# If $a^4+64=0$ and $a^2 \ne-4a-8$, what is the value of $a^2-4a$?

If $a^4+64=0$ and $a^2 \ne-4a-8$, what is the value of $a^2-4a$?

I tried to add $-16a^2$ to both sides and doing some algebra but it didnt help much.

• If $a^4=-64$, then $a^2=\pm8i$. What is $(\pm8i)!$ supposed to mean? The extension per $\Gamma$ function? – Hagen von Eitzen Jul 26 '13 at 21:55
• This is a high school question,I think it is supposed to be simple. – guest Jul 26 '13 at 21:58
• By "$!=$", do you mean $\ne$? – 6005 Jul 26 '13 at 21:58
• @HagenvonEitzen: I believe $!=$ is supposed to be $\neq$. OP, please confirm. – Ross Millikan Jul 26 '13 at 21:58
• Yes it means "not equal".Sorry for that. – guest Jul 26 '13 at 21:59

Note that $a^4+64=(a^2+4a+8)(a^2-4a+8)$. So if one of the items on the right is not $0$, then $\dots$
• You are welcome. You can check the factorization by multiplying out. You can maybe discover it by $a^4+64=(a^2+8)^2 -(4a)^2$, difference of squares. Your adding $16a^2$ to both sides was a great idea, would have worked. – André Nicolas Jul 26 '13 at 22:05
Here's how to find the factorization, it is very similar to completing the square, we have \begin{align*}a^4 + 64 &= a^4 + 16a^2 + 64 - 16a^2\\ &= (a^2 + 8)^2 - 16a^2 \\ &= (a^2 - 4a + 8)(a^2 + 4a + 8)\end{align*}