Question

How do you factor `x^4 - 13x^2 - 30`?

Answer 1

`x^4-13x^2-30 = (x^2-15)(x^2+2) = (x-sqrt(15))(x+sqrt(15))(x^2+2)`

Explanation:

`x^4-13x^2-30` is quadratic in `x^2`.

To factor it, first find a pair of factors of `30` whose difference is `13`. Such a pair is `15` and `2`.

That gives us:

`x^4-13x^2-30 = (x^2-15)(x^2+2)`

The first of these quadratic factors can be considered a difference of squares:

`x^2-15 = x^2 - (sqrt(15))^2 = (x-sqrt(15))(x+sqrt(15))`

using the difference of squares identity:

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

The second factor `x^2+2` has no linear factors with real coefficients. If you were allowed complex coefficients then you could write:

`x^2+2 = (x-sqrt(2)i)(x+sqrt(2)i)`

Answer 2

`(x+sqrt(15))(x-sqrt(15))(x^2+2)` in `RR`

`(x+sqrt(15))(x-sqrt(15))(x+sqrt(2)i)(x-sqrt(2)i)` in `CC`

(If you have never used `i` or `CC`, just ignore this last answer)

Explanation:

I'll call `P(x)` the polynomial.

We notice that `P(x)=Q(x^2)` for some `Q` quadratic polynomial. (In fact, P(x) has only even powers of `x`)

let `X=x^2`

So the equation becomes:

`X^2 -13X-30`, which I know how to factor:

`a,b=(13+-sqrt(169+120))/2=(13+-17)/2 => a=15,b=-2`

(if you had eyes wide open, you could notice that `-13=-(15-2)` and `-30=(-2)15`, so you could avoid tedious calculations)

i.e. `Q(X)=(X-15)(X+2)`

So we factored the polynomial in

`P(x)=(x^2-15)(x^2+2)`

We notice that `(x^2-15)=(x+sqrt(15))(x-sqrt(15))`

If you work in `RR`, then `(x^2+2)` is irreducible (it has no roots)

So `P(x)=(x+sqrt(15))(x-sqrt(15))(x^2+2)`

If you work in `CC`, `x^2 +2 =(x+sqrt(2)i)(x-sqrt(2)i)`

So `P(x)=(x+sqrt(15))(x-sqrt(15))(x+sqrt(2)i)(x-sqrt(2)i)`

(If you have never used `i` or `CC`, just ignore this last two sentences)