Question

Expressing x^3+1/x^2+x-2 in partial fraction

Answer 1

Quotient=x-1
Remainder=3x-1

Answer 2

x^3+1/(x-1)(x+2)=(x-1)+3x-1/(x-1)(x+2)
=(x-1)+A/x-1+B/x+2
X^3+1=(x-1)^2(x+2)+A(x+2)+B(x-1)

When x=1,
A=2/3
When x=-2,
B=7/3
Thus, x^3+1/x^2+x-2= x-1+ 2/3(x-1)+7/3(x+2)

Answer 3

Can those two be the final answer ?

Answer 4

@amistre64

Answer 5

x^3+1
-------
x^2+x-2

can we factor them?


(x+1)(x^2-x+1)
---------------
(x+2) (x-1)

hmm, not unless you had a typo

long division then ..... assuming youve covered this



x - 1
----------------
x^2+x-2 | x^3+1
-(x^3+x^2-2x)
---------------
-x^2+2x+1
-(-x^2 -x+2)
--------------
3x-1 is our remainder

Answer 6

partial fraction decomposition .... is prolly what your refering to :/

Answer 7

yes

Answer 8

so factored form is great lol

(x+1)(x^2-x+1) A B
--------------- = ------- + -----
(x+2) (x-1) x+2 x-1

when x=-2 we can solve for A with coverup

(-2+1)(4+2+1)
--------------- = A = 7/3
(-2-1)



when x=1 we can solve for B with coverup

(1+1)(1-1+1)
--------------- = B = 2/3
(1+2)


assuming ive done it correctly lol

Answer 9

okay

Answer 10

if we simply work the remainder

3x-1
---------
(x+2)(x-1)



3(-2)-1
--------- = A = 7/3
(-2-1)


3(1)-1
--------- = B = 2/3 all the same
(1+2)

Answer 11

Okay... I'm confused with this....
Express Partial fraction
2x^2+3x+2/x^2+3x+2

Answer 12

The quotient is 2
Remainder is -3x-2

Answer 13

and the factors of the denominator?

Answer 14

Nxt, I need to simplify rite

Answer 15

x^2/(x+1)(x+2)= A/(x+1)+ B(x+2)

x^2/(x+2)(x+1)=A(x+2) + B(x+1)
Are these considered the same
Will it affect the final answer?

Answer 16

those are not considered the same

Answer 17

\[\frac{-3x-2}{(x+2)(x+1)}=\frac A{x+2}+\frac B{x+1}\]
now lets mutliply thru by (x+2)(x+1)
\[\bcancel{\cancel{\frac{(x+2)(x+1)}{(x+2)(x+1)}}}(-3x-2)=\frac {A\cancel{(x+2)}(x+1)}{\cancel{x+2}}+\frac {B(x+2)\cancel{(x+1)}}{\cancel{x+1}}\]
\[-3x-2=A(x+1)+B(x+2)\]

Answer 18

Based on this working...
-3x-2/(x+2)(x+1)= A/x+2 + B/x+1
What if I write like this.....
-3x-2/(x+1)(x+2)=A/x+1 + B/x+2
Is this considered the same?

Answer 19

its the same generality, but you wont get the same A=A and B=B results

Answer 20

A and B are just some unknowns that you are working to find. their names are immaterial really. But if your expecting that the name A has some magical mystical property that it must maintain in all expressions then your thought process would be a little off

Answer 21

So, if I swith them, the final answer might be different?

Answer 22

it wont be 'different'. As long as your working it consistently

Answer 23

Ok, thanks Amistre!

Answer 24

youre welcome.