = 3\cdot3 = 9$


Now we let $P(x)^4 = R(x)$ and $Q(x)^3 = S(x)$

We want to find $deg(R(x)S(x)) = deg(R(x))+deg(S(x)) = 9+8 = 17$.

So the answer is (D) 17.

  • Solution 2 by mihirb

See also

1980 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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 26 27 28 29 30
All AHSME Problems and Solutions

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