How do you divide (3x^4+22x^3+ 15 x^2+26x+8)/(x-2) 3x4+22x3+15x2+26x+8x−2?
1 Answer
Long divide the coefficients to find:
(3x^4+22x^3+15x^2+26x+8)/(x-2) = 3x^3+28x^2+71x+1683x4+22x3+15x2+26x+8x−2=3x3+28x2+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+1683x4+22x3+15x2+26x+8x−2=3x3+28x2+71x+168
with remainder
Or you can write:
3x^4+22x^3+15x^2+26x+83x4+22x3+15x2+26x+8
= (x-2)(3x^3+28x^2+71x+168) + 344=(x−2)(3x3+28x2+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+8f(2)=3⋅16+22⋅8+15⋅4+26⋅2+8
=48+176+60+52+8=48+176+60+52+8
=344=344
