How do you multiply #(x^4-x^3+x^2-x)/(2x^3+2x^2+x+1)div (x^3-4x^2+x-4)/(2x^3-8x^2+x-4)#?

2 Answers
Sep 5, 2016

Answer:

#(x(x-1))/(x+1)#

Explanation:

Factorise each expression first. Each has 4 terms - use grouping.

#(x^4-x^3+x^2-x)/(2x^3+2x^2+x+1)div (x^3-4x^2+x-4)/(2x^3-8x^2+x-4)#

Group into pairs.
#((x^4-x^3)+(x^2-x))/((2x^3+2x^2)+(x+1))div ((x^3-4x^2)+(x-4))/((2x^3-8x^2)+(x-4))#

Take out the common factors
#(x^3(x-1)+x(x-1))/(2x^2(x+1)+(x+1))div color(red)((x^2(x-4)+(x-4)))/(color(blue)((2x^2(x-4)+(x-4))#

change #div# to #xx# and invert the fraction.
Take out the common brackets and factors

#(x(x-1)(x^2+1))/((x+1)(2x^2 +1)) xx (color(blue)((x-4)(2x^2+1)))/color(red)((x-4)(x^2+1))#l

Cancel like factors

#(x(x-1)cancel(x^2+1))/((x+1)cancel(2x^2 +1)) xx ((cancel(x-4)cancel(2x^2+1)))/(cancel(x-4)cancel(x^2+1))#l

#(x(x-1))/(x+1)#

Sep 5, 2016

Answer:

#(x^4-x^3+x^2-x)/(2x^3+2x^2+x+1) -: (x^3-4x^2+x-4)/(2x^3-8x^2+x-4)#

#=(x(x-1))/(x+1)#

#=x-2+2/(x+1)#

with exclusion: #x != 4#

Explanation:

#(x^4-x^3+x^2-x)/(2x^3+2x^2+x+1) -: (x^3-4x^2+x-4)/(2x^3-8x^2+x-4)#

#=(x^4-x^3+x^2-x)/(2x^3+2x^2+x+1) xx (2x^3-8x^2+x-4)/(x^3-4x^2+x-4)#

#=(x((x^3-x^2)+(x-1)))/((2x^3+2x^2)+(x+1)) xx ((2x^3-8x^2)+(x-4))/((x^3-4x^2)+(x-4))#

#=(x(x^2(x-1)+1(x-1)))/(2x^2(x+1)+1(x+1)) xx (2x^2(x-4)+1(x-4))/(x^2(x-4)+1(x-4))#

#=(xcolor(red)(cancel(color(black)((x^2+1))))(x-1))/(color(orange)(cancel(color(black)((2x^2+1))))(x+1)) xx (color(orange)(cancel(color(black)((2x^2+1))))color(purple)(cancel(color(black)((x-4)))))/(color(red)(cancel(color(black)((x^2+1))))color(purple)(cancel(color(black)((x-4)))#

#=color(blue)((x(x-1))/(x+1))#

#=(x^2+x-2x-2+2)/(x+1)#

#=(x(x+1)-2(x+1)+2)/(x+1)#

#=((x-2)(x+1)+2)/(x+1)#

#=color(blue)(x-2+2/(x+1))#

with exclusion: #x != 4#

This value needs to be excluded because when #x=4# the left hand side boils down to #0/0# which is undefined, while the right hand side is #12/5# which is defined.

If #x# is allowed to take Complex values, then there are also exclusions:

#x != i" "# and #" "x != sqrt(2)/2i#

These values would result in #x^2+1 = 0# or #2x^2+1 = 0# respectively, making the left hand side expression #0/0# undefined.