# Factoring help

• Apr 28th 2010, 06:47 PM
sinjid9
Factoring 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 PM
Prove It
Quote:

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 \$\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 PM
harish21
Quote:

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 \$\displaystyle x^2-y^2\$ where x = a+b and y = (2a-b)

you should know that \$\displaystyle a^2-b^2=(a-b)(a+b)\$
• Apr 28th 2010, 08:01 PM
sinjid9
Thanks for the help, both of you.