How do you factor #y^3-8y+2y^2-16#?

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Answer 1 · 2015-06-21T07:48:49

#=color(red)((y+2) (y^2-8) #

#y^3-8y+2y^2-16#

Factorising by grouping:
Forming two groups of two terms:

#color(red)(y^3-8y)+color(blue)(2y^2-16#
#=y (y^2-8) + 2(y^2-8)#
#=color(red)((y+2) (y^2-8) #

Answer 2 · 2015-06-21T08:39:34

#y^3-8y+2y^2-16#

#=(y+2)(y^2-8)#

#=(y+2)(y-2sqrt(2))(y+2sqrt(2))#

First factor by grouping

#f(y) = y^3-8y+2y^2-16#

#=(y^3-8y)+(2y^2-16)#

#=y(y^2-8)+2(y^2-8)#

#=(y+2)(y^2-8)#

The remaining quadratic term can be factorized by treating it as a difference of squares with irrational coefficients:

#f(y) =(y+2)(y^2-(sqrt(8))^2)#

#=(y+2)(y-sqrt(8))(y+sqrt(8))#

...using the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

Finally note that #sqrt(8) = sqrt(2^2*2) = 2sqrt(2)#

#f(y)=(y+2)(y-2sqrt(2))(y+2sqrt(2))#