# How do you simplify  (3x^2 +4x -4) /( 3x^4 -2x^3)?

Sep 18, 2015

$\frac{3 {x}^{2} + 4 x - 4}{3 {x}^{4} - 2 {x}^{3}} = \frac{x + 2}{{x}^{3}}$

#### Explanation:

We factor both the numerator and denominator, then cut whatever's on both.

The denominator is simple enough, just put the highest common term in evidence.

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

The numerator is more complicated, let's assume that there are real numbers $a$ and $b$ such that $\left(3 x + a\right) \left(x - b\right) = 3 {x}^{2} + 4 x - 4$. (Or, we can put that 3 in evidence, and work with $\left(x + a\right) \left(x + b\right)$ but then we'd have to work with rationals.)

By expanding that product we have that
$\left(3 x + a\right) \left(x + b\right) = 3 {x}^{2} + a x + 3 b x + a b$
$3 {x}^{2} + a x + 3 b x + a b = 3 {x}^{2} + \left(a + 3 b\right) x + a b$

So we need to find two numbers that
$a + 3 b = 4$
$a b = - 4$

And that $f \left(- \frac{a}{3}\right)$ and $f \left(- b\right)$ are roots, i.e.: zero the function (you can check that by plugging either value on the parenthesis and seeing it will make a zero).

$a = - 2$ and $b = 2$ are numbers that fit all three criterias. (It's a bit of a trial and error and that's why it's easier to do it this way only when you have nice numbers. If the going gets tough use the quadratic equation to find the roots, and test the positions.)

So know we have:
$\frac{3 {x}^{2} + 4 x - 4}{3 {x}^{4} - 2 {x}^{3}} = \frac{\left(3 x - 2\right) \left(x + 2\right)}{{x}^{3} \left(3 x - 2\right)}$

We cancel the common term
$\frac{3 {x}^{2} + 4 x - 4}{3 {x}^{4} - 2 {x}^{3}} = \frac{x + 2}{{x}^{3}}$