# How do you factor x^4-1?

Apr 18, 2018

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

using complex numbers

${x}^{4} - 1 = \left(x + i b\right) \left(x - i b\right) \left(x + 1\right) \left(x - 1\right)$

#### Explanation:

we make use of the difference of squares

${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$

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

we can use dos for the second bracket once more

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

for real numbers we can proceed no further, but if we use complex numbers

note$\text{ } {i}^{2} = - 1$

we see

${a}^{2} + {b}^{2} = {a}^{2} - {\left(i b\right)}^{2} = \left(a + i b\right) \left(a - i b\right)$

$\left(1\right) \rightarrow {x}^{4} - 1 = \left(x + i b\right) \left(x - i b\right) \left(x + 1\right) \left(x - 1\right)$

Apr 18, 2018

$\left(x - 1\right) \left(x + 1\right) \left(x + i\right) \left(x - i\right)$

#### Explanation:

${x}^{4} - 1 \text{ is a "color(blue)"difference of squares}$

$\text{which factors in general as}$

•color(white)(x)a^2-b^2=(a-b)(a+b)

$\text{here "a=x^2" and } b = 1$

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

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

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

$\text{we can factor "x^2+1" by equating to zero and solving}$

${x}^{2} + 1 = 0 \Rightarrow {x}^{2} = - 1 \Rightarrow x = \pm \sqrt{- 1} = \pm i$

$\text{factors are } \left(x - \left(+ i\right)\right) \left(x - \left(- i\right)\right)$

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