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

Expanding each factor yields $(x^{8}+...)(x^{9}+...)$ . The degree of the first factor is $8$ , while the degree of the second factor is $9$ . Therefore, the degree of the polynomial is $8 + 9 = \boxed{17\ (\textbf{D})}$

Solution 2

Let $\deg{P(x)}$ be the degree of a polynomial in $x$ . Recall that for any polynomials $P(x)$ and $Q(x)$ , and nonnegative integer $n$ , $\deg{(P(x)^n)} = n \deg{P(x)}$ and $\deg{(P(x)Q(x))} = \deg {P(x)} + \deg {Q(x)}$ .

So $\deg{((x^2+1)^4(x^3+1)^3)}$ $= \deg{((x^2+1)^4)} + \deg{((x^3+1)^3)}$ $= 4 \deg{(x^2+1)} + 3 \deg{(x^3+1)}$ $= 4 \cdot 2 + 3 \cdot 3$ $= \boxed{(\textbf{D})\ 17}$ .

-mihirb, edited by j314andrews

1980 AHSME (Problems • Answer Key • Resources)
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1980 AHSME Problems/Problem 2

Contents

Problem

Solution 1

Solution 2

See also