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

##### 1 Answer

#### 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)#