How do you divide #(2x^4+5x^3-3x+1)/(x^2-x+2)#?

1 Answer
Jun 12, 2018

#2x^2+7x+3# with a remainder of #-14x-5#.

Explanation:

#(2x^4+5x^3-3x+1)/(x^2-x+2)#

We can use the polynomial long division method, begin by writing:

#x^2-x+2|overline(2x^4+5x^3-3x+1)#

The first question we have to ask is how many times does #x^2# go into #2x^4#. The answer is obviously #2x^2#. We put this factor on top, so the next line of our working will look like:

#color(white)(...................)2x^2#
#x^2-x+2|overline(2x^4+5x^3-3x+1)#

We now have to multiply the divisor #x^2-x+2# by #2x^2# to get
#2x^4-2x^3+4x^2# and subtract this from the dividend, the next line of working will look like:

#color(white)(...................)2x^2#
#x^2-x+2|overline(2x^4+5x^3-0x^2-3x+1)#
#color(white)(...............)ul(-(2x^4-2x^3+4x^2))#

Notice in the dividend their is no #x^2# so I have added a #0x^2# term in place of where the #x^2# term should be to make the working clearer. We do the subtraction to get:

#color(white)(...................)2x^2#
#x^2-x+2|overline(2x^4+5x^3-0x^2-3x+1)#
#color(white)(...............)ul(-(2x^4-2x^3+4x^2+0x+0))#
#color(white)(......................)0+7x^3-4x^2-3x+1#

We now repeat the process, treating: #7x^3-4x^2+3x+1# as the "new dividend." Keep repeating until the degree of the dividend is less than the degree of the divisor. The working, when written out in full will look like:

#color(white)(...................)2x^2+7x+3#
#x^2-x+2|overline(2x^4+5x^3-0x^2-3x+1)#
#color(white)(...............)ul(-(2x^4-2x^3+4x^2+0x+0))#
#color(white)(......................)0+7x^3-4x^2-3x+1#
#color(white)(.......................)ul(-(7x^3-7x^2+14x+0))#
#color(white)(................................)0+3x^2-17x+1#
#color(white)(....................................)ul(-(3x^2-3x+6))#
#color(white)(.............................................)-14x-5#

So the result is:

#2x^2+7x+3# with a remainder of #-14x-5#. We can also use this to rewrite the original fraction as:

#(2x^4+5x^3-3x+1)/(x^2-x+2)=2x^2+7x+3+(-14x-5)/(x^2-x+2)#