How do you factor 15y3−19y2−30y+7 ?

1 Answer
George C.
May 20, 2017

15y3−19y2−30y+7 has no factorisation with rational coefficients.

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

15y3−19y2−30y+7=15(y−y1)(y−y2)(y−y3)

Explanation:

Given (rearranged into standard order):

f(y)=15y3−19y2−30y+7


Rational Roots Theorem

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

±115,±15,±13,±1,±1915,±195,±193,±19

We find:

f(−1915)=−3593225

f(−1)=3

f(15)=925

f(13)=−419

f(1915)=−31

f(195)=795118

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:

(−1915,−1), (15,13) and (1915,195)


Where do we go from here?

If we can find the zeros y1, y2 and y3 then the given polynomial factors as:

15y3−19y2−30y+7=15(y−y1)(y−y2)(y−y3)

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:

cos3θ=4cos3θ−3cosθ

and the substitution:

y=1945+kcosθ

for some k chosen to cause the terms in cos3θ and cosθ to match the cos3θ formula.

This probably gets messier than you would like.

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