How do you factor completely #24y^3+56y^2-6y-14 #?
1 Answer
Mar 10, 2016
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 ) #
