How do you list all possible roots and find all factors of #x^3+27#?

1 Answer
Dec 10, 2016

Answer:

#x^3+27 = (x+3)(x-3/2-(3sqrt(3))/2 i)(x-3/2+(3sqrt(3))/2 i)#

with corresponding zeros:

#x = -3# and #x = 3/2+-(3sqrt(3))/2 i#

Explanation:

The sum of cubes identity can be written:

#a^3+b^3 = (a+b)(a^2-ab+b^2)#

We can use this with #a=x# and #b=3# as follows:

#x^3+27 = x^3+3^3#

#color(white)(x^3+27) = (x+3)(x^2-3x+9)#

So one linear factor is #(x+3)# with corresponding zero #x=-3#.

We can factor the remaining quadratic using Complex coeffients by completing the square and using the difference of squares identity, which can be written:

#a^2-b^2 = (a-b)(a+b)#

with #a=(x-3/2)# and #b = (3sqrt(3))/2 i# as follows:

#x^2-3x+9 = (x^2-3x+9/4)+27/4#

#color(white)(x^2-3x+9) = (x-3/2)^2-((3sqrt(3))/2 i)^2#

#color(white)(x^2-3x+9) = ((x-3/2)-(3sqrt(3))/2 i)((x-3/2)+(3sqrt(3))/2 i)#

#color(white)(x^2-3x+9) = (x-3/2-(3sqrt(3))/2 i)(x-3/2+(3sqrt(3))/2 i)#

Hence the remaining zeros are:

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

and the full factorisation can be written:

#x^3+27 = (x+3)(x-3/2-(3sqrt(3))/2 i)(x-3/2+(3sqrt(3))/2 i)#

Structly speaking, we have not listed all of the factors of #x^3+27#.

The ones we have identified so far are:

#x^3+27#

#x+3#

#x^2-3x+9#

#x-3/2-(3sqrt(3))/2i#

#x-3/2+(3sqrt(3))/2i#

#1#

In addition the results of multiplying these linear factors give us the other factors:

#(x+3)(x-3/2-(3sqrt(3))/2i) = x^2+(3/2-(3sqrt(3))/2i)x-(9/2+(9sqrt(3))/2i)#

#(x+3)(x-3/2+(3sqrt(3))/2i) = x^2+(3/2+(3sqrt(3))/2i)x-(9/2-(9sqrt(3))/2i)#

I don't think it was intended that you should list all of these factors, but there they are in case.