# How do you solve (10h^2 + 21h - 10) /( 12h^2 - 7h - 12 )div( 2h^2 + 9h + 10)/( 4h^2 + 11h + 6 )?

Aug 30, 2016

The expression can be simplified to $\frac{5 h - 2}{3 h - 4}$, with the restrictions being $h \ne - \frac{3}{4} , \frac{4}{3} , - 2 , - \frac{5}{2}$

#### Explanation:

Make it into a multiplication:

$\implies \frac{10 {h}^{2} + 21 h - 10}{12 {h}^{2} - 7 h - 12} \times \frac{4 {h}^{2} + 11 h + 6}{2 {h}^{2} + 9 h + 10}$

Factor everything:

$\implies \frac{10 {h}^{2} + 25 h - 4 h - 10}{12 {h}^{2} - 16 h + 9 h - 12} \times \frac{4 {h}^{2} + 8 h + 3 h + 6}{2 {h}^{2} + 4 h + 5 h + 10}$

$\implies \frac{\left(5 h \left(2 h + 5\right) - 2 \left(2 h + 5\right)\right)}{\left(4 h \left(3 h - 4\right) + 3 \left(3 h - 4\right)\right)} \times \frac{\left(4 h \left(h + 2\right) + 3 \left(h + 2\right)\right)}{2 h \left(h + 2\right) + 5 \left(h + 2\right)}$

$\implies \frac{\left(5 h - 2\right) \left(2 h + 5\right)}{\left(4 h + 3\right) \left(3 h - 4\right)} \times \frac{\left(4 h + 3\right) \left(h + 2\right)}{\left(2 h + 5\right) \left(h + 2\right)}$

Cancel using the property $\frac{a}{a} = 1 , a \ne 0$

$\implies \frac{5 h - 2}{3 h - 4}$

Finally, state restrictions on the variable. They are: $h \ne - \frac{3}{4} , \frac{4}{3} , - 2 \mathmr{and} - \frac{5}{2}$.

Hopefully this helps!