How do you divide #(3x^4+22x^3+ 15 x^2+26x+8)/(x-2) #?
1 Answer
Long divide the coefficients to find:
#(3x^4+22x^3+15x^2+26x+8)/(x-2) = 3x^3+28x^2+71x+168#
with remainder
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
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
As a check, let
#f(2) = 3*16+22*8+15*4+26*2+8#
#=48+176+60+52+8#
#=344#