Question

How do you factor completely `24y^3+56y^2-6y-14`?

Answer 1

2(3y+7)(2y-1)(2y+1)

Explanation:

Begin by 'grouping' the expression

`[ 24y^3 + 56y^2 ] + [-6y - 14 ]`

now factor each group

`rArr 8y^2(3y+7 ) -2(3y + 7)`

there is a common factor of (3y + 7 )

`rArr (3y + 7 )(8y^2 - 2)`

common factor of 2 in `(8y^2 - 2 ) = 2(4y^2 - 1 )`

`4y^2 - 1 " is a difference of squares "`

and `4y^2 -1 = (2y - 1 )(2y + 1 )`

Finally it all comes together as `2(3y + 7 )(2y - 1 )(2y + 1 )`