how to divide the following binomial

Question

(2y^2-15+28)/ (y-4)

Answer 1

Assuming you mean (2y^2 - 15y + 28) / (y - 4)

It only divides if y - 4 is a factor of 2y^2 - 15y + 28

(y - 4)(?y + ??) = 2y^2 - 15y + 28

The ? has to be 2 if we want y * 2y = 2y^2

(y - 4)(2y + ??) = 2y^2 - 15y + 28

The ?? has to be -7 if we want -4 * -7 = 28

(y - 4)(2y - 7) = 2y^2 - 15y + 28

But will it make the right middle term?  FOIL to find out.

2y^2 - 7y - 8y + 28

2y^2 - 15y + 28

Yes, it works.

Now the original problem looks like:

(y - 4)(2y - 7) / (y - 4)

The y - 4's cancel, leaving us with:

2y - 7