Question

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

Answer 1

`=(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)`

Answer 2

`(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)`