How do you multiply (9x^2-18x-7)/(2x-4)*(x-2)/(9x^2-49) and simplify?

Aug 16, 2016

Start by factoring everything:

$= \frac{9 {x}^{2} + 3 x - 21 x - 7}{2 \left(x - 2\right)} \times \frac{x - 2}{\left(3 x - 7\right) \left(3 x + 7\right)}$

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

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

Eliminate what you can:

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

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

Note the restrictions on the variable. These are found by setting the denominator to $0$ and solving in the original expression.

$2 x - 4 = 0$

$2 x = 4$

$x = 2$

AND

$9 {x}^{2} - 49 = 0$

$\left(3 x + 7\right) \left(3 x - 7\right) = 0$

$x = - \frac{7}{3} \mathmr{and} \frac{7}{3}$

Hence, $\frac{9 {x}^{2} - 18 x - 7}{2 x - 4} \times \frac{x - 2}{9 {x}^{2} - 49} = \frac{3 x + 1}{2 \left(3 x + 7\right)} , x \ne 2 , \frac{7}{3} , - \frac{7}{3}$

Hopefully this helps!