# How do you factor  z^3 + 7z + 7^2 + 7?

Jun 13, 2017

It can be factored in two different ways. Check out the Explanation! :)

#### Explanation:

Apply the power function when it appears on a constant value:
${z}^{3} + 7 z + {7}^{2} + 7 = {z}^{3} + 7 z + 49 + 7 = {z}^{3} + 7 z + 56$.

We can now do two different factorings:

$\to$ by $z$:
$z \left({z}^{2} + 7\right) + 56$

or

$\to$ by $7$:
${z}^{3} + 7 \left(z + 8\right)$

Jul 21, 2017

${z}^{3} + 7 z + {z}^{2} + 7 = \left(z + 1\right) \left({z}^{2} + 7\right)$

$\textcolor{w h i t e}{{z}^{3} + 7 z + {z}^{2} + 7} = \left(z + 1\right) \left(z - \sqrt{7} i\right) \left(z + \sqrt{7} i\right)$

#### Explanation:

I suspect the question has been mistranscribed somewhere along the line. A more plausible cubic that we can factor by grouping would be:

${z}^{3} + 7 z + {z}^{2} + 7 = \left({z}^{3} + 7 x\right) + \left({z}^{2} + 7\right)$

$\textcolor{w h i t e}{{z}^{3} + 7 z + {z}^{2} + 7} = z \left({z}^{2} + 7\right) + 1 \left({z}^{2} + 7\right)$

$\textcolor{w h i t e}{{z}^{3} + 7 z + {z}^{2} + 7} = \left(z + 1\right) \left({z}^{2} + 7\right)$

This can only be factored further using complex coefficients, since ${z}^{2} + 7 > 0$ for any real values of $z$ ...

$\textcolor{w h i t e}{{z}^{3} + 7 z + {z}^{2} + 7} = \left(z + 1\right) \left({z}^{2} - {\left(\sqrt{7} i\right)}^{2}\right)$

$\textcolor{w h i t e}{{z}^{3} + 7 z + {z}^{2} + 7} = \left(z + 1\right) \left(z - \sqrt{7} i\right) \left(z + \sqrt{7} i\right)$