How do you factor #3y^2+7y-6#?

1 Answer
Aug 6, 2015

Answer:

#3y^2+7y-6 = (3y-2)(y+3)#

Explanation:

You could use the quadratic formula for this
but (hoping for integer coefficient factors):

Since #(ay+b)(cy+d) = (ac)y^2 + (ad+bc)y +bd#

and for the given example
#color(white)("XXXX")#(ac) = 3# and #(bd) = (-6)#

we want to find #(a,c)#, factors of 3
#color(white)("XXXX")#only obvious factors are [(1,3), (3,1)]
and factors, #(b,d)#,of (-6)
#color(white)("XXXX")#[(1,-6), (-1,6), (2,-3), (-2,3), (6, -1), (-6,1), (3,-2), (-3,2)]
which satisfy
#color(white)("XXXX")#(ad+bc) = 7#

with a small amount of effort we find
#color(white)("XXXX")##(a,c) = (3,1)# and #(b,d) = (-2,3)#