Question

How do you factor `2x^4-9x^3+19x^2-15x`?

Answer 1

`2x^4-9x^3+19x^2-15x`

`=x(2x-3)(x^2-3x+5)`

`=x(2x-3)(x-3/2-sqrt(11)/2i)(x-3/2+sqrt(11)/2i)`

Explanation:

`f(x) = 2x^4-9x^3+19x^2-15x = x(2x^3-9x^2+19x-15)`

Let `g(x) = 2x^3-9x^2+19x-15`

By the rational roots theorem, any rational zeros of `g(x)` (and therefore of `f(x)`) are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-15` and `q` a divisor of the coefficient `2` of the leading term.

That means that the only possible rational zeros are:

`+-1/2, +-1, +-3/2, +-5/2, +-3, +-5, +-15`

In addition note using Descartes' rule of signs, we find that this cubic has `1` or `3` positive Real zeros and no negative Real zeros.

So the only possible rational zeros are:

`1/2, 1, 3/2, 5/2, 3, 5, 15`

Trying each in turn, we find:

`g(3/2) = 2(3/2)^3-9(3/2)^2+19(3/2)-15`

`=27/4-81/4+114/4-60/4 = 0`

So `x=3/2` is a zero and `(2x-3)` a factor:

`2x^3-9x^2+19x-15 = (2x-3)(x^2-3x+5)`

The zeros of the remaining quadratic can be found using the quadratic formula:

`x = (3+-sqrt((-3)^2-4(1)(5)))/(2*1)`

`=(3+-sqrt(9-20))/2`

`=3/2+-sqrt(11)/2i`

Hence factors:

`(x-3/2-sqrt(11)/2i)(x-3/2+sqrt(11)/2i)`