# How do you multiply (-b^2 + 2) /(-b^4 - 6b^2 + 16) * 20/(35b + 70)?

Mar 4, 2018

$= \frac{4}{7 \left({b}^{2} + 8\right) \left(b + 2\right)}$

#### Explanation:

$\textcolor{b l u e}{\frac{\left(- {b}^{2} + 2\right)}{\left(- {b}^{4}\right) - 6 {b}^{2} + 16}} \cdot \frac{20}{35 b + 70}$?

Divide $- 1$ out of the brackets to change the signs

$= \textcolor{b l u e}{\frac{- \left({b}^{2} - 2\right)}{- \left({b}^{4} + 6 {b}^{2} - 16\right)}} \cdot \frac{20}{35 b + 70}$

Negative divided by a negatives gives a positive

Factorise

=+((b^2 - 2)) /((b^4 + 6b^2 - 16)) * 20/(35(b + 2)

$= \frac{\left({b}^{2} - 2\right)}{\left({b}^{2} + 8\right) \left({b}^{2} - 2\right)} \cdot \frac{20}{35 \left(b + 2\right)} \text{ } \leftarrow$

$= \frac{\cancel{\left({b}^{2} - 2\right)}}{\left({b}^{2} + 8\right) \cancel{\left({b}^{2} - 2\right)}} \cdot {\cancel{20}}^{4} / \left({\cancel{35}}^{7} \left(b + 2\right)\right) \text{ } \leftarrow$ cancel

$= \frac{4}{7 \left({b}^{2} + 8\right) \left(b + 2\right)}$