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)#?

Question

Kennedy Adams · Accepted 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# #(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.

Genesis Allen · Answer

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

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)#

Eli Assouline · Answer

undefined To divide the given expressions, we can use the division algorithm for polynomials. First, we need to factorize the denominators and rewrite the division as multiplication by the reciprocal.

The expression becomes:

[(x^4 - x^3 + x^2 - x) / (2x^3 + 2x^2 + x + 1)] * [(2x^3 - 8x^2 + x - 4) / (x^3 - 4x^2 + x - 4)]

Next, we can cancel out common factors between the numerators and denominators:

[(x^4 - x^3 + x^2 - x) / (2x^3 + 2x^2 + x + 1)] * [(2(x^3 - 4x^2 + x - 4)) / (x^3 - 4x^2 + x - 4)]

Simplifying further:

[(x^4 - x^3 + x^2 - x) / (2x^3 + 2x^2 + x + 1)] * [2(x^3 - 4x^2 + x - 4) / (x^3 - 4x^2 + x - 4)]

Now, we can cancel out the common factors:

[(x^4 - x^3 + x^2 - x) / (2x^3 + 2x^2 + x + 1)] * [2 / 1]

Simplifying the multiplication:

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

This is the simplified expression after dividing the given expressions.