4.2.1 Resolution of an algebraic fraction into partial fractions, when D(x) consists of non- repeated linear factors.
If linear factor (ax+b) occurs as a factor of D(x) , then there is a partial fraction of the form ax+bA , where AAA is a constant to be found.
In N(x)D(x)N(x)D(x)\frac{N(x)}{D(x)}D(x)N(x)D(x)N(x) , the polynomial D(x)D(x)D(x) may be written as,
We have, N(x)D(x)=A1a1x+b1+A2a2x+b2+A3a3x+b3+⋯+Ananx+bn,N(x)D(x)=A1a1x+b1+A2a2x+b2+A3a3x+b3+⋯+Ananx+bn,\frac{N(x)}{D(x)} = \frac{A_1}{a_1x + b_1} +\frac{A_2}{a_2x + b_2} +\frac{A_3}{a_3x + b_3} +\dots +\frac{A_n}{a_nx + b_n},D(x)N(x)=a1x+b1A1+a2x+b2A2+a3x+b3A3+⋯+anx+bnAn,D(x)N(x)=a1x+b1A1+a2x+b2A2+a3x+b3A3+⋯+anx+bnAn,
where A1A2…AnA_{1}A_{2}\dots A_{n}A1A2…An are constants to be determined. The following examples illustrate how we can find these constants:
Resolve 5x+4(x−4)(x+2)5x+4(x−4)(x+2)\frac{5x + 4}{(x - 4)(x + 2)}(x−4)(x+2)5x+4(x−4)(x+2)5x+4 into partial fractions.
Let 5x+4(x−4)(x+2)=Ax−4+Bx+25x+4(x−4)(x+2)=Ax−4+Bx+2\frac{5x + 4}{(x - 4)(x + 2)} = \frac{A}{x - 4} +\frac{B}{x + 2}(x−4)(x+2)5x+4=x−4A+x+2B(x−4)(x+2)5x+4=x−4A+x+2B (i)
Multiplying throughout by (x−4)(x+2)(x - 4)(x + 2)(x−4)(x+2) , we get
Equation (ii) is an identity, which holds good for all values of xxx and hence for
x=4andx=−2.x=4andx=−2.x = 4\mathrm{and}x = -2.x=4andx=−2.x=4andx=−2.
Put x−4=0x - 4 = 0x−4=0 i.e., x=4x = 4x=4 (factor corresponding to AAA ) on both sides of the equation (ii),
we get 5(4)+4=A(4+2)⇒A=45(4)+4=A(4+2)⇒A=45(4) + 4 = A(4 + 2)\Rightarrow \boxed {A = 4}5(4)+4=A(4+2)⇒A=45(4)+4=A(4+2)⇒A=4
Put x+2=0x + 2 = 0x+2=0 i.e., x=−2x = - 2x=−2 (factor corresponding to BBB ), we get
Thus required partial fractions are 4x−4+1x+24x−4+1x+2\frac{4}{x - 4} +\frac{1}{x + 2}x−44+x+21x−44+x+21
Hence, 5x+4(x−4)(x+2)=4x−4+1x+25x+4(x−4)(x+2)=4x−4+1x+2\frac{5x + 4}{(x - 4)(x + 2)} = \frac{4}{x - 4} +\frac{1}{x + 2}(x−4)(x+2)5x+4=x−44+x+21(x−4)(x+2)5x+4=x−44+x+21
This method is called the zero's method. This method is especially useful with linear factors in the denominator D(x)D(x)D(x) .