How do you factor completely #2y^3 + 6y^2 - 5y - 15#?

2 Answers
Aug 1, 2018

Answer:

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

Explanation:

There are #4# terms. Make two groups which each have a common factor.

#(2y^3+6y^2) + (-5y -15)#

#=2y^2(y+3) -5(y+3)" "larr# there is a common bracket

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

Aug 1, 2018

Answer:

#(y+3)(sqrt2y-sqrt5)(sqrt2y+sqrt5)#

Explanation:

#"factor first/second and third/fourth terms by "color(blue)"grouping"#

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

#"take out the "color(blue)"common factor "(y+3)#

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

#"factor "(2y^2-5)" as a "color(blue)"difference of squares"#

#•color(white)(x)a^2-b^2=(a-b)(a+b)#

#"with "a=sqrt2y" and "b=sqrt5#

#=(y+3)(sqrt2y-sqrt5)(sqrt2y+sqrt5)#