# How do you factor 15y^3-19y^2-30y+7 ?

May 20, 2017

#### Answer:

$15 {y}^{3} - 19 {y}^{2} - 30 y + 7$ has no factorisation with rational coefficients.

It is "possible" to find a factorisation of the form:

$15 {y}^{3} - 19 {y}^{2} - 30 y + 7 = 15 \left(y - {y}_{1}\right) \left(y - {y}_{2}\right) \left(y - {y}_{3}\right)$

#### Explanation:

Given (rearranged into standard order):

$f \left(y\right) = 15 {y}^{3} - 19 {y}^{2} - 30 y + 7$

$\textcolor{w h i t e}{}$
Rational Roots Theorem

By the rational roots theorem, any rational zeros of $f \left(y\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $7$ and $q$ a divisor of the coefficient $15$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{15} , \pm \frac{1}{5} , \pm \frac{1}{3} , \pm 1 , \pm \frac{19}{15} , \pm \frac{19}{5} , \pm \frac{19}{3} , \pm 19$

We find:

$f \left(- \frac{19}{15}\right) = - \frac{3593}{225}$

$f \left(- 1\right) = 3$

$f \left(\frac{1}{5}\right) = \frac{9}{25}$

$f \left(\frac{1}{3}\right) = - \frac{41}{9}$

$f \left(\frac{19}{15}\right) = - 31$

$f \left(\frac{19}{5}\right) = \frac{7951}{18}$

$f \left(y\right)$ is not zero at any of the possible rational values, but changes sign between them as we have found.

So $f \left(y\right)$ has no rational roots, but it has $3$ irrational roots, in:

$\left(- \frac{19}{15} , - 1\right)$, $\left(\frac{1}{5} , \frac{1}{3}\right)$ and $\left(\frac{19}{15} , \frac{19}{5}\right)$

$\textcolor{w h i t e}{}$
Where do we go from here?

If we can find the zeros ${y}_{1}$, ${y}_{2}$ and ${y}_{3}$ then the given polynomial factors as:

$15 {y}^{3} - 19 {y}^{2} - 30 y + 7 = 15 \left(y - {y}_{1}\right) \left(y - {y}_{2}\right) \left(y - {y}_{3}\right)$

Since all three zeros are real but irrational, it falls into Cardano's "casus irreducibilis". His method will involve taking cube roots of non-real complex numbers.

A better option for an "algebraic" solution is a probably a trigonometric one, using the fact that:

$\cos 3 \theta = 4 {\cos}^{3} \theta - 3 \cos \theta$

and the substitution:

$y = \frac{19}{45} + k \cos \theta$

for some $k$ chosen to cause the terms in ${\cos}^{3} \theta$ and $\cos \theta$ to match the $\cos 3 \theta$ formula.

This probably gets messier than you would like.