# How do you multiply (8x^2)/(x^2-9)*(x^2+6x+9)/(16x^3)?

Mar 7, 2016

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

#### Explanation:

$\frac{8 {x}^{2}}{{x}^{2} - 9} \cdot \frac{{x}^{2} + 6 x + 9}{16 {x}^{3}}$

1. Factorising $\left({x}^{2} + 6 x + 9\right)$

We can Split the Middle Term of this expression to factorise it.

(x^2 +6x +9) = x^2 +3x +3x+9= x (x+3) + 3 (x+3)= color(purple)((x+3) *(x+3)

2. Factorising $\left({x}^{2} - 9\right)$:
The above expression is of the form ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$
So,(x^2-9) = (x^2 -3^2) = color(blue)((x+3)(x-3)

The expression now becomes:
$\left(\frac{8 {x}^{2}}{\textcolor{b l u e}{\left(x + 3\right) \left(x - 3\right)}}\right) \cdot \left(\frac{\textcolor{p u r p \le}{\left(x + 3\right) \cdot \left(x + 3\right)}}{16 {x}^{3}}\right)$

$= \left(\frac{8 {x}^{2}}{\textcolor{b l u e}{\cancel{\left(x + 3\right)} \left(x - 3\right)}}\right) \cdot \left(\frac{\textcolor{p u r p \le}{\cancel{\left(x + 3\right)} \cdot \left(x + 3\right)}}{16 {x}^{3}}\right)$

$= \left(\frac{8 {x}^{2}}{\left(x - 3\right)} \cdot \frac{\left(x + 3\right)}{16 {x}^{3}}\right)$

$= \left(\frac{\cancel{8 {x}^{2}}}{\left(x - 3\right)} \cdot \frac{\left(x + 3\right)}{\cancel{16 {x}^{3}}}\right)$

=((x+3)) / ((x-3) *(2x)

=((x+3)) / (2x * (x) + 2x * (-3)

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