How do you factor out gcf from polynomial x^3 + 15?

Nov 16, 2016

${x}^{3} + 15 = \left({15}^{\frac{1}{3}} + x\right) \left({15}^{\frac{2}{3}} - {15}^{\frac{1}{3}} x + {x}^{2}\right)$

Explanation:

All odd order real polynomial has at least one real root. So we expect to obtain at least one real solution to the proposal

${x}^{3} + 15 = \left({x}^{2} + a x + b\right) \left(x + c\right)$

After equating coefficients we arrive at

$\left\{\begin{matrix}15 - b c = 0 \\ b + a c = 0 \\ a + c = 0\end{matrix}\right.$

Solving for $a , b , c$ we obtain

$a = - \sqrt[3]{15} , b = {\left(\sqrt[3]{15}\right)}^{2} , c = \sqrt[3]{15}$

so

${x}^{3} + 15 = \left({15}^{\frac{1}{3}} + x\right) \left({15}^{\frac{2}{3}} - {15}^{\frac{1}{3}} x + {x}^{2}\right)$