How do you use polynomial long division to divide #(2x^3-x+1)div(x^2+x+1)# and write the polynomial in the form #p(x)=d(x)q(x)+r(x)#?

1 Answer
Jun 25, 2017

#2(x-1) -(x-2)/(x^2+x-1)#
#color(white)()#

#d(x)=2#
#q(x)=(x-1)#
#r(x)=-(x-2)/(x^2+x-1)#

Explanation:

Note that #r(x)# will be the remainder and that #d(x)q(x)# will be a factorisation.

Note that I am using a place keeper for convenience of formatting:
#0x^2# which has no value.

#" "2x^3+0x^2-x+1#
#color(magenta)(2x)(x^2+x+1)-> ul(2x^3+2x^2+2xlarr" Subtract")#
#" "0-2x^2-3x+1#
#color(magenta)(-2)(x^2+x+1)->" "ul(-2x^2-2x-2larr" Subtract")#
#" "0-x+2larr" Remainder"#

#color(magenta)((2x-2)+(-x+2)/(x^2+x+1))#

Write as: #2(x-1) -(x-2)/(x^2+x-1)#

Note that I have changed the remainder's sin (for the whole) from positive to negative to change #-x# to #+x#

Thus we have:

#d(x)=2#
#q(x)=(x-1)#
#r(x)=-(x-2)/(x^2+x-1)#