How do you factor #y= 3x^5 -6x^2 -27x-18# ?
1 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
#y=3x^5-6x^2-27x-18=3(x^5-2x^2-9x-6)#
Next let
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
#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^4-x^3+x^2-3x-6 = (x+1)(x^3-2x^2+3x-6)#
Let
By the rational root theorem, any rational zeros of
That means that its only possible rational zeros are:
#+-1# ,#+-2# ,#+-3# ,#+-6#
By examining the sum of the coefficients we can rule out
#g(2) = 8-8+6-6 = 0#
So
#x^3-2x^2+3x-6 = (x-2)(x^2+3)#
The remaining quadratic factor has no linear factors with Real coefficients since
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)#