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

1 Answer
George C.
May 1, 2016

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

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

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

Explanation:

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)#

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