Question

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

Answer 1

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

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`

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Rational Roots Theorem

By the rational roots theorem, any rational zeros of `f(y)` are expressible in the form `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:

`+-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`

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

So `f(y)` has no rational roots, but it has `3` irrational roots, in:

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

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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:

`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 `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.