# How do you factor #15x^3-18x^6-6x+9x^4#?

##### 1 Answer

#### Explanation:

Given:

#15x^3-18x^6-6x+9x^4#

Let's arrange into standard form - putting the terms in descending order of degree to get:

#-18x^6+9x^4+15x^3-6x#

Note that all of the terms are divisible by

#-18x^6+9x^4+15x^3-6x=-3x(6x^5-3x^3-5x^2+2)#

Focusing on the remaining quintic factor, note that the sum of the coefficients is

#6-3-5+2=0#

We can deduce that

#6x^5-3x^3-5x^2+2 = (x-1)(6x^4+6x^3+3x^2-2x-2)#

Let:

#f(x) = 6x^4+6x^3+3x^2-2x-2#

By the rational zeros theorem, any rational zeros of

That means that the only possible rational zeros are:

#+-1/6, +-1/3, +-1/2, +-2/3, +-1, +-2#

Further note that the pattern of the signs of the coefficients of

Note also that the pattern of the signs of the coefficients of

To cut a long story a little shorter, we find that none of the "possible" rational zeros are zeros of

*possible* to find algabraic expressions for these zeros and hence linear factors, but they are very messy.

So it's probably best to stick with the rational factorisation we have found:

#15x^3-18x^6-6x+9x^4 = -3x(x-1)(6x^4+6x^3+3x^2-2x-2)#