1.

We run through the steps for long division of polynomials. First divide the first term in the divisor into the first term in the dividend.

That gives us the first term in the quotient.

We multiply this term in the quotient by the divisor and subtract it from the dividend.

Remember; we are subtracting signed numbers, so change the sign and add.

When we subtract, we could bring down all of the terms from the dividend, but, as we shall see, only the next term will come into play at the next step, so, mathematicians, being as notoriously lazy as they are, have developed the technique of bringing down only one term at a time from the dividend.

Repeat the process with the remainder. Divide the first term in the divisor into the first term in the remainder. That gives us the next term in the quotient.

We multiply this term in the quotient by the divisor and subtract it from the remainder. When we change the sign and add and bring down the next term from the dividend we get

Repeat the process with the remainder. Divide the first term in the divisor into the first term in the remainder. That gives us the next term in the quotient. We multiply this term in the quotient by the divisor and subtract it from the remainder.

When we change the sign and add we get

At this point the remainder has smaller degree than the divisor. When that happens, we stop and express our answer as a quotient and a remainder.

We form a fraction by putting the remainder over the divisor and adding it to the quotient.

The reason we do this is because the answer to a division problem is the thing we need to multiply times the divisor to get the dividend. Notice that if we multiply

it will distribute as

or

so we can check our work by multiplying the quotient by the divisor and adding the remainder to see if we get the dividend.

and it checks.

Notice how the check corresponds to the division problem.

The terms in the first row of the multiplication problem are the first terms in the polynmials which are being subtracted in the division problem, and the terms on the second line are the second terms in the division problem. Since addition is commutative, the fact that the terms come out in different places does not make a difference. What matters is that they are all there.