Question

hey can anybody tell me how to solve the improper fraction
(8-4x+11x^2-2X^3)/(1-(1/4)X)(1-x+(1/2)x^2) using partial fractions

Answer 1

if i were you i would multiply top and bottom by 4 to clear those stupid fraction before i began

Answer 2

\[\frac{(8-4x+11x^2-2x^3)}{(1-\frac{1}{4}x)(1-x+\frac{1}{2}x^2)}\] is that it?

Answer 3

actually i wold multiply top and bottom by 8. i hate working with fractions

Answer 4

ya right

Answer 5

\[\frac{64-32x+88x^2-16x^3}{(4-x)(2-2x-x^2)}\] yes?

Answer 6

no that is not right

Answer 7

\[\frac{64−32x+88x^2−16x^3}{(4−x)(2−2x+x^2)}\]

Answer 8

but this is not an improper fraction because numerator and denominator both have degree 3

Answer 9

whenever degrees r equal it is said to b improper fraction

Answer 10

oh crap, i guess you can do it, but it will be a real pain. denominators will be
\[2-2x+x^2\] and
\[(4-x)^2\]

Answer 11

u just tell me which method i shd use here without multiplying 8 to the numerator....bcoz i wanna retain those fracttions of denominator as it is

Answer 12

oh maybe not. ok we can start. first write
\[\frac{A}{1-\frac{1}{4}x}+\frac{Bx+C}{1-x+\frac{1}{2}x^2}=\frac{8-4x+11x^2-2x^3}{(1+\frac{1}{4}x)(1-x-\frac{1}{2}x^2)}\]

Answer 13

now cross multiply on the left to add the fractions. you will have
\[A(1-x+\frac{1}{2}x^2)+(Bx+C)(1-\frac{1}{4}x)=x-4x-=+11x^2-2x^3\]

Answer 14

ya k fine got it ....thank u

Answer 15

typo on that line. the last = sign should not be there

Answer 16

then multiply out, and combine like terms. then equation like coefficients. it is a pain

Answer 17

ya k fine got it ....thank u

Answer 18

yw