# How do you factor completely 2x^4+x^3+2x+1?

Apr 21, 2018

$\left(2 x + 1\right) \left(x + 1\right) \left(x - \frac{1}{2} - \frac{1}{2} \sqrt{3} i\right) \left(x - \frac{1}{2} + \frac{1}{2} \sqrt{3} i\right)$

#### Explanation:

$\textcolor{b l u e}{\text{factor by grouping}}$

$= \textcolor{red}{{x}^{3}} \left(2 x + 1\right) \textcolor{red}{+ 1} \left(2 x + 1\right)$

$\text{take out the "color(blue)"common factor } \left(2 x + 1\right)$

$= \left(2 x + 1\right) \left(\textcolor{red}{{x}^{3} + 1}\right)$

${x}^{3} + 1 \text{ is a "color(blue)"sum of cubes}$

•color(white)(x)a^3+b^3=(a+b)(a^2-ab+b^2)

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

$\text{we can factor "x^2-x+1" by solving } {x}^{2} - x + 1 = 0$

$\text{using the "color(blue)"quadratic formula}$

$\text{with "a=1,b=-1" and } c = 1$

$\Rightarrow x = \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1 \pm \sqrt{3} i}{2} = \frac{1}{2} \pm \frac{1}{2} \sqrt{3} i$

$\left(x - \left(\frac{1}{2} + \frac{1}{2} \sqrt{3} i\right)\right) \left(x - \left(\frac{1}{2} - \frac{1}{2} \sqrt{3} i\right)\right)$

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

$\Rightarrow 2 {x}^{4} + {x}^{3} + 2 x + 1$

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