Question

How do you factor `x^4 - x^2 - 20`?

Answer 1

`(x-2)(x+2)(x-i sqrt(5))(x+i sqrt (5))`

Explanation:

create new variable equal to `x^2`. Call it y.

Your equation now is `y^2 - y - 20` whose roots are `y=4` and `y=-5`
`y=4` is ok. That is `x=pm 2`. `y=-5` leads to imaginary roots `x= pm i sqrt(5)`

Answer 2

`x^4-x^2-20=(x^2+4)(x^2-5)`
= `(x+sqrt5)(x-sqrt5)(x^2+4)`
= `(x+sqrt5)(x-sqrt5)(x+2i)(x-2i)`

Explanation:

`x^4-x^2-20`

= `x^4-5x^2+4x^2-20`

= `x^2(x^2-5)+4(x^2-5)`

= `(x^2+4)(x^2-5)`

These are factors with rational coefficients. If we are allowed to have monomials with irrational coefficients too, `x^2-5=(x+sqrt5)(x-sqrt5)` and

`x^4-x^2-20=(x+sqrt5)(x-sqrt5)(x^2+4)`

If we also allow monomials with complex numbers, then factors of `x^2+4` are `(x+2i)(x-2i)` and

`x^4-x^2-20=(x+sqrt5)(x-sqrt5)(x+2i)(x-2i)`

Answer 3

`Exp.=(x^2-5)(x^2+4).`

Explanation:

Expression `= x^4-x^2-20.`

Note that `5xx4=20, 5-4=1`

`:. Exp.=x^4-5x^2+4x^2-20=x^2(x^2-5)+4(x^2-5)=(x^2-5)(x^2+4).`

If, at the first glance, it may not strike that the given Exp. is a qudr. poly, of `x^2`, then, we may use substitution `x^2=t` in it and see that it is `t^2-t-20 = (t-5)(t+4)=(x^2-5)(x^2+4),` as before!.