How do you factor completely: #4x^3 + 12x^2 + 3x + 9#?

1 Answer
George C.
Jul 11, 2015

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

This has no simpler linear factors with real coefficients, since #4x^2+3 >= 3 > 0# for all #x in RR#.

Explanation:

#4x^3+12x^2+3x+9#

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

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

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

The remaining quadratic factor #(4x^2+3)# cannot be factored into linear factors with real coefficients, since #4x^2+3 >= 3 > 0# for all #x in RR#.

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