Decompose \( \frac{x^{3}-3 x+1}{x^{2}-x-2} \) inds its perthal frections
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To decompose the expression \( \frac{x^{3}-3x+1}{x^{2}-x-2} \) into partial fractions, we first factor the denominator. The quadratic \( x^{2}-x-2 \) factors into \( (x-2)(x+1) \). Now we can express the original fraction as: \[ \frac{x^{3}-3x+1}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1} \] where \( A \) and \( B \) are constants to be determined. Multiplying both sides by the denominator \( (x-2)(x+1) \), we have: \[ x^{3}-3x+1 = A(x+1) + B(x-2) \] Expanding the right side gives: \[ x^{3}-3x+1 = Ax + A + Bx - 2B = (A+B)x + (A-2B) \] Next, we equate the coefficients from both sides: 1. From the \( x^{3} \) term: There is no \( x^{3} \) term on the right, so we set the coefficient to 0. 2. From the \( x^{1} \) term: \( A + B = 0 \) 3. From the constant term: \( A - 2B = 1 \) Now, we can solve the system of equations: From \( A + B = 0 \), we get \( B = -A \). Substituting \( B = -A \) into \( A - 2B = 1 \) gives: \[ A - 2(-A) = 1 \implies A + 2A = 1 \implies 3A = 1 \implies A = \frac{1}{3} \] Thus, \( B = -\frac{1}{3} \). Now, substituting back into the partial fraction form: \[ \frac{x^{3}-3x+1}{(x-2)(x+1)} = \frac{1/3}{x-2} - \frac{1/3}{x+1} \] So, the partial fraction decomposition is: \[ \frac{x^{3}-3x+1}{x^{2}-x-2} = \frac{1/3}{x-2} - \frac{1/3}{x+1} \]