Find the constants A, B, C

\(\frac{2x^2 + 2x - 2}{x(x^2-1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}\)

 May 4, 2023

To find the constants A, B, and C, we can multiply both sides of the equation by x(x^2−1) and then solve for A, B, and C.

Multiplying both sides of the equation by x(x^2−1), we get:


Expanding both sides of the equation, we get:


Combining like terms, we get:


For this equation to be true for all values of x, the coefficients of x2, x, and the constant term must all be equal to 0.

Therefore, we have the following system of equations:

2+A+B+C=0 A+B-1=0 A+C-2=0

Solving this system of equations, we get:


Therefore, the constants A, B, and C are −2, 1, and 1, respectively.

 May 4, 2023

I agree with you up until you solved the system of equations.

Using the second equation, I get A+B=1.

Substitute into the first:

2+1+C=0, so C=-3.

Then use the third equation:

A-3-2=0, so A=5.

Then B=-4. So (A,B,C)=(5,-4,-3)

gb1falcon  May 4, 2023

\(\displaystyle \frac{2x^{2}+2x-2}{x(x^{2}-1)}\equiv \frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}\)

Begin by putting the terms on the rhs over a common denominator.

\(\displaystyle \frac{2x^{2}+2x-2}{x(x^{2}-1)} \equiv \frac{A(x-1)(x+1)+Bx(x+1)+Cx(x-1)}{x(x^{2}-1)}\)

For the rhs to be identical to the lhs we need

\(\displaystyle 2x^{2}+2x-2 \equiv A(x-1)(x+1)+Bx(x+1)+Cx(x-1),\)

(the bottom lines are already identical).

There are two ways of proceeding, either collect up terms on the rhs and equate coefficients, 

\(\displaystyle 2x^{2}+2x-2 \equiv x^{2}(A+B+C)+x(B-C)+(-A) \\ 2 = A+B+C \\2 = B-C \\ -2 = -A \\ A=2, \qquad B = 1, \qquad C=-1,\)

or, quicker, substitute convenient values for x,

\(\displaystyle x=0: -2=-A , \qquad A=2,\\ x=1 : \;\;\;2=2B, \qquad \;\;\;B=1 \\ x=-1: -2=2C, \qquad C=-1.\)

 May 5, 2023

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