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

Oct 9, 2017

$3 {x}^{3} \left(24 {x}^{3} + 15 {x}^{2} + 3\right)$

#### Explanation:

$3 {x}^{3} \left(24 {x}^{3} + 5 {x}^{2} + 3\right)$

Nov 5, 2017

$72 {x}^{9} + 15 {x}^{6} + 9 {x}^{3} = 3 {x}^{3} \left(24 {x}^{6} + 5 {x}^{3} + 3\right)$

with no simpler factors with rational coefficients.

#### Explanation:

Given:

$72 {x}^{9} + 15 {x}^{6} + 9 {x}^{3}$

First note that all of the terms are divisible by $3 {x}^{3}$, 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:

$24 {x}^{6} + 5 {x}^{3} + 3$

is in the form:

$a {t}^{2} + b t + c$

with $t = {x}^{3}$, $a = 24$, $b = 5$ and $c = 3$

This quadratic has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {\textcolor{b l u e}{5}}^{2} - 4 \left(\textcolor{b l u e}{24}\right) \left(\textcolor{b l u e}{3}\right) = 25 - 288 = - 253$

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.