Short Answer. Find the following product:(y^2-5y-2)(y+8)
try and use FOIL
(y^2-5y-2)(y+8) = (y+8) (y^2 - 5y -2) Take the y term and multiply it by each term in the trinomial (y^2 - 5y -2) and ..
@mth3v4 --> FOIL is for the product of binomials which is not the case here.
factor the trinomial before you do anything
y(y^2) + y (-5y) + y (-2) =y^3 - 5y^2 -2y. That will be added to the next set of multiplications so..
Multiply 8 times each term of the trinomial (y^2 -5y -2) to get ...
how would the steps looking
8(y^2) + 8 ( -5y) + 8(-2) = 8y^2 -40y -16.
ok would the other things you wrote be in the steps
y^3 - 5y^2 -2y + 8y^2 -40y -16 = y^3 + 3y^2 -42y - 16. (check it, please) If this is not the method you like, then how do you do these in class?
this is my first time doing it in class so i don't really know of anyother way
Questions? What I did was to take each term of the binomial (y + 8) and multiply it by each term of the trinomial (y^2 - 5y -2).
ok
I can do it another way if this method doesn't make sense.
no that method helped me
Okay. Thanks.
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