How do you factor $81 p^{2} - 144 y^{2}$ $81p^2 - 144y^2$ ?

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Answer 1 · 2016-09-27T19:24:51

$(9 p + 12 y) (9 p - 12 y)$ $(9p+12y)(9p-12y)$

If you look carefully at the two terms in the polynomial, you'll quickly figure out that they are square numbers.

Thus, you can write them like this:

$81 p^{2} - 144 y^{2}$ $81p^2-144y^2$ is the same as ${(9 p)}^{2} - {(12 y)}^{2}$ $(9p)^2-(12y)^2$

This actually looks like $a^{2} - b^{2}$ $a^2-b^2$ , where $a = 9 p$ $a=9p$ and $b = 12 y$ $b=12y$

Recall the identity, $a^{2} - b^{2} = (a + b) (a - b)$ $a^2-b^2=(a+b)(a-b)$

Now you just have to put it in that form!

${(9 p)}^{2} - {(12 y)}^{2} = (9 p + 12 y) (9 p - 12 y)$ $(9p)^2-(12y)^2=(9p+12y)(9p-12y)$

How do you factor 81p2−144y281p^2 - 144y^2?