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

Aug 14, 2018

$15 {x}^{3} - 18 {x}^{6} - 6 x + 9 {x}^{4} = - 3 x \left(x - 1\right) \left(6 {x}^{4} + 6 {x}^{3} + 3 {x}^{2} - 2 x - 2\right)$

#### Explanation:

Given:

$15 {x}^{3} - 18 {x}^{6} - 6 x + 9 {x}^{4}$

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

$- 18 {x}^{6} + 9 {x}^{4} + 15 {x}^{3} - 6 x$

Note that all of the terms are divisible by $3$ and by $x$, so we can separate out a common factor $3 x$. Let us instead separate out a common factor $- 3 x$ to make the coefficient of the leading term of the remaining factor posistive:

$- 18 {x}^{6} + 9 {x}^{4} + 15 {x}^{3} - 6 x = - 3 x \left(6 {x}^{5} - 3 {x}^{3} - 5 {x}^{2} + 2\right)$

Focusing on the remaining quintic factor, note that the sum of the coefficients is $0$. That is:

$6 - 3 - 5 + 2 = 0$

We can deduce that $x = 1$ is a zero and $\left(x - 1\right)$ a factor:

$6 {x}^{5} - 3 {x}^{3} - 5 {x}^{2} + 2 = \left(x - 1\right) \left(6 {x}^{4} + 6 {x}^{3} + 3 {x}^{2} - 2 x - 2\right)$

Let:

$f \left(x\right) = 6 {x}^{4} + 6 {x}^{3} + 3 {x}^{2} - 2 x - 2$

By the rational zeros theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 2$ and $q$ a divisor of the coefficient $6$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{6} , \pm \frac{1}{3} , \pm \frac{1}{2} , \pm \frac{2}{3} , \pm 1 , \pm 2$

Further note that the pattern of the signs of the coefficients of $f \left(x\right)$ is $+ + + - -$. With one change of signs, Descartes' Rule of Signs tells us that $f \left(x\right)$ has exactly one positive real zero.

Note also that the pattern of the signs of the coefficients of $f \left(- x\right)$ is $+ - + + -$. With three changes of sign, Descartes' Rule of Signs tells us that $f \left(x\right)$ has $3$ or $1$ negative real zeros.

To cut a long story a little shorter, we find that none of the "possible" rational zeros are zeros of $f \left(x\right)$ so there are no more factors with rational or indeed integer coefficients.

$f \left(x\right)$ has two irrational zeros, which are approximately $- 0.65188$ and $0.60347$. It also has a complex conjugate pair of non-real zeros. It is 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:

$15 {x}^{3} - 18 {x}^{6} - 6 x + 9 {x}^{4} = - 3 x \left(x - 1\right) \left(6 {x}^{4} + 6 {x}^{3} + 3 {x}^{2} - 2 x - 2\right)$