Question

How do you factor `3u^4 - 19u^2 + 28`?

Answer 1

`3u^4-19u^2+28`

`= 3(u^2)^2-19(u^2)+28`

`= (3u^2-7)(u^2-4)`

`=(3u^2-7)(u-2)(u+2)`

If we allow irrational factors:

`=(sqrt(3)u-sqrt(7))(sqrt(3)u+sqrt(7))(u-2)(u+2)`

To find the factors for the factorization into `(3u^2-7)(u^2-4)`,
I basically tried a few combinations of the form `(3u^2-a)(u^2-b)` with `ab=28`.

Having recognised that the quartic is a quadratic in `u^2` I could have used any of the normal methods (quadratic formula, AC Method, etc.) for factoring quadratics, but it was simple enough to try a couple of guesses.

The factorizations into linear factors are a couple of examples of the general difference of squares identity:

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