# 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.

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

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

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

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