How do you factor #64x^3+27#?

Question

19571 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-23T22:56:25

Use the identity: # a^3 + b^3 = (a +b)(a^2 - ab + b^2)#

f(x) = (4x + 3)(16x^2 - 12x + 9) =

Answer 2 · 2015-05-23T22:58:29

Using the identity #a^3+b^3 = (a+b)(a^2-ab+b^2)# we find:

#64x^3+27 = (4x)^3 + 3^3#

#= (4x+3)((4x)^2-(4x)*3+3^2)#

#= (4x+3)(16x^2-12x+9)#

#16x^2-12x+9# has no simpler factors with real coefficients.

To check this, evaluate its discriminant:

#Delta(16x^2-12x+9) = (-12)^2-(4xx16xx9)#

#= 144-576 = -432#

Since #Delta < 0# the quadratic equation #16x^2-12x+9 = 0# has no real roots. It has two distinct complex roots.

In case you are curious, the complex factors of #16x^2-12x+9# are:

#(4x+3omega)# and #(4x+3omega^2)#

where #omega = -1/2 + sqrt(3)/2i#