How do you factor #13a^3 - 16a - 72#?

1 Answer
Apr 1, 2016

Answer:

#13a^3-16a-72#

#= (a-2)(13a^2+26a+36)#

#= (a-2)(sqrt(13)a+sqrt(13)-sqrt(23)i)(sqrt(13)a+sqrt(13)+sqrt(23)i)#

Explanation:

By the rational root theorem, any rational zeros of this quadratic will be expressible in the form #a=p/q# for integers #p, q# with #p# a divisor of the constant term #-72# and #q# a divisor of the coefficient #13# of the leading term.

So the possible rational zeros are:

#+-1/13, +-2/13, +-3/13, +-4/13, +-6/13, +-8/13, +-9/13, +-12/13, +-1, +-18/13, +-24/13, +-2, +-36/13, +-3, +-6, +-8, +-9, +-12, +-18, +-24, +-36, +-72#

Let #f(a) = 13a^3-16a-72#

We can rule out any of the non-integral fractions, since for such fractions, #13a^3# will have a denominator #13^2# in lowest terms, which will not be possible to cancel out by the other terms.

So try the integer possibilities in sequence.

We find #f(2) = 13*8-16*2-72 = 104-32-72 = 0#

So #a=2# is a zero and #(a-2)# a factor:

#13a^3-16a-72 = (a-2)(13a^2+26a+36)#

The remaining quadratic factor has negative discriminant:

#Delta = 26^2-(4*13*36) = 676-1872 = -1196#

so no linear factors with Real coefficients.

We can factor using Complex coefficients as follows:

#(a-2)(13a^2+26a+36)#

#= (a-2)(13(a^2+2a+1)+23)#

#= (a-2)(13(a+1)^2+23)#

#= (a-2)((sqrt(13)(a+1))^2-(sqrt(23)i)^2)#

#= (a-2)(sqrt(13)(a+1)-sqrt(23)i)(sqrt(13)(a+1)+sqrt(23)i)#

#= (a-2)(sqrt(13)a+sqrt(13)-sqrt(23)i)(sqrt(13)a+sqrt(13)+sqrt(23)i)#