5 2 Repeated Linear Factors – Class 11TH Math

Case II: Repeated Linear Factors

If $Q(x)$ contains a repeated linear factor $(x-a)^n$ with $n \ge 2$ , then

\dfrac{P(x)}{Q(x)} = \dfrac{A_1}{x-a} + \dfrac{A_2}{(x-a)^2} + \cdots + \dfrac{A_n}{(x-a)^n}

Resolve $\dfrac{x^2+x-1}{(x+2)^3}$ into partial fractions.

Solution

\dfrac{x^2+x-1}{(x+2)^3} = \dfrac{A}{x+2} + \dfrac{B}{(x+2)^2} + \dfrac{C}{(x+2)^3}

x^2 + x - 1 = A(x+2)^2 + B(x+2) + C = Ax^2 + (4A+B)x + (4A+2B+C)

Equating: $A = 1$ , $4A + B = 1$ $\Rightarrow$ $B = -3$ . Put $x = -2$ : $C = 1$ .

Hence $\dfrac{1}{x+2} - \dfrac{3}{(x+2)^2} + \dfrac{1}{(x+2)^3}$ .

Resolve $\dfrac{1}{(x+1)^2(x^2-1)}$ into partial fractions.

Solution

$(x+1)^2(x^2-1) = (x+1)^3(x-1)$ , so

\dfrac{1}{(x-1)(x+1)^3} = \dfrac{A}{x-1} + \dfrac{B}{x+1} + \dfrac{C}{(x+1)^2} + \dfrac{D}{(x+1)^3}

1 = A(x+1)^3 + B(x+1)^2(x-1) + C(x-1)(x+1) + D(x-1)

$x = 1$ $\Rightarrow$ $A = \dfrac{1}{8}$ ; $x = -1$ $\Rightarrow$ $D = -\dfrac{1}{2}$ .

Equating $x^3$ and $x^2$ coefficients: $B = -\dfrac{1}{8}$ , $C = -\dfrac{1}{4}$ .

Hence

\dfrac{1}{8(x-1)} - \dfrac{1}{8(x+1)} - \dfrac{1}{4(x+1)^2} - \dfrac{1}{2(x+1)^3}