# How do you factor completely a^2 - 2ab - 15b^2?

Apr 18, 2016

${a}^{2} - 2 a b - 15 {b}^{2} = \left(a - 5 b\right) \left(a + 3 b\right)$

#### Explanation:

Since this is a homogeneous polynomial of degree $2$, you can treat it like a quadratic in one variable.

Find a pair of factors of $15$ which differ by $2$. The pair $5 , 3$ works, hence:

${a}^{2} - 2 a b - 15 {b}^{2} = \left(a - 5 b\right) \left(a + 3 b\right)$

Alternatively, you can complete the square and use the difference of squares identity:

${A}^{2} - {B}^{2} = \left(A - B\right) \left(A + B\right)$

with $A = \left(a - b\right)$ and $B = 4 b$ as follows:

${a}^{2} - 2 a b - 15 {b}^{2}$

$= {\left(a - b\right)}^{2} - {b}^{2} - 15 {b}^{2}$

$= {\left(a - b\right)}^{2} - {\left(4 b\right)}^{2}$

$= \left(\left(a - b\right) - 4 b\right) \left(\left(a - b\right) + 4 b\right)$

$= \left(a - 5 b\right) \left(a + 3 b\right)$
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