Asked by Anonymous
Sorry. The problem should have been:
(x^2)/(x^3-8)(x-2)
Answers
Answered by
Steve
note that x^3-8 = (x-2)(x^2+2x+4)
So, your fraction is now
x^2
--------------------
(x-2)^2 (x^2+2x+4)
As partial fractions, that becomes
A/(x-2) + B(x-2)^2 + (Cx+D)/(x^2+2x+4)
Placing all that over the common denominator, we have
A(x-2)(x^2+2x+4) + B(x^2+2x+4) + (Cx+D)(x-2)^2
= A(x^3-8)+B(x^2+2x+4) + Cx^3+(-4C+D)x^2+(4C-4D)x+4D
Expand that all out and equate coefficients with the left side (0x^3+x^2+0x+0) and we have
A+C = 0
B-4C+D = 1
2B+4C-4D = 0
-8A+4B+4D = 0
A = 1/6
B = 1/3
C = -1/6
D = 0
and you can form the fractions now.
So, your fraction is now
x^2
--------------------
(x-2)^2 (x^2+2x+4)
As partial fractions, that becomes
A/(x-2) + B(x-2)^2 + (Cx+D)/(x^2+2x+4)
Placing all that over the common denominator, we have
A(x-2)(x^2+2x+4) + B(x^2+2x+4) + (Cx+D)(x-2)^2
= A(x^3-8)+B(x^2+2x+4) + Cx^3+(-4C+D)x^2+(4C-4D)x+4D
Expand that all out and equate coefficients with the left side (0x^3+x^2+0x+0) and we have
A+C = 0
B-4C+D = 1
2B+4C-4D = 0
-8A+4B+4D = 0
A = 1/6
B = 1/3
C = -1/6
D = 0
and you can form the fractions now.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.