Question

How do you factor `x^4-1`?

Answer 1

`x^4-1=(x^2+1)(x+1)(x-1)`

using complex numbers

`x^4-1=(x+ib)(x-ib)(x+1)(x-1)`

Explanation:

we make use of the difference of squares

`a^2-b^2=(a+b)(a-b)`

`x^4-1=(x^2+1)(x^2-1)`

we can use dos for the second bracket once more

`x^4-1=(x^2+1)(x+1)(x-1)--(1)`

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

note`" "i^2=-1`

we see

`a^2+b^2=a^2-(ib)^2=(a+ib)(a-ib)`

`(1)rarrx^4-1=(x+ib)(x-ib)(x+1)(x-1)`

Answer 2

`(x-1)(x+1)(x+i)(x-i)`

Explanation:

`x^4-1" is a "color(blue)"difference of squares"`

`"which factors in general as"`

`•color(white)(x)a^2-b^2=(a-b)(a+b)`

`"here "a=x^2" and "b=1`

`rArrx^4-1=(x^2-1)(x^2+1)`

`x^2-1" is a "color(blue)"difference of squares"`

`rArrx^4-1=(x-1)(x+1)(x^2+1)`

`"we can factor "x^2+1" by equating to zero and solving"`

`x^2+1=0rArrx^2=-1rArrx=+-sqrt(-1)=+-i`

`"factors are "(x-(+i))(x-(-i))`

`rArrx^4-1=(x-1)(x+1)(x-i)(x+i)`