How do you factor #y= 3x^5 -6x^2 -27x-18# ?

1 Answer
Mar 11, 2016

Answer:

#3x^5-6x^2-27x-18#

#= 3(x+1)(x+1)(x-2)(x^2+3)#

#= 3(x+1)(x+1)(x-2)(x-sqrt(3)i)(x+sqrt(3)i)#

Explanation:

First separate out the common scalar factor #3# to find:

#y=3x^5-6x^2-27x-18=3(x^5-2x^2-9x-6)#

Next let #f(x) = x^5-2x^2-9x-6#

Notice that if the signs of the coefficients are reversed on the terms of odd degree then their sum is zero. That is:

#-1-2+9-6 = 0#

Hence #f(-1) = 0#, so #x = -1# is a zero and #(x+1)# a factor:

#x^5-2x^2-9x-6 = (x+1)(x^4-x^3+x^2-3x-6)#

Notice that if we reverse the signs of the coefficients of the terms of odd degree in the remaining quartic factor then the sum is zero. That is:

#1+1+1+3-6 = 0#

Hence we have another factor #(x+1)#:

#x^4-x^3+x^2-3x-6 = (x+1)(x^3-2x^2+3x-6)#

Let #g(x) = x^3-2x^2+3x-6#

By the rational root theorem, any rational zeros of #g(x)# must be expressible in the form #p/q# for integers #p#, #q# with #p# a divisor of the constant term #-6# and #q# a divisor of the coefficient #1# of the leading term.

That means that its only possible rational zeros are:

#+-1#, #+-2#, #+-3#, #+-6#

By examining the sum of the coefficients we can rule out #+-1#, but we find:

#g(2) = 8-8+6-6 = 0#

So #(x-2)# is a factor:

#x^3-2x^2+3x-6 = (x-2)(x^2+3)#

The remaining quadratic factor has no linear factors with Real coefficients since #x^2+3 >= 3 > 0# for all #x in RR#. If we allow Complex coefficients then it factors as #(x-sqrt(3)i)(x+sqrt(3)i)#

Putting it all together:

#3x^5-6x^2-27x-18#

#= 3(x+1)(x+1)(x-2)(x^2+3)#

#= 3(x+1)(x+1)(x-2)(x-sqrt(3)i)(x+sqrt(3)i)#