How do you factor $(4y-2)^2-(3y)^2$ ?

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Answer 1 · 2015-06-27T07:37:06

This is a difference of squares:

$(4y-2)^2-(3y)^2$

$= ((4y-2)-3y)((4y-2)+3y)$

$=(y-2)(7y-2)$

For any $a$ and $b$ , we have $a^2-b^2 = (a-b)(a+b)$

Let $a=4y-2$ and $b=3y$

Then

$(4y-2)^2-(3y)^2$

$=a^2 - b^2$

$=(a-b)(a+b)$

$= ((4y-2)-3y)((4y-2)+3y)$

$=(y-2)(7y-2)$

How do you factor (4y-2)^2-(3y)^2(4y-2)^2-(3y)^2?