Question

How do you factor `x^4-x^2-12`?

Answer 1

`x^4-x^2-12=(x-2)(x+2)(x^2+3)`

Explanation:

Use the difference of squares identity:

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

with `a=x` and `b=2` below.

Treat the quartic as a quadratic in `x^2`, noting that `4*3 = 12` and `4-3 = 1`, so:

`x^4-x^2-12`

`=(x^2)^2-(x^2)-12`

`=(x^2-4)(x^2+3)`

`=(x^2-2^2)(x^2+3)`

`=(x-2)(x+2)(x^2+3)`

which has no simpler factors with Real coefficients, since `x^2+3 >= 3 > 0` for any Real value of `x`. With Complex coefficients we can factor one step more...

`=(x-2)(x+2)(x-3i)(x+3i)`