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

Question

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

Nathan Axel · Accepted 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 #344#

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#

Ethan Adkins · Answer

undefined To divide (3x^4+22x^3+15x^2+26x+8) by (x-2), you can use long division. Here are the steps:

1. Divide the first term of the dividend (3x^4) by the first term of the divisor (x), which gives you 3x^3.
2. Multiply the divisor (x-2) by the quotient obtained in step 1 (3x^3), which gives you 3x^4-6x^3.
3. Subtract the result obtained in step 2 from the original dividend (3x^4+22x^3+15x^2+26x+8) to get the new dividend: (22x^3+15x^2+26x+8)-(3x^4-6x^3) = -3x^4+22x^3+15x^2+26x+8+6x^3 = -3x^4+28x^3+15x^2+26x+8.
4. Repeat steps 1-3 with the new dividend (-3x^4+28x^3+15x^2+26x+8) until the degree of the new dividend is less than the degree of the divisor.

Continuing the process, you will find that the quotient is 3x^3+16x^2+47x+94 and the remainder is 190.

Therefore, (3x^4+22x^3+15x^2+26x+8)/(x-2) = 3x^3+16x^2+47x+94 with a remainder of 190.