Solve using FOIL

(y-2) (3y3-5y+7)

Generally FOIL refers only to binomials.

For general polynomial multiplication/division, visit calc101.com

Over on the right there are links for long multiplication and long division.

That should help you understand the details

To solve the expression (y-2)(3y^3-5y+7) using the FOIL method, follow these steps:

Step 1: Identify the terms in the two sets of parentheses.
The first set of parentheses is y-2, and the second set of parentheses is 3y^3-5y+7.

Step 2: Apply FOIL (First, Outer, Inner, Last) method.
Multiply the terms in the first positions, the outer positions, the inner positions, and the last positions, respectively, using FOIL.

First: Multiply the first terms in each parentheses:
(y * 3y^3) = 3y^4

Outer: Multiply the terms on the outside:
(y * -5y) = -5y^2

Inner: Multiply the terms on the inside:
(-2 * 3y^3) = -6y^3

Last: Multiply the last terms in each parentheses:
(-2 * -5y) = 10y

Step 3: Combine the like terms.
Combine all the terms obtained in the previous step to simplify the expression.

3y^4 - 5y^2 - 6y^3 + 10y + 7

Now, the expression (y-2)(3y^3-5y+7) has been simplified using the FOIL method.