How do you factor completely $15y^3+12y^2+5y+4$ ?

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Answer 1 · 2016-05-01T10:54:03

$15y^3+12y^2+5y+4)$

$=(3y^2+1)(5y+4)$

$=(sqrt(3)y-i)(sqrt(3)y+i)(5y+4)$

Factor by grouping:

$15y^3+12y^2+5y+4)$

$=(15y^3+12y^2)+(5y+4)$

$=3y^2(5y+4)+1(5y+4)$

$=(3y^2+1)(5y+4)$

If $y in RR$ then $y^2 >= 0$ , hence $3y^2+1 > 0$ . So this quadratic factor has no linear factors with Real coefficients.

If we allow Complex coefficients we can factor as a difference of squares:

$3y^2+1 = (sqrt(3)y)^2-i^2 = (sqrt(3)y-i)(sqrt(3)y+i)$

How do you factor completely 15y^3+12y^2+5y+4 15y^3+12y^2+5y+4 ?