# How do you factor y= x^4-2x^3-8x+16 ?

Jan 6, 2016

$y = {\left(x - 2\right)}^{2} \left({x}^{2} + 2 \cdot x + 4\right)$
or $y = {\left(x - 2\right)}^{2} \left(x + 1 - i \cdot \sqrt{3}\right) \left(x + 1 + i \cdot \sqrt{3}\right)$

#### Explanation:

Since the last term of the expression is $16 = {2}^{4}$ it's a good idea to try these possible roots: $\pm 1 , \pm 2 , \pm 4 , \pm 8 , \pm 16$ and its inverses, $\pm \frac{1}{2} , \pm \frac{1}{4} , \pm \frac{1}{8} \mathmr{and} \pm \frac{1}{16}$.

Luckily ${x}_{1} = 2$ is a root of the equation:
$f \left(x = 2\right) = {2}^{4} - 2 \cdot {2}^{3} - 8 \cdot 2 + 16 = 16 - 16 - 16 + 16 = 0$

If at first this case of factorization (common factor) was not immediately seen, now we can do as follows
$y = {x}^{4} - 2 {x}^{3} - 8 x + 16$
$y = {x}^{3} \cdot \left(x - 2\right) - 8 \cdot \left(x + 2\right)$ => $y = \left(x - 2\right) \left({x}^{2} - 8\right)$

It remains to factor the polynomial $\left({x}^{3} - 8\right)$ that has one real root and two complex conjugate roots:
${x}^{3} - 8 = 0$ => $x = {\left(8\right)}^{\frac{1}{3}}$
$x = 2 \cdot \left(\cos \left(k \cdot {360}^{\circ} / 3\right) + i \cdot \sin \left(k \cdot {360}^{\circ} / 3\right)\right)$, $k = 0 , 1 , 2$

For $k = 0 \to {x}_{2} = 2 \cdot \left(\cos {0}^{\circ} + i \cdot {\sin}^{\circ}\right) = 2 \cdot \left(1 + i \cdot 0\right)$ => ${x}_{2} = 2$
For $k = 1 \to {x}_{3} = 2 \cdot \left(\cos {120}^{\circ} + i \cdot \sin {120}^{\circ}\right) = 2 \cdot \left(- \frac{1}{2} + i \cdot \frac{\sqrt{3}}{2}\right)$ => ${x}_{3} = - 1 + \sqrt{3}$
For $k = 2 \to {x}_{4} = 2 \cdot \left(\cos {240}^{\circ} + i \cdot \sin {240}^{\circ}\right) = 2 \cdot \left(- \frac{1}{2} - i \cdot \frac{\sqrt{3}}{2}\right)$ => ${x}_{3} = - 1 - \sqrt{3}$

So
$y = \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) \left(x - {x}_{3}\right) \left(x - {x}_{4}\right)$
$y = \left(x - 2\right) \left(x - 2\right) \left(x + 1 - i \cdot \sqrt{3}\right) \left(x + 1 + i \cdot \sqrt{3}\right)$ => $y = {\left(x - 2\right)}^{2} \left(x + 1 - i \cdot \sqrt{3}\right) \left(x + 1 + i \cdot \sqrt{3}\right)$

Or, since $\left(x + 1 - i \cdot \sqrt{3}\right) \left(x + 1 + i \cdot \sqrt{3}\right) = \left({x}^{2} + 2 x + 1 - \left(- 3\right)\right) = {x}^{2} + 2 x + 4$:
$y = {\left(x - 2\right)}^{2} \left({x}^{2} + 2 x + 4\right)$