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Advanced Math/On partial fraction
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QUESTION: Express the following in partial fraction:
(x^6-x^5-4x^2+x)/(x^4+3x^2+2)
i have tried it many times for hours, again & again still cant get it right. Thanks you kindly.
ANSWER: The bottom factors in to (x²+1)(x²+2).
First we must divide the bottom into the top to get a remainder term to make the largets power on the top to be less than the largest power on the bottom.
First, factor out an x on the top so that you have x(x^5-x^4-4x+1).
Divide x^4+3x^2+2 into the x^5-x^4-4x+1, giving an x with a remainder of -x^4-3x^3-6x+1.
Divide x^4+3x^2+2 into -x^4-3x^3-6x+1 , giving a -1 with a remainder of -3x^3+3x^2-6x+3.
We now have the problem x[x-1 + (-3x^3+3x²-6x+3)/((x²+1)(x²+2))].
What needs to be partial fractions is the
(-3x^3+3x²-6x+3)/((x²+1)(x²+2)), which I will write as
(Ax+B)/(x²+1) + (Cx+D)/(x²+2).
Setting these two equal gives
(Ax+B)(x²+2) + (Cx+D)(x²+1) = (-3x^3+3x²-6x+3).
Multiplying the left side out gives
Ax^3+Bx²+2Ax+2B + Cx^3+Dx²+Cx+D = -3x^3+3x²-6x+3.
Grouping like terms gives A+C=-3, B+D=3, 2A+C=-6, and 2B+D=3.
Note that all you have to do is solve
[A+C=-3, 2A+C=6] and [B+D=3, 2B+D=3=3].
You get (without to much work at all) A=9, C=-12, B=3, and D=0.
Looking back at the problem, (Ax+B)/(x²+1) + (Cx+D)/(x²+2), we now have x[x-1 + (9x+3)/(x²+1) - 12x/(x²+2)].
---------- FOLLOW-UP ----------
QUESTION: I use elimination for A+C=-3 & 2A+C=-6 ,by subtracting the both equation. I got A=-3 & C=0. The same also i do for B+D=3 & 2B+D=3 and i got B=0 & D=3. (x^6-x^5-4x^2+x)/(x^4+3x^2+2) = x(x-1-(3x)/(x^2+1) + 3/(x^+2))
(x^6-x^5-4x^2+x)/(x^4+3x^2+2)=x^2-x-(3x^2)/(x^2+1) + (3x)/(x^+2). But the real answers are x^2 - x - 3 + 3/(x^2+1) + (3x)/(x^2+2) . Can you have a look what is wrong? Bless you.
Yes, the values you got for A, B, C, and D are right.
Note that the final answer is x[x-1 + (-3x^3+3x²-6x+2)/(x^4+3x²+2)].
When I mutlipy the (x-1) by (x^4+3x²+2)/(x^4+3x²+2) and add it on, I get the correct answer after combining like terms and multplying the entire numerator by x.
Thank you for blessing me - may you be blessed as well.
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