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One of the things I've struggled with most in algebra/calculus is all the "factoring tricks". When I take time away from doing math I inevitably forget most if not all of them. The old proverb "use it or lose it" definitely comes into play.

I'm a huge fan of Khan Academy and have started to keep a list of tutorials which demonstrate factoring, but I haven't found them all:

Was wondering if anyone could help me fill in the gaps by providing their own list of factoring tips be them in the thread itself, or to online resources like Khan Academy, this site, or elsewhere

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  • $\begingroup$ It is not possible to answer your question without reading through all your links and lessons, because who knows what is covered in "Factoring Special Products". So it would be fine to indicate shortly what kind of expressions are addressed. $\endgroup$ – Phira May 10 '11 at 18:28
  • $\begingroup$ Also you have to be aware that you are also asking about applications of factoring methods in your examples. $\endgroup$ – Phira May 10 '11 at 18:36
  • $\begingroup$ "flagged" for community wiki. $\endgroup$ – Rasmus May 10 '11 at 20:22
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Remembering the geometric series helps for factoring things like $a^n-b^n$:

$$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1})$$

A similar factorization applies for $a^\ell+b^\ell$ for $\ell$ odd:

$$\begin{align*} a^3+b^3&=(a+b)(a^2-ab+b^2)\\ a^5+b^5&=(a+b)(a^4-a^3 b+a^2 b^2-ab^3+b^4) \end{align*}$$

whose general pattern I'll leave for you to tease out.

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Completing a square with a middle term that is a square itself:

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

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    $\begingroup$ Also $x^4-x^2+16=x^4+8x^2+16-9x^2=(x^2+4)^2-(3x)^2=(x^2+4-3x)(x^2+4+3x)$. $\endgroup$ – Isaac May 10 '11 at 19:59
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Completing the rectangle:

$$ab+7a+11b=ab+7a+11b+77-77=(a+11)(b+7)-77.$$

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As indicated in my comment above, you did not really say what material is covered in your links, but in any case I suggest you look at the algebra section of alcumus:

http://www.artofproblemsolving.com/Alcumus/Introduction.php

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Consider also what trinomials look like when squared:

$(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)$

This allows you to take the general two variable quadratic to a linear function of two squares with integer coefficients $(a,b,c,d,e,f)\in \mathbb{Z}^6$:

$ax^2+bxy+cy^2+dx+ey=f \qquad \to$

$\bigg[(4ac-b^2)y+2ae-bd\bigg]+(4ac-b^2)\bigg[2ax+by+d\bigg]^2=(4ac-b^2)(4af+d^2)+(2ae-bd)^2$

Or if you let $$ D=4ac-b^2$$ $$f(x,y)=2ax+by+d$$ $$g(y)=Dy+2ae-bd$$ $$A=D\cdot(4af+d^2)+(2ae-bd)^2 \in \mathbb{Z}$$

it looks like

$\bigg[g(y)\bigg]^2+D\cdot \bigg[f(x,y)\bigg]^2=A$

The above can help when you study diophantine equations

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