Question

How do you long divide `(3x^2 - 8x + 2x^3 +3) div (x + 3)`?

Answer 1

You use Ruffini's rule.

`(2x^3 + 3x^2 - 8x + 3)/(x+3) = (2x^2 -3x + 1) + 0`

Explanation:

First thing we put the numerator in order of degrees, so
`2x^3 + 3x^2 - 8x + 3`

Then we put the divisor in the form of `(x-k)`,
`x+3 = x-(-3)`

So we list the coefficients out and the `k` we found like this

       |        2     3     -8    3
 -3   | _______________
       |

The first coefficient we repeat, and put under the rule

       |        2     3     -8    3
 -3   | _______________
       |        2

Now multiply that value by `k` and put it under the next coefficient

       |        2     3     -8    3
 -3   | ______-6_________
       |        2

Add it up and put the result under the rule

       |        2     3     -8    3
 -3   | ______-6_________
       |        2    -3

Repeat it until there's no more

       |        2     3     -8    3
 -3   | ______-6__ 9_ -3____
       |        2    -3      1     0

Now, since we're dividing a third degree polinomial by a first degree, the result must be a second degree polinomial. Under normal long division we have that `a/b = q + r` where r is the remainder, so for polynomials we have `f(x)/g(x) = h(x) + c`. Same concept.

The last number is the remainder, and the rest are the coefficients in order of degree, so

`(2x^3 + 3x^2 - 8x + 3)/(x+3) = (2x^2 -3x + 1) + 0`

Answer 2

The answer that I got was the same as the previous solution however, I think that you asked for it to be done using the long division algorithm.

I hope this helps :)

Explanation: