Partial Fractions: Decomposition (Type-3)


When one of the factors in the denominator is quadratic: Let the quadratic factor in the denominator is \((ax^2 + bx + c)\) then we can write \(\frac{Ax + B}{ax^2 + bx + c}\) equal to the main fraction. Hence for every quadratic factor the numerator will be in the form of \((Ax + B)\). Here A, and B are the unknown constants.

Example(1): \(\frac{x}{(1 + x^3)}\)

Solution: The equation can be simplified as \(\frac{x}{(1 + x) (1 - x + x^2)}\) and according to the method the given equation can also be written as.

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

$$ \frac{x}{(1 + x) (1 - x + x^2)} = \frac{A}{(1 + x)} + \frac{Bx + C}{(1 - x + x^2)}....(1) $$

Now taking the LCM of RHS (Right Hand Side) terms.

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

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

The terms in the denominators on both sides are the same so they will be canceled.

\(x = A (1 - x + x^2) +\) \((1 + x) (Bx + C)\)

$$ x = A (1 - x + x^2) + (1 + x) (Bx + C) $$

Write down the terms of \(x^2\), \(x\), and constants separately as written below.

\(x = A - Ax + Ax^2 +\) \(Bx + C + Bx^2 + Cx\)

$$ x = A - Ax + Ax^2 + Bx + C + Bx^2 + Cx $$

\(x = x^2 \ (A + B) +\) \(x \ (- A + B + C) + (A + C)\)

$$ x = x^2 \ (A + B) + x \ (- A + B + C) + (A + C) $$

After comparing the values of \(x^2\), \(x\), and constants of both sides we get. $$ A + B = 0....(2) $$ $$ - A + B + C = 1....(3) $$ $$ A + C = 0....(4) $$ by subtracting the equation (3) from the equation (4). $$ A + C + A - B - C = 0 - 1 $$ $$ 2A - B = -1 $$ from the equation (2) put the value of \(B = -A\). $$ 2A + A = -1 $$ $$ 3A = -1 $$ $$ A = \frac{-1}{3} $$ Now put the value of A in equation (2). $$ A + B = 0 $$ $$ \frac{-1}{3} + B = 0 $$ $$ B = \frac{1}{3} $$ Now put the value of A and B in equation (3) to get the value of C. $$ - A + B + C = 1 $$ $$ \frac{1}{3} + \frac{1}{3} + C = 1 $$ $$ C = 1 - \frac{1}{3} - \frac{1}{3} $$ $$ C = \frac{1}{3} $$ After putting the values of A, B, and C in equation (1), we get the final equation of decomposition.

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

$$ \frac{x}{(1 + x) (1 - x + x^2)} = \frac{-1}{3 \ (1 + x)} + \frac{\frac{1}{3} \ x + \frac{1}{3}}{(1 - x + x^2)} $$

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

$$ \frac{x}{(1 + x) (1 - x + x^2)} = \frac{-1}{3 \ (1 + x)} + \frac{x + 1}{3 \ (1 - x + x^2)} $$



Example(2): \(\frac{3 + 2x}{(1 + x) \ (1 + x^2)}\)

Solution: According to the method the given equation can also be written as.

\(\frac{3 + 2x}{(1 + x) \ (1 + x^2)} =\) \(\frac{A}{(1 + x)} + \frac{Bx + C}{(1 + x^2)}....(1)\)

$$ \frac{3 + 2x}{(1 + x) \ (1 + x^2)} = \frac{A}{(1 + x)} + \frac{Bx + C}{(1 + x^2)}....(1) $$

Now taking the LCM of RHS (Right Hand Side) terms.

\(\frac{3 + 2x}{(1 + x) \ (1 + x^2)} =\) \(\frac{A (1 + x^2) + (1 + x) (Bx + C)}{(1 + x) \ (1 + x^2)}\)

$$ \frac{3 + 2x}{(1 + x) \ (1 + x^2)} = \frac{A (1 + x^2) + (1 + x) (Bx + C)}{(1 + x) \ (1 + x^2)} $$

The terms in the denominators on both sides are the same so they will be canceled.

\(3 + 2x = A \ (1 + x^2) +\) \((1 + x) \ (Bx + C)\)

$$ 3 + 2x = A \ (1 + x^2) + (1 + x) \ (Bx + C) $$

Write down the terms of \(x^2\), \(x\), and constants separately as written below.

\(3 + 2x = A + Ax^2\) \(+ Bx + C + Bx^2 + Cx\)

$$ 3 + 2x = A + Ax^2 + Bx + C + Bx^2 + Cx $$

\(3 + 2x = x^2 \ (A + B) +\) \(x \ (B + C) + (A + C)\)

$$ 3 + 2x = x^2 \ (A + B) + x \ (B + C) + (A + C) $$

After comparing the values of \(x^2\), \(x\), and constants of both sides we get. $$ A + B = 0....(2) $$ $$ B + C = 2....(3) $$ $$ A + C = 3....(4) $$ by subtracting the equation (3) from the equation (4). $$ A + C - B - C = 3 - 2 $$ $$ A - B = 1 $$ from the equation (2) put the value of \(A = -B\). $$ - B - B = 1 $$ $$ -2B = 1 $$ $$ B = \frac{-1}{2} $$ Now put the value of B in equation (2). $$ A + B = 0 $$ $$ A - \frac{1}{2} = 0 $$ $$ A = \frac{1}{2} $$ Now put the value of A in equation (4) to get the value of C. $$ A + C = 3 $$ $$ \frac{1}{2} + C = 3 $$ $$ C = 3 - \frac{1}{2} $$ $$ C = \frac{5}{2} $$ After putting the values of A, B, and C in equation (1), we get the final equation of decomposition.

\(\frac{3 + 2x}{(1 + x) \ (1 + x^2)} =\) \(\frac{1}{2 \ (1 + x)} + \frac{\frac{-1}{2} \ x + \frac{5}{2}}{(1 + x^2)}\)

$$ \frac{3 + 2x}{(1 + x) \ (1 + x^2)} = \frac{1}{2 \ (1 + x)} + \frac{\frac{-1}{2} \ x + \frac{5}{2}}{(1 + x^2)} $$

\(\frac{3 + 2x}{(1 + x) \ (1 + x^2)} =\) \(\frac{1}{2 \ (1 + x)} + \frac{5 - x}{2 \ (1 + x^2)}\)

$$ \frac{3 + 2x}{(1 + x) \ (1 + x^2)} = \frac{1}{2 \ (1 + x)} + \frac{5 - x}{2 \ (1 + x^2)} $$




Partial Fractions

Chapter 4: Decomposition (Type-3)


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