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

2 Answers
Oct 9, 2017

Answer:

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

Explanation:

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

Nov 5, 2017

Answer:

#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.