How do you factor the expression #4x^3 + 6x^2 + 6x + 9#?

Question

1520 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-24T00:36:34

Factor by grouping to find:

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

Factor by grouping:

#4x^3+6x^2+6x+9#

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

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

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

The remaining quadratic factor #(2x^2+3)# has no simpler linear factors with Real coefficients since it is non-zero for any Real number #x#:

#2x^2+3 >= 3 > 0#

If we resort to Complex numbers then it factorises further:

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