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Factor the denominator completely.

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Answer

1−32(x−1)+1(x−1)2+32(x+1)1 - \frac{3}{2(x-1)} + \frac{1}{(x-1)^2} + \frac{3}{2(x+1)}

Step-by-step solution

7 steps

  1. Factor the denominator completely.

    The given expression is:

    x3−x2−3x+5(x−1)(x2−1)\frac{x^3 - x^2 - 3x + 5}{(x-1)(x^2-1)}

    We know that x2−1=(x−1)(x+1)x^2-1 = (x-1)(x+1).
    Substitute this into the denominator:

    (x−1)(x2−1)=(x−1)(x−1)(x+1)=(x−1)2(x+1)(x-1)(x^2-1) = (x-1)(x-1)(x+1) = (x-1)^2(x+1)

    So the expression becomes:

    x3−x2−3x+5(x−1)2(x+1)\frac{x^3 - x^2 - 3x + 5}{(x-1)^2(x+1)}

  2. Check if the fraction is proper.

    The degree of the numerator (x3−x2−3x+5x^3 - x^2 - 3x + 5) is 3.
    The degree of the denominator ((x−1)2(x+1)=(x2−2x+1)(x+1)=x3−x2−x+1(x-1)^2(x+1) = (x^2-2x+1)(x+1) = x^3 - x^2 - x + 1) is 3.
    Since the degree of the numerator is equal to the degree of the denominator, this is an improper rational fraction. We must perform polynomial long division first.

  3. Perform polynomial long division.

    Divide x3−x2−3x+5x^3 - x^2 - 3x + 5 by x3−x2−x+1x^3 - x^2 - x + 1.

    x3−x2−3x+5x3−x2−x+1=1+(x3−x2−3x+5)−(x3−x2−x+1)x3−x2−x+1\frac{x^3 - x^2 - 3x + 5}{x^3 - x^2 - x + 1} = 1 + \frac{(x^3 - x^2 - 3x + 5) - (x^3 - x^2 - x + 1)}{x^3 - x^2 - x + 1} =1+−2x+4x3−x2−x+1= 1 + \frac{-2x + 4}{x^3 - x^2 - x + 1}

    So, the expression can be written as:

    1+4−2x(x−1)2(x+1)1 + \frac{4 - 2x}{(x-1)^2(x+1)}

  4. Set up the partial fraction decomposition for the proper fraction.

    Let f(x)=4−2x(x−1)2(x+1)f(x) = \frac{4 - 2x}{(x-1)^2(x+1)}.
    The denominator has a repeated linear factor (x−1)2(x-1)^2 and a distinct linear factor (x+1)(x+1). The form of the partial fraction decomposition is:

    4−2x(x−1)2(x+1)=Ax−1+B(x−1)2+Cx+1\frac{4 - 2x}{(x-1)^2(x+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+1}

  5. Clear the denominators to solve for A, B, and C.

    Multiply both sides by (x−1)2(x+1)(x-1)^2(x+1):

    4−2x=A(x−1)(x+1)+B(x+1)+C(x−1)24 - 2x = A(x-1)(x+1) + B(x+1) + C(x-1)^2 4−2x=A(x2−1)+B(x+1)+C(x2−2x+1)4 - 2x = A(x^2-1) + B(x+1) + C(x^2-2x+1)

  6. Solve for A, B, and C by substituting convenient values for xx.

    • Let x=1x=1:
    4−2(1)=A(12−1)+B(1+1)+C(1−1)24 - 2(1) = A(1^2-1) + B(1+1) + C(1-1)^2 2=A(0)+B(2)+C(0)2 = A(0) + B(2) + C(0) 2=2B  ⟹  B=12 = 2B \implies B = 1
    • Let x=−1x=-1:
    4−2(−1)=A((−1)2−1)+B(−1+1)+C(−1−1)24 - 2(-1) = A((-1)^2-1) + B(-1+1) + C(-1-1)^2 4+2=A(0)+B(0)+C(−2)24 + 2 = A(0) + B(0) + C(-2)^2 6=4C  ⟹  C=64=326 = 4C \implies C = \frac{6}{4} = \frac{3}{2}
    • Let x=0x=0:
    4−2(0)=A(02−1)+B(0+1)+C(0−1)24 - 2(0) = A(0^2-1) + B(0+1) + C(0-1)^2 4=A(−1)+B(1)+C(1)4 = A(-1) + B(1) + C(1) 4=−A+B+C4 = -A + B + C

    Substitute the values of B=1B=1 and C=32C=\frac{3}{2}:

    4=−A+1+324 = -A + 1 + \frac{3}{2} 4=−A+22+324 = -A + \frac{2}{2} + \frac{3}{2} 4=−A+524 = -A + \frac{5}{2} A=52−4A = \frac{5}{2} - 4 A=52−82A = \frac{5}{2} - \frac{8}{2} A=−32A = -\frac{3}{2}

  7. Write the complete partial fraction decomposition.

    Substitute the values of A, B, and C back into the expression from Step 3:

    1+−32x−1+1(x−1)2+32x+11 + \frac{-\frac{3}{2}}{x-1} + \frac{1}{(x-1)^2} + \frac{\frac{3}{2}}{x+1} 1−32(x−1)+1(x−1)2+32(x+1)1 - \frac{3}{2(x-1)} + \frac{1}{(x-1)^2} + \frac{3}{2(x+1)}
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