# How do you factor 243(3x - 1)^2 - 48(2y + 3)^2?

Aug 18, 2016

Use the difference of squares property to get $3 \left(27 x + 8 y + 3\right) \left(27 x - 8 y - 21\right)$.

#### Explanation:

What should always jump out at you in a factoring question containing a minus sign and stuff squared is difference of squares:
${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

But the 243 and 48 kind of kill that idea, because they aren't perfect squares. However, if we factor out a $3$, we have:
$3 \left(81 {\left(3 x - 1\right)}^{2} - 16 {\left(2 y + 3\right)}^{2}\right)$

Which can be rewritten as:
$3 \left({\left(9 \left(3 x - 1\right)\right)}^{2} - {\left(4 \left(2 y + 3\right)\right)}^{2}\right)$

Now we can apply difference of squares, with:
$a = 9 \left(3 x - 1\right)$
$b = 4 \left(2 y + 3\right)$

Doing so gives:
$3 \left({\left(9 \left(3 x - 1\right)\right)}^{2} - {\left(4 \left(2 y + 3\right)\right)}^{2}\right)$
=3((9(3x-1)+4(2y+3))(9(3x-1)-4(2y+3))

Let's get rid of some parentheses by distributing:
3((9(3x-1)+4(2y+3))(9(3x-1)-4(2y+3))
$= 3 \left(27 x - 9 + 8 y + 12\right) \left(27 x - 9 - 8 y - 12\right)$

Finally, collect terms:
$3 \left(27 x - 9 + 8 y + 12\right) \left(27 x - 9 - 8 y - 12\right)$
$= 3 \left(27 x + 8 y + 3\right) \left(27 x - 8 y - 21\right)$