# How do you factor completely x^2-2xy-15y^2?

Jul 3, 2016

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

#### Explanation:

${x}^{2} - 2 x y - 15 {y}^{2}$
Looking at the given algebraic expression we recognize from the first two terms that to factor the expression we have to apply the property:
$\textcolor{b l u e}{{\left(x - y\right)}^{2} = {x}^{2} - 2 x y + {y}^{2}}$
But in the given expression we need the term ${y}^{2}$ so we can add it and subtract so that as if $0$ is added to the expression.
Let's add ${y}^{2}$ then subtract it
$= {x}^{2} - 2 x y - 15 {y}^{2} + {y}^{2} - {y}^{2}$
$= {x}^{2} - 2 x y + {y}^{2} - 15 {y}^{2} - {y}^{2}$
$= {\left(x - y\right)}^{2} - 16 {y}^{2}$
$= {\left(x - y\right)}^{2} - {\left(4 y\right)}^{2}$
Checking the last step reached it is the difference of two squares that says:
$\textcolor{b l u e}{{a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)}$
where in our case:$a = \left(x - y\right)$ and $b = 4 y$
Then,
${\left(x - y\right)}^{2} - {\left(4 y\right)}^{2}$
$= \left(x - y - 4 y\right) \left(x - y + 4 y\right)$
$= \left(x - 5 y\right) \left(x + 3 y\right)$