Decompose \( \frac{x^{3}-3 x+1}{x^{2}-x-2} \) inds its perthal frections

We wish to decompose   (x³ – 3x + 1)/(x² – x – 2) into partial fractions. Notice that the degree of the numerator (3) is higher than that of the denominator (2), so we must first perform polynomial long division. Step 1. Divide x³ – 3x + 1 by x² – x – 2.   • Divide x³ by x² to get x.   • Multiply the divisor by x: x(x² – x – 2) = x³ – x² – 2x.   • Subtract:     [x³ + 0x² – 3x + 1] – [x³ – x² – 2x] = x² – x + 1.   • Now, divide x² by x² to get 1.   • Multiply the divisor by 1: 1 · (x² – x – 2) = x² – x – 2.   • Subtract:     [x² – x + 1] – [x² – x – 2] = 3. The quotient is (x + 1) and the remainder is 3. Therefore, we can write   (x³ – 3x + 1)/(x² – x – 2) = x + 1 + 3/(x² – x – 2). Step 2. Factor the denominator of the remaining fraction.   x² – x – 2 factors as (x – 2)(x + 1). Thus,   3/(x² – x – 2) = 3/[(x – 2)(x + 1)]. Step 3. Express 3/[(x – 2)(x + 1)] as a sum of partial fractions. Assume   3/[(x – 2)(x + 1)] = A/(x – 2) + B/(x + 1). Multiply both sides by (x – 2)(x + 1) to clear denominators:   3 = A(x + 1) + B(x – 2). Now, equate coefficients by expanding the right side:   A(x + 1) + B(x – 2) = (A + B)x + (A – 2B). Since the left side is the constant 3 (which can be viewed as 0·x + 3), we have:   Coefficient of x: A + B = 0,   Constant term:  A – 2B = 3. Solve for A and B:   From A + B = 0, we have A = –B.   Substitute into A – 2B = 3:     (–B) – 2B = 3 ⟹ –3B = 3 ⟹ B = –1.   Then, A = –(–1) = 1. Thus, the partial fractions are:   3/(x² – x – 2) = 1/(x – 2) – 1/(x + 1). Step 4. Write the final decomposition. Substitute back into the expression from the long division:   (x³ – 3x + 1)/(x² – x – 2) = x + 1 + [1/(x – 2) – 1/(x + 1)]. This is the desired partial fraction decomposition.