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

##### 1 Answer

#### Answer:

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

#15y^3-19y^2-30y+7 = 15(y-y_1)(y-y_2)(y-y_3)#

#### Explanation:

Given (rearranged into standard order):

#f(y) = 15y^3-19y^2-30y+7#

**Rational Roots Theorem**

By the rational roots theorem, any rational zeros of

That means that the only possible *rational* zeros are:

#+-1/15, +-1/5, +-1/3, +-1, +-19/15, +-19/5, +-19/3, +-19#

We find:

#f(-19/15) = -3593/225#

#f(-1) = 3#

#f(1/5) = 9/25#

#f(1/3) = -41/9#

#f(19/15) = -31#

#f(19/5) = 7951/18#

So

#(-19/15, -1)# ,#(1/5, 1/3)# and#(19/15, 19/5)#

**Where do we go from here?**

If we can find the zeros

#15y^3-19y^2-30y+7 = 15(y-y_1)(y-y_2)(y-y_3)#

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 3theta = 4cos^3theta - 3cos theta#

and the substitution:

#y = 19/45 + k cos theta#

for some

This probably gets messier than you would like.