How do you factor

(a+b)^2-(2a-b)^2

and

(a+b)^2-4(b+c)^2

without simplifying the expression?

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- Apr 28th 2010, 06:47 PMsinjid9Factoring help
How do you factor

(a+b)^2-(2a-b)^2

and

(a+b)^2-4(b+c)^2

without simplifying the expression? - Apr 28th 2010, 06:51 PMProve It
Each of them is a difference of two squares.

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

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

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

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

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

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

$\displaystyle = (a + 3b + 2c)(a - b - 2c)$. - Apr 28th 2010, 06:51 PMharish21
- Apr 28th 2010, 08:01 PMsinjid9
Thanks for the help, both of you.