# 1980 AHSME Problems/Problem 2

## 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

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