How do you simplify (2z^3+128)/(16+8z+z^2)?

Oct 2, 2016

The expression can be simplified to $\frac{2 \left({z}^{2} - 4 z + 16\right)}{z + 4}$ with a restriction of $z \ne - 4$.

Explanation:

Factor

$= \frac{2 \left({z}^{3} + 64\right)}{\left(z + 4\right) \left(z + 4\right)}$

Use synthetic division to factor the expression ${z}^{3} + 64$. We know that $z + 4$ is a factor, because by the remainder theorem $f \left(- 4\right) = {\left(- 4\right)}^{3} + 64 = 0$, if $f \left(x\right) = {z}^{3} + 64$.

$- 4 \text{_|"1" "0" "0" "64}$
$\text{ " -4" "16" "-64}$
$\text{--------------------------------------------------}$
$\text{ "1" " -4" "16" } 0$

Hence, when ${z}^{3} + 64$ is divided by $z + 4$, the quotient is ${z}^{2} - 4 z + 16$ with a remainder of $0$. The expression ${z}^{2} - 4 z + 16$ is not factorable, however, because no two numbers multiply to $+ 16$ and add to $- 4$.

So, our initial expression becomes:

$= \frac{2 \left(z + 4\right) \left({z}^{2} - 4 z + 16\right)}{\left(z + 4\right) \left(z + 4\right)}$

Now, eliminate using the property $\frac{a}{a} = 1 , a \ne 0$

$= \frac{2 \left({z}^{2} - 4 z + 16\right)}{z + 4}$

Finally, state your restrictions on the variable. This can be done by setting the original expression to $0$ and solving.

${z}^{2} + 8 x + 16 = 0$

$\left(z + 4\right) \left(z + 4\right) = 0$

$z = - 4$

Hence, $z \ne - 4$.

Hopefully this helps!