1
$\begingroup$

I have tried to understand the last step in this question, but I don't quite see the logic

Factor $ x^2+4x+4-9y^2$

The first thing I thought was that I could factor by grouping, however this is not successful. So, the next thing I tried to do was factor the

$ x^2+4x+4$ part to get $ (x+2)^2-9y^2$

Then factor out the common terms in the y-portion:

$ (x+2)^2-(3y)^2$ But this is as far as I get. I eventually looked up the answer:

$(x+2-3y)(x+2+3y)$Is this some difference distributive rule?

  • 1
    Or $\rm\ X^2 = B^2\iff X = \pm B\ $ and these roots yield the factorization $\rm X^2 - B^2\ =\ (X-B)\:(X+B) $ which works even if you have no knowledge of difference of squares factorization.2012-04-28

1 Answers 1

2

You are probably familiar with the "difference of squares" rule: $a^2-b^2=(a-b)(a+b).$ Note that your situation with $(x+2)^2-(3y)^2$ is just a difference of squares, where we have $a=x+2$ and $b=3y$. Thus, we get the factorization $(x+2-3y)(x+2+3y).$

The difference of squares rule can be proven by repeated use of the distributive property: $\begin{align*}(a-b)(a+b)&=((a-b)\times a)+((a-b)\times b)\\&=(a^2-ba)+(ab-b^2)\\&=a^2-ab+ab-b^2\\&=a^2-b^2\end{align*}$