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

1 Answer
Dec 28, 2015

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:

enter image source here

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#