# How do you factor 3x^4-2x^2-1=0?

May 4, 2018

$\left(3 {x}^{2} + 1\right) \left(x - 1\right) \left(x + 1\right) = 0$

#### Explanation:

$\text{using the a-c method of factoring the quadratic in } {x}^{4}$

$\text{the factors of the product } 3 \times - 1 = - 3$

$\text{which sum to - 2 are - 3 and + 1}$

$\text{split the middle term using these factors}$

$3 {x}^{4} - 3 {x}^{2} + {x}^{2} - 1 \leftarrow \textcolor{b l u e}{\text{factor by grouping}}$

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

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

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

${x}^{2} - 1 \text{ is a "color(blue)"difference of squares}$

$\text{and factors in general as}$

•color(white)(x)a^2-b^2=(a-b)(a+b)

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

$\Rightarrow 3 {x}^{4} - 2 {x}^{2} - 1 = \left(3 {x}^{2} + 1\right) \left(x - 1\right) \left(x + 1\right)$

May 4, 2018

$\left(x - 1\right) \left(x + 1\right) \left(3 {x}^{2} + 1\right) = 0$

#### Explanation:

$3 {x}^{4} - 2 {x}^{2} - 1 = 0$ or

$3 {x}^{4} - 3 {x}^{3} + 3 {x}^{3} - 3 {x}^{2} + {x}^{2} - 1 = 0$ or

$3 {x}^{3} \left(x - 1\right) + 3 {x}^{2} \left(x - 1\right) + \left(x + 1\right) \left(x - 1\right) = 0$ or

$\left(x - 1\right) \left(3 {x}^{3} + 3 {x}^{2} + x + 1\right) = 0$ or

$\left(x - 1\right) \left\{3 {x}^{2} \left(x + 1\right) + 1 \left(x + 1\right)\right\} = 0$ or

$\left(x - 1\right) \left(x + 1\right) \left(3 {x}^{2} + 1\right) = 0$ [Ans]