# How do you combine (4a)/(a^2-ab-2b^2) -( 6b)/(a^2+4ab+3b)?

Mar 15, 2016

$\frac{4 a}{{a}^{2} - a b - 2 {b}^{2}} - \frac{6 b}{{a}^{2} + 4 a b + \textcolor{red}{3 {b}^{2}}}$

$= \frac{2 \left(2 {a}^{2} + 3 a b + 6 {b}^{2}\right)}{{a}^{3} + 2 {a}^{2} b - 5 a {b}^{2} - 6 {b}^{3}}$

#### Explanation:

This problem makes more sense if the expression is:

$\frac{4 a}{{a}^{2} - a b - 2 {b}^{2}} - \frac{6 b}{{a}^{2} + 4 a b + \textcolor{red}{3 {b}^{2}}}$

$= \frac{4 a}{\left(a + b\right) \left(a - 2 b\right)} - \frac{6 b}{\left(a + b\right) \left(a + 3 b\right)}$

$= \frac{\left(4 a\right) \left(a + 3 b\right) - \left(6 b\right) \left(a - 2 b\right)}{\left(a + b\right) \left(a - 2 b\right) \left(a + 3 b\right)}$

$= \frac{4 {a}^{2} + 12 a b - 6 a b + 12 {b}^{2}}{\left(a + b\right) \left(a - 2 b\right) \left(a + 3 b\right)}$

$= \frac{2 \left(2 {a}^{2} + 3 a b + 6 {b}^{2}\right)}{\left(a + b\right) \left(a - 2 b\right) \left(a + 3 b\right)}$

$= \frac{2 \left(2 {a}^{2} + 3 a b + 6 {b}^{2}\right)}{\left({a}^{2} - a b - 2 {b}^{2}\right) \left(a + 3 b\right)}$

$= \frac{2 \left(2 {a}^{2} + 3 a b + 6 {b}^{2}\right)}{{a}^{3} + 2 {a}^{2} b - 5 a {b}^{2} - 6 {b}^{3}}$

Note that I did not try to factor $\left(2 {a}^{2} + 3 a b + 6 {b}^{2}\right)$ since it has a negative discriminant $\Delta = {3}^{2} - \left(4 \cdot 2 \cdot 6\right) = 9 - 48 = - 39$, so no linear factors with Real coefficients.