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1954 AHSME Problems/Problem 32

Revision as of 15:56, 29 January 2020 by Dividend (talk | contribs) (Created page with "== Problem 32== The factors of <math>x^4+64</math> are: <math> \textbf{(A)}\ (x^2+8)^2\qquad\textbf{(B)}\ (x^2+8)(x^2-8)\qquad\textbf{(C)}\ (x^2+2x+4)(x^2-8x+16)\\ \textbf{...")
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Problem 32

The factors of $x^4+64$ are:

$\textbf{(A)}\ (x^2+8)^2\qquad\textbf{(B)}\ (x^2+8)(x^2-8)\qquad\textbf{(C)}\ (x^2+2x+4)(x^2-8x+16)\\ \textbf{(D)}\ (x^2-4x+8)(x^2-4x-8)\qquad\textbf{(E)}\ (x^2-4x+8)(x^2+4x+8)$

Solution

Notice that: \begin{align*} x^4+64 &= x^4+2^6 \\ &= x^4+4\cdot 2^4 \end{align*}

So, using the Sophie Germain Identity, we can factor the expression as: \begin{align*} x^4+64 & = x^4+4\cdot 2^4\\ & = (x^2+2\cdot2^2-4x)(x^2+2\cdot2^2+4x)\\ & = (x^2-4x+8)(x^2+4x+8) \end{align*} So it is clear that our answer is $\boxed{\textbf{(E)}}$.

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