Question

How do you divide `(3x^4+22x^3+ 15 x^2+26x+8)/(x-2)`?

Answer 1

Long divide the coefficients to find:

`(3x^4+22x^3+15x^2+26x+8)/(x-2) = 3x^3+28x^2+71x+168`

with remainder `344`

Explanation:

There are other ways, but I like to long divide the coefficients like this:

A more compact form of this is called synthetic division, but I find it easier to read the long division layout.

Reassembling polynomials from the coefficients, we find:

`(3x^4+22x^3+15x^2+26x+8)/(x-2) = 3x^3+28x^2+71x+168`

with remainder `344`.

Or you can write:

`3x^4+22x^3+15x^2+26x+8`

`= (x-2)(3x^3+28x^2+71x+168) + 344`

Note that if you are long dividing polynomials that have a 'missing' term, then you need to include a `0` for the coefficient of that term. We don't need to do that for our example.

As a check, let `f(x) = 3x^4+22x^3+15x^2+26x+8` and evaluate `f(2)`, which should be the remainder:

`f(2) = 3*16+22*8+15*4+26*2+8`

`=48+176+60+52+8`

`=344`