How do you simplify (2x^2+6x+4)/(4x^2-12x-16)?

Mar 13, 2018

$\frac{x + 2}{2 x - 8}$

Explanation:

To make progress, attempts should be made to factorise the expressions on the top of the fraction (the numerator) and the bottom of the fraction (the denominator) to see if there are any in common that might be cancelled.

As a first observation, all of the coefficients are even so it is going to be possible at least to extract $2$ as a factor. In fact, the coefficients of the quadratic in the denominator are multiples of $4$ so it will be possible to extract that particular factor twice in that expression. That is

$\frac{2 {x}^{2} + 6 x + 4}{4 {x}^{2} - 12 x - 16}$

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

I will leave cancelling of any common factors until the end.

Next, it is possible to factorise both of the quadratics in the numerator and denominator as follows:

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

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

So, now the fun starts cancelling common factors from the numerator and the denominator.

I can cancel the common factors $2$ and $\left(x + 1\right)$, that is

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

implies

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

which may be written

$\frac{x + 2}{2 x - 8}$

There are no more common factors to be cancelled.