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Mathematics 14 Online
OpenStudy (anonymous):

Short Answer. Find the following product:(y^2-5y-2)(y+8)

OpenStudy (anonymous):

try and use FOIL

Directrix (directrix):

(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 ..

Directrix (directrix):

@mth3v4 --> FOIL is for the product of binomials which is not the case here.

OpenStudy (anonymous):

factor the trinomial before you do anything

Directrix (directrix):

y(y^2) + y (-5y) + y (-2) =y^3 - 5y^2 -2y. That will be added to the next set of multiplications so..

Directrix (directrix):

Multiply 8 times each term of the trinomial (y^2 -5y -2) to get ...

OpenStudy (anonymous):

how would the steps looking

Directrix (directrix):

8(y^2) + 8 ( -5y) + 8(-2) = 8y^2 -40y -16.

OpenStudy (anonymous):

ok would the other things you wrote be in the steps

Directrix (directrix):

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?

OpenStudy (anonymous):

this is my first time doing it in class so i don't really know of anyother way

Directrix (directrix):

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).

OpenStudy (anonymous):

ok

Directrix (directrix):

I can do it another way if this method doesn't make sense.

OpenStudy (anonymous):

no that method helped me

Directrix (directrix):

Okay. Thanks.

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