All methods are supposed to be correct in most if not all situations and it should not matter if you use a different method given the options
Given:

To find:
The product of the polynomials.
Solution:
1.

Multiply the numerical coefficient and add the powers of x.

2. 
Multiply each term of first polynomial with each term of 2nd polynomial.
Multiply the numerical coefficient and add the powers of x.


3. 
Multiply each term of first polynomial with each term of 2nd polynomial.
Multiply the numerical coefficient and add the powers of x.

Add or subtract like terms together.

The answer for multiplying polynomials:



Cost less salvage value = 970,000 - 4500 = 965,500
Capacity of machine = 1,000,000 units.
units consumed at the end of second year = 200,000 + 300,000 = 500,000 units.
Capacity remaining = 1,000,000 - 500,000 = 500,000 units
Book value at end of second year = (500,000/1,000,000)*965,500 + 4500
= $487,250