Simplify the product
( 2y-5)(3y+1)
a6y^2+13y-5
b6y^2-13y-5
c6y^2+13y+5
d6y^2-13y+5

It's B. If you FOIL you get 6y^2+2y-15y-5

   To simplify (2y-5)(3y+1), you may use FOIL to multiply the binomials or the distributive property. I'll show you how to simplify this expression using the distributive property. Since FOIL and multiplying any polynomials by each other is essentially multiplying each term in one of the parentheses to the every other term in the other parentheses, you can do this just by using the distributive property by distributing each term to another. For instance, you could distribute each term in the first parentheses by each term in the 2nd parentheses by writing 2y(3y+1)+(-5)(3y+1). Note how you can do 3y(2y-5)+1(2y-5) as well, and note how this whole method works in the first place because if you want to check this by going back to the original expression, you can combine like terms. In 2y(3y+1)+(-5)(3y+1), (3y+1) is the like term because it is present as it is being multiplied to two different numbers that will eventually be added. This is analogous to 2x-5x, where x=(3y+1). Now, you can just distribute each term into the parentheses, whereas I'll write out this step as 2y(3y)+2y(1)+(-5)(3y)+(-5)(1). Now, just complete the multiplication and then combine like terms for the "y" terms. You will end up with a quadratic equation. :D

I'll say b) log(- that if not let me know what i can do