# How do you simplify (3z-6)/(3z^2-12)?

Sep 18, 2016

$\frac{1}{z + 2}$

#### Explanation:

The first step in simplifying is to factorise the numerator/denominator.

Numerator:

3z - 6 has a $\textcolor{b l u e}{\text{common factor}}$ 0f 3.

$\Rightarrow 3 z - 6 = 3 \left(z - 2\right) \leftarrow \text{ factorised form}$

Denominator:

$3 {z}^{2} - 12 \text{ also has a common factor of 3}$

$\Rightarrow 3 {z}^{2} - 12 = 3 \left({z}^{2} - 4\right)$

The factor ${z}^{2} - 4 \text{ is a difference of squares}$ and, in general is factorised as follows.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

now ${\left(z\right)}^{2} = {z}^{2} \text{ and " (2)^2=4rArra=z" and } b = 4$

Thus ${z}^{2} - 4 = \left(z - 2\right) \left(z + 2\right)$

$\Rightarrow 3 {z}^{2} - 12 = 3 \left(z - 2\right) \left(z + 2\right) \leftarrow \text{ in factorised form}$

Transferring these results into the original expression.

$\Rightarrow \frac{3 \left(z - 2\right)}{3 \left(z - 2\right) \left(z + 2\right)}$

cancelling common factors on numerator/denominator gives.

$\frac{{\cancel{3}}^{1} \cancel{\left(z - 2\right)}}{{\cancel{3}}^{1} \cancel{\left(z - 2\right)} \left(z + 2\right)} = \frac{1}{z + 2}$