Question

How do you factor the trinomial `9a^5b + 33ab^5`?

Answer 1

`9a^5b+33ab^5= 3ab(3a^4+11b^4)`

`= 3ab(sqrt(3)a^2+root(4)(132)ab+sqrt(11)b^2)(sqrt(3)a^2-root(4)(132)ab+sqrt(11)b^2)`

Explanation:

This is actually a binomial (having two terms), not a trinomial (three terms).

That aside, first we identify that both of the terms are divisible by `3ab`, so separate that factor out first:

`9a^5b+33ab^5= 3ab(3a^4+11b^4)`

Since the coefficients of the quartic are both positive, this has no linear factors with Real coefficients, but it does have quadratic factors with Real coefficients:

Consider:

`(c^2+2cd+2d^2)(c^2-2cd+2d^2)=c^4+4d^4`

If we let `c = root(4)(3)a` and `d = root(4)(11/4)b` then:

`3a^4+11b^4`

`= c^4+4d^4`

`=(c^2+2cd+2d^2)(c^2-2cd+2d^2)`

`=(sqrt(3)a^2+2root(4)(33/4)ab+2sqrt(11/4)b^2)(sqrt(3)a^2-2root(4)(33/4)ab+2sqrt(11/4)b^2)`

`=(sqrt(3)a^2+root(4)(132)ab+sqrt(11)b^2)(sqrt(3)a^2-root(4)(132)ab+sqrt(11)b^2)`

Putting this all together:

`9a^5b+33ab^5`

`= 3ab(sqrt(3)a^2+root(4)(132)ab+sqrt(11)b^2)(sqrt(3)a^2-root(4)(132)ab+sqrt(11)b^2)`

Footnote

The identity:

`(c^2+2cd+2d^2)(c^2-2cd+2d^2)=c^4+4d^4`

came from thinking about `4`th roots of `-1`

`-1` has square roots `+-i`, which in turn have square roots:

`1/sqrt(2) (+-1+-i)`

Rather than mess about with a `1/sqrt(2)` factor, we can look at `root(4)(-4)` instead.

`-4` has square roots `+-2i`, which in turn have square roots:

`+-1+-i`

So:

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

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

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

Hence:

`c^4+4d^4 = (c^2+2cd+2d^2)(c^2-2cd+2d^2)`