maths:x^4+3x-2/(x^2+1)^3(x-4)=(Ax+B)/(x^2+1)+(Cx+D)/(x^2+1)^2+(Ex+F)/(x^2+1)^3+G/(x-4)+H/(x-4)^2.fine the value of ABCDEFGH,it a partial fraction o plz i need the solution urgently
This is DEFINITELY NOT "government"!!
(x^4+3x-2)/((x^2+1)^3(x-4))
=
266 / 4913(x-4)
- 266(x+4) / (x^2+1)
+ 23(x+4) / 289(x^2+1)^2
+ (7-11x) / 17(x^2+1)^3
Problems like this are so tedious that they really test only patience, not knowledge.
and there is no (x-4)^2 term, since your original fraction had none. If you really want that to be included, I can only refer you to
http://www.wolframalpha.com/input/?i=%28x^4%2B3x-2%29%2F%28%28x^2%2B1%29^3%28x-4%29^2%29
Scroll down a bit to the partial fraction breakdown.
To find the value of ABCDEFGH in the partial fraction decomposition of the given expression, we need to equate the numerator of the original expression to the sum of the numerators in the decomposed form.
Let's start by multiplying both sides of the equation by the denominator, (x^2 + 1)^3(x - 4). This will eliminate the denominators and allow us to solve for the numerators.
(x^4 + 3x - 2) = (Ax + B)(x^2 + 1)^2(x - 4) + (Cx + D)(x - 4)(x^2 + 1) + (Ex + F)(x^2 + 1) + G(x^2 + 1)^3 + H(x^2 + 1)^3(x - 4)
Now, we can multiply out the terms on the right side and gather like terms:
x^4 + 3x - 2 = (Ax + B)(x^4 + 2x^2 + 1)(x - 4) + (Cx + D)(x^3 - 4x^2 + x - 4) + (Ex + F)(x^2 + 1) + G(x^6 + 3x^4 + 3x^2 + 1) + H(x^6 + 3x^4 + 3x^2 + 1)(x - 4)
Expanding each term further:
x^4 + 3x - 2 = (Ax^5 - 4Ax^4 + Ax - 4Bx^3 + 8Bx^2 - 4B)(x - 4) + (Cx^4 - 4Cx^3 + Cx - 4Dx^2 + 16Dx - 4D) + (Ex^2 + Ex + Fx^2 + F) + Gx^6 + 3Gx^4 + 3Gx^2 + G + Hx^7 - 4Hx^6 + Hx^5 - 4Hx^5 + 16Hx^4 - 4Hx^3
Now, we can collect terms with the same powers of x:
x^4 + 3x - 2 = (Ax^5 - 4Ax^4 + Ax - 4Bx^3 + 8Bx^2 - 4B)(x - 4) + (Cx^4 - 4Cx^3 + Cx - 4Dx^2 + 16Dx - 4D) + (Ex^2 + Fx^2) + (Ex + F) + Gx^6 + (3Gx^4 - 4Hx^6 + Hx^5 - 4Hx^5 + 16Hx^4) + (- 4Hx^3)
Now, equating coefficients on both sides of the equation, we can solve for A, B, C, D, E, F, G, and H.
The coefficient of x^5 on the left side is 0, so the coefficient of x^5 on the right side must also be 0. This gives us:
H = 0
The coefficient of x^6 on the left side is 0, so the coefficient of x^6 on the right side must also be 0. This gives us:
G - 4H = 0
G = 4H
Since we know H = 0, G = 4(0) = 0
The coefficient of x^4 on the left side is 1, so the coefficient of x^4 on the right side must also be 1. This gives us:
3G - 4H + 16H = 1
3(0) - 4(0) + 16(0) = 1
0 - 0 + 0 = 1
0 = 1
This equation leads to a contradiction, indicating that there is no solution for ABCDEFGH in the given partial fraction decomposition.