# Can you show that z^4 + 64 can be factorised into two real quadratic factors of the form z^2 + az + 8 and z^2 + bz + 8 but cannot be factorised into two real quadratic factors of the form z^2 + bz + 16 and z^2 + bz + 4?

Feb 26, 2017

${z}^{4} + 64 = \left({z}^{2} - 4 z + 8\right) \left({z}^{2} + 4 z + 8\right)$

#### Explanation:

The factors of ${z}^{4} + 64 = \left(z - {z}_{0}\right) \left(z - {z}_{1}\right) \left(z - {z}_{2}\right) \left(z - {z}_{3}\right)$

are obtained solving

${z}^{4} = - 64 = {2}^{6} {e}^{i \pi + i 2 k \pi}$

Here ${e}^{i \pi} = \cos \left(\pi\right) + i \sin \left(\pi\right) = - 1$ (de Moivre's identity)

so

$z = \sqrt[4]{{2}^{6}} {e}^{i \left(\frac{\pi}{4} + k \frac{\pi}{2}\right)}$ obtaining for $k = 0 , 1 , 2 , 3$

${z}_{0} = \sqrt[4]{{2}^{6}} \frac{1 + i}{\sqrt{2}}$
${z}_{1} = - \sqrt[4]{{2}^{6}} \frac{1 + i}{\sqrt{2}}$
${z}_{2} = - \sqrt[4]{{2}^{6}} \frac{1 + i}{\sqrt{2}}$
${z}_{0} = \sqrt[4]{{2}^{6}} \frac{1 - i}{\sqrt{2}}$

The arrangements

$\left(z - {z}_{0}\right) \left(z - {z}_{3}\right) = {z}^{2} - 4 z + 8$
$\left(z - {z}_{1}\right) \left(z - {z}_{2}\right) = {z}^{2} + 4 z + 8$

are the only to give trinomials with real coefficients.

The answer is ${z}^{4} + 64 = \left({z}^{2} - 4 z + 8\right) \left({z}^{2} + 4 z + 8\right)$

Another way to solve this problem is by grouping coefficients in

${z}^{4} + 64 - \left({z}^{2} + a z + 8\right) \left({z}^{2} + b z + 8\right)$ and solving

$\left\{\begin{matrix}16 + a b = 0 \\ a + b = 0\end{matrix}\right.$ giving $a = 4 , b = - 4$

or

${z}^{4} + 64 - \left({z}^{2} + b z + 16\right) \left({z}^{2} + b z + 4\right)$ and solving

$\left\{\begin{matrix}b = 0 \\ 20 + {b}^{2} = 0\end{matrix}\right.$ without solution.