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

1 Answer
Mar 10, 2016

Answer:

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 ) #