How do I factor this?

Question

How do I factor $p^2+8pq+16q^2-9r^2$? I know how to group the first two terms, but I dont know what to do with the other half. Can someone help me with this problem?

Answer 1

$$(p+4q)^2=p^2+8pq+16q^2\qquad a^2-b^2=(a-b)(a+b)\qquad 9r^2=(3r)^2$$

Answer 2

julien’s already given you the key algebraic facts, so I’ll show how you might think about the problem.

The fact that you have $p^2$, $pq$, and $q^2$ terms suggests that you should try to combine them as a product of the form $(ap+bq)(cp+dq)$. The $9r^2$ term clearly can’t be combined with the other three directly, but it’s an obvious square, $(3r)^2$; if the first three turn out actually to be the square of something of the form $ap+bq$, you’ll have a difference of two squares, which you can then factor.

Answer 3

Using Quadratic Equation formula

$$p=\frac{-8q\pm\sqrt{(8q)^2-4(16q^2-9r^2)}}2=-4q\pm 3r$$

$$\implies p^2+8pq+16q^2-9r^2=\{p-(-4q+3r)\}\{p-(-4q-3r)\}=(p+4q-3r)(p+4q+3r)$$