How do you write $\frac{z^{3} - 8}{z^{2} + 2 z + 4}$ $(z^3-8)/(z^2+2z+4)$ in simplest form?

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How do you write $\frac{z^{3} - 8}{z^{2} + 2 z + 4}$ $(z^3-8)/(z^2+2z+4)$ in simplest form?

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Answer 1 · 2015-02-05T04:30:01

You first need to factor both the numerator and denominator completely. You will notice that the denominator is factored. We now need to factor the numerator. You should recognize that the numerator is a difference of cubes.

$(\frac{(z - 2) (z^{2} + 2 z + 4)}{(z^{2} + 2 z + 4)})$ $(((z-2)(z^2+2z+4))/((z^2+2z+4)))$

$\frac{(z^{2} + 2 z + 4)}{(z^{2} + 2 z + 4)} = 1$ $((z^2+2z+4))/((z^2+2z+4))=1$

Giving us $(z - 2) \cdot 1$ $(z-2)*1$ which is

$z - 2$ $z-2$

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How do you write $\frac{z^{3} - 8}{z^{2} + 2 z + 4}$ $(z^3-8)/(z^2+2z+4)$ in simplest form?

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How do you write $\frac{z^{3} - 8}{z^{2} + 2 z + 4}$ $(z^3-8)/(z^2+2z+4)$ in simplest form?