How do you factor completely #6x^3 - 10x^2 + 9x - 15#?

1 Answer
George C.
Nov 19, 2015

Factor by grouping to find:

#6x^3-10x^2+9x-15#

#=(2x^2+3)(3x-5)#

#=(sqrt(2)x-sqrt(3)i)(sqrt(2)x+sqrt(3)i)(3x-5)#

Explanation:

#6x^3-10x^2+9x-15#

#=(6x^3-10x^2)+(9x-15)#

#=(2x^2(3x-5)+3(3x-5)#

#=(2x^2+3)(3x-5)#

#2x^2+3# has no simpler factors with Real coefficients, since #x^2>=0# for all #x in RR#, hence #2x^2+3 >= 3 > 0# for all #x in RR#.

If we allow Complex coefficients then #2x^2+3# can be treated as a difference of squares(!), hence:

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

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