How do you divide #(x^4 - 7x^3 + 2x^2 + 9x)/(x^3-x^2+2x+1)#?

1 Answer
Feb 10, 2016

Answer:

#(x^4 - 7x^3 + 2x^2 + 9x) -: (x^3 - x^2 + 2x + 1) = x - 6#
with the remainder #-6x^2 + 20x + 6#.

Explanation:

I'm aware that in some countries, polynomial long division is written in a different format. I will use the format that I'm most used to, and I hope that it will be no problem for you to convert this into your prefered format.

As you have asked how to do the division, let me explain the basic steps of polynomial long division:

1) Divide the highest power of your numerator by the highest power of the denominator.

In your case, that's #x^4 -: x^3 = x#

2) Multiply the result from 1) with your denominator.

In your case, that's # x * (x^3 - x^2 + 2x + 1) = x^4 - x^3 + 2x^2 + x#

3) Subtract the result from 2) from the numerator.

In your case, that's #(x^4 - 7x^3 + 2x^2 + 9x) - (x^4 - x^3 + 2x^2 + x) = -6x^3 + 8x#

4) Consider the result from 3) to be your new numerator.
Repeat the steps 1) - 3) as long as the highest power of your numerator is greater or equal to the highest power of your denominator.

Here's the total calculation:

# color(white)(xii) (x^4 - 7x^3 + 2x^2 + 9x) -: (x^3 - x^2 + 2x + 1) = x - 6#
# -(x^4 - x^3 color(white)(x)+ 2x^2 + x)#
# color(white)(x) color(white)(xxxxxxxxxxxxx)/ #
# color(white)(xxx) -6x^3 color(white)(xxxxx) + 8x#
# color(white)(x) -(-6x^3 + 6x^2- 12x - 6)#
# color(white)(xxx) color(white)(xxxxxxxxxxxxxxxx)/ #
# color(white)(xxxxxxxx) -6x^2 + 20x + 6#

Thus,

#(x^4 - 7x^3 + 2x^2 + 9x) -: (x^3 - x^2 + 2x + 1) = x - 6#

with the remainder #-6x^2 + 20x + 6#.

Or, if you prefer a different notation,

#(x^4 - 7x^3 + 2x^2 + 9x) -: (x^3 - x^2 + 2x + 1) = x - 6 + (-6x^2 + 20x + 6)/(x^3 - x^2 + 2x + 1)#