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

##### 2 Answers

#### Answer:

#### Explanation:

#### Answer:

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

#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

This quadratic has discriminant

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

Since

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.