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

1 Answer
May 20, 2017

Answer:

#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.