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The part of the problem I'm doing has me factoring this: $a^2-2a$ and I'm at a loss on how to factor it. Would I be right in saying:

$(a-1)^2$

Okay so I just ran across this part now: $a^4-16$ I'm pretty bad at factoring and my teacher neglected to teach how to do this so I'm in the process of trying to figure stuff out for myself lol.

Would I need to take the square root of $a^4$ and $16$ to get $a^2 -4$ and then factor from there?

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closed as off-topic by José Carlos Santos, Cesareo, Paul Frost, ancientmathematician, Taroccoesbrocco Feb 15 at 21:47

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  • $\begingroup$ $a^2-2a$ is a difference of two squares ! Indeed, $a^2-2a=a^2-2a+1-1=(a-1)^2-1$. Then the factorization is $((a-1)+1)((a-1)-1)=a(a-2)$. $\endgroup$ – Yves Daoust Apr 6 '16 at 19:44
  • $\begingroup$ @YvesDaoust, that's neat, but why go in such a roundabout way? $\endgroup$ – Yuriy S May 31 '16 at 7:33
  • $\begingroup$ @YuriyS: just continuing on the OP's path, showing it is not a dead-end. (Also, the obvious factorization wasn't worth a comment.) $\endgroup$ – Yves Daoust May 31 '16 at 10:06
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Both terms contain a factor of $a$: $$a^2 - 2a = \color{red}{a}\cdot \color{blue}{a} - \color{red}{2}\cdot \color{blue}{a} = (\color{red}{a - 2})\color{blue}{a} = a(a-2){}{}{}{}{}{}$$

Compare $a^2 - 2a$ to $(a - 1)^2 = a^2 - 2a + 1$.

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  • $\begingroup$ You're welcome, Anonymous! $\endgroup$ – Namaste Jan 28 '14 at 17:00
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First, you can always check a result by substituting particular values, like

$$0^2-2\cdot0\stackrel{?}=(0-1)^2,\\ 1^2-2\cdot0\stackrel{?}=(1-1)^2.$$

This often detects gross errors.

Second, the expression $a^4-16$ is indeed a difference of squares. But which squares ? Those of $a^2$ and $4$. So you can write

$$(a^2)^2-4^2=(a^2+4)(a^2-4).$$

The second factor is itself a difference of squares so that

$$(a^2+4)(a^2-4)=(a^2+4)(a+2)(a-2).$$

Now can we factor $a^2+4$ ? The answer is negative because if that was possible, there would be two factors of the first degree, like

$$a^2+4=(a-r)(a-s)$$

and the expression would cancel for $a=r$ with

$$r^2+4=(r-r)(r-s)=0.$$

Obviously, this is impossible as $r^2+4$ is a strictly positive number.

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No it would not be right since $(a-1)^2 = (a-1)(a-1) = a^2-a-a+1 = a^2-2a+1$.

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To factor, you don't always need 2 binomials. We can have 2 trinomials for some and even a nominal and binomial. In your case. To factor, you have to divide the gcf out. In quadratic equations, the gcf are usually binomials. In your case just divide by the gcf to get a(a-2).

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