Question

How do you factor `72x^9+15x^6+9x^3`?

Answer 1

`3x^3(24x^3 + 15x^2 + 3)`

Explanation:

`3x^3(24x^3 + 5x^2 + 3)`

Answer 2

`72x^9+15x^6+9x^3 = 3x^3(24x^6+5x^3+3)`

with no simpler factors with rational coefficients.

Explanation:

Given:

`72x^9+15x^6+9x^3`

First note that all of the terms are divisible by `3x^3`, so we can separate that out as a factor to find:

#72x^9+15x^6+9x^3 = 3x^3(24x^6+5x^3+3)#

Note that:

`24x^6+5x^3+3`

is in the form:

`at^2+bt+c`

with `t = x^3`, `a=24`, `b=5` and `c=3`

This quadratic has discriminant `Delta` given by the formula:

`Delta = b^2-4ac = color(blue)(5)^2-4(color(blue)(24))(color(blue)(3)) = 25-288 = -253`

Since `Delta < 0` this quadratic cannot be factored as a product of linear terms with real coefficients and the sextic in `x` has no real roots either, so no linear factors with real coefficients.

Note however that it does have a factorisation as a product of three quadratic factors with real coefficients. However, the coefficients of those factors are somewhat messy to express: They are expressible in terms of trigonometric functions or in terms of irreducible cube roots of Complex numbers.

So let's stop with our rational factorisation.