How do you factor $4(x+3)^2-9(x-1)^2$ ?

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Answer 1 · 2016-02-14T00:55:26

This can be factored as a difference of squares.

Differences of squares are of the form $a^2-b^2=(a - b)(a + b)$ . We can find the factors by finding the square roots of each part of the expression:

$sqrt(4(x+3)^2)=2(x+3)$

$sqrt(9(x-1)^2)=3(x-1)$

Thus, the expression can be factored into:

$= (2(x + 3) - 3(x - 1))(2(x + 3) + 3(x - 1))$

$=(2x+6-3x+3)(2x+6+3x-3)$

$=(9-x)(3+5x)$

Hopefully this helps!

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