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

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Answer 1 · 2015-11-19T23:36:51

Factor by grouping to find:

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

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

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

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

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

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

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

$2x^2+3$ 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:

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

How do you factor completely 6x^3 - 10x^2 + 9x - 156x^3 - 10x^2 + 9x - 15?