# How do you divide and simplify \frac{9x^2-4}{2x-2} -: \frac{21x^2-2x-8}{1} ?

Jun 5, 2018

$\frac{3 x + 2}{2 \left(x - 1\right) \left(7 x + 4\right)}$

#### Explanation:

$\text{begin by factoring numerators/denominators}$

$9 {x}^{2} - 4 \text{ is a "color(blue)"difference of squares}$

•color(white)(x)a^2-b^2=(a-b)(a+b)

$9 {x}^{2} - 4 = \left(3 x - 2\right) \left(3 x + 2\right)$

$2 x - 2 = 2 \left(x - 1\right) \leftarrow \text{ common factor of 2}$

$21 {x}^{2} - 2 x - 8 \leftarrow \textcolor{b l u e}{\text{factor using a-c method}}$

$\text{the factors of the product } 21 \times - 8 = - 168$

$\text{which sum to - 2 are - 14 and + 12}$

$\text{split the middle term using these factors}$

$21 {x}^{2} - 14 x + 12 x - 8$

$= 7 x \left(3 x - 2\right) + 4 \left(3 x - 2\right)$

$= \left(3 x - 2\right) \left(7 x + 4\right)$

$\text{the original can now be expressed as}$

$\frac{\left(3 x - 2\right) \left(3 x + 2\right)}{2 \left(x - 1\right)} \div \frac{\left(3 x - 2\right) \left(7 x + 4\right)}{1}$

$\text{to divide the 2 fractions change division to multiply}$
$\text{and turn the second fraction upside down}$
$\text{cancel common factors on numerator/denominator}$

$= \frac{\cancel{\left(3 x - 2\right)} \left(3 x + 2\right)}{2 \left(x - 1\right)} \times \frac{1}{\cancel{\left(3 x - 2\right)} \left(7 x + 4\right)}$

$= \frac{3 x + 2}{2 \left(x - 1\right) \left(7 x + 4\right)}$