Hi again Bridget;
Now let's do the equation!
y^{3}14y^{2}+48y=
For the FOIL...
FIRST must be...(y^{2})(y)=y^{3}
OUTER and INNER must addup to 14y^{2}
LAST must be (8)(6y) or (8y)(6) or (24)(2y) or (24y)(2) or (16y)(3) or (16)(3), etc. and both numbers must be negative.
(y^{2}6y)(y8)
Let's FOIL...
FIRST...(y^{2})(y)=y^{3}^{
}
OUTER...(y^{2})(8)=8y^{2}
INNER...(6y)(y)=6y^{2}
LAST...(6y)(8)=48y
y^{3}8y^{2}6y^{2}+48y
y^{3}14y^{2}+48y
Nov 25

Vivian L.