Factoring help

**sinjid9** · April 28th 2010, 06:47 PM

How do you factor
(a+b)^2-(2a-b)^2
and
(a+b)^2-4(b+c)^2
without simplifying the expression?

**Prove It** · April 28th 2010, 06:51 PM

Originally Posted by sinjid9

How do you factor
(a+b)^2-(2a-b)^2
and
(a+b)^2-4(b+c)^2
without simplifying the expression?

Each of them is a difference of two squares.

Remember that $(x + y)(x - y) = x^2 - y^2$ .

$(a + b)^2 - (2a - b)^2 = [(a + b) + (2a - b)][(a + b) - (2a - b)]$

$= 3a(2b - a)$ .

$(a+b)^2-4(b+c)^2 = (a + b)^2 - [2(b + c)]^2$

$= (a + b)^2 - (2b + 2c)^2$

$= [(a + b) + (2b + 2c)][(a + b) - (2b + 2c)]$

$= (a + 3b + 2c)(a - b - 2c)$ .

**harish21** · April 28th 2010, 06:51 PM

Originally Posted by sinjid9

How do you factor
(a+b)^2-(2a-b)^2
and
(a+b)^2-4(b+c)^2
without simplifying the expression?

(a+b)^2-(2a-b)^2 is in the form of $x^2-y^2$ where x = a+b and y = (2a-b)

you should know that $a^2-b^2=(a-b)(a+b)$

**sinjid9** · April 28th 2010, 08:01 PM

Thanks for the help, both of you.

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