# How do you multiply (6x^2 + x - 12 )/(2x^2 - 5x - 12 )*(3x^2 - 14x + 8)/( 9x^2 - 18x + 8)?

May 25, 2015

You can factorize all these quadratic equations first.

In order to factorize them, all tou need to do is find its roots and then turn each root into a factor by equaling it to zero.

For example, let's say you've found $x = \frac{1}{2}$. Then, your factor will be $\left(2 x - 1\right) = 0$.

-

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

Now, we can proceed to cancel some binomials that are both dividing and multiplying this equation.

$\frac{\cancel{3 x - 4} \left(2 x + 3\right)}{\left(2 x + 3\right) \left(x - 4\right)} \cdot \frac{\left(3 x - 2\right) \left(x - 4\right)}{\left(3 x - 2\right) \cancel{3 x - 4}}$

$\frac{\cancel{3 x - 4} \cancel{2 x + 3}}{\cancel{2 x + 3} \left(x - 4\right)} \cdot \frac{\left(3 x - 2\right) \left(x - 4\right)}{\left(3 x - 2\right) \cancel{3 x - 4}}$

$\frac{\cancel{3 x - 4} \cancel{2 x + 3}}{\cancel{2 x + 3} \cancel{x - 4}} \cdot \frac{\left(3 x - 2\right) \cancel{x - 4}}{\left(3 x - 2\right) \cancel{3 x - 4}}$

$\frac{\cancel{3 x - 4} \cancel{2 x + 3}}{\cancel{2 x + 3} \cancel{x - 4}} \cdot \frac{\cancel{3 x - 2} \cancel{x - 4}}{\cancel{3 x - 2} \cancel{3 x - 4}}$

Thus, your result is $1$. :)

May 25, 2015

It may help to factor each of the quadratics into linear factors first - There may be some terms we can eliminate :-)

$6 {x}^{2} + x - 12 = \left(3 x - 4\right) \left(2 x + 3\right)$
$3 {x}^{2} - 14 x + 8 = \left(3 x - 2\right) \left(x - 4\right)$
$2 {x}^{2} - 5 x - 12 = \left(2 x + 3\right) \left(x - 4\right)$
$9 {x}^{2} - 18 x + 8 = \left(3 x - 4\right) \left(3 x - 2\right)$

So

$\frac{6 {x}^{2} + x - 12}{2 {x}^{2} - 5 x - 12} \cdot \frac{3 {x}^{2} - 14 x + 8}{9 {x}^{2} - 18 x + 8}$

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

$= \frac{3 x - 4}{x - 4} \cdot \frac{x - 4}{3 x - 4}$

$= 1$

Note that this equation is only valid when the denominator is non-zero, which will be when $x$ is not one of the values:
$- \frac{3}{2}$, $4$, $\frac{4}{3}$ or $\frac{2}{3}$.