# How do you factor 9xy^2 - 25?

Jun 3, 2015

If you look at 9 and 25, they are both perfect squares. So I suppose if you wanted to factor this, you can call this equation a "difference of two squares". Let's take this for example:

$\left({x}^{\text{*"A)-2)(x^("*} B} + 2\right) = \left(x - 2\right) \left(x + 2\right)$ $= {x}^{2} \cancel{+ 2 x - 2 x} - 4 = {x}^{2} - 4$

where ${x}^{\text{*A}}$ and ${x}^{\text{*B}}$ are just labels to distinguish the first $x$ from the second $x$.

Notice that ${x}^{\text{*"A)x^("*B}} = {x}^{2}$ and $- 2 \cdot 2 = - \left(2 \cdot 2\right) = - \left(4\right)$. Similarly, we can work backwards.

First, we can take the square root of $9 x {y}^{2}$ to get $3 {x}^{\frac{1}{2}} y$. Then, we can take the positive square root of $| 25 |$ to get $5$. Afterwards we can just follow the example above to get:

$\left(3 {x}^{\frac{1}{2}} y - 5\right) \left(3 {x}^{\frac{1}{2}} y + 5\right)$

or

$\left(3 \sqrt{x} y - 5\right) \left(3 \sqrt{x} y + 5\right)$

Normally though, it's unusual to put square roots in factored answers.