Question 1.38: Express 6x+8/x²+3x+2 as its partial fractions....

The denominator is factorized as

The linear factor, x + 1, produces a partial fraction ofthe form \frac{A}{x+1}. The linear factor, x + 2, produces a partial fraction of the form \frac{B}{x+2}. 4 and B are unknown constants whose values have to be found. So we have

\frac{6 x+8}{x^2+3 x+2}=\frac{6 x+8}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}             (1.8)

Multiplying both sides of Equation (1.8) by (x + 1) and (x + 2) we obtain

6 x+8=A(x+2)+B(x+1)           (1.9)

We now evaluate A and B. There are two techniques which enable us to do this; evaluation using a specific value of x and equating coefficients. Each is illustrated in turn.

Evaluation using a specific value of x
We examine Equation (1.9). We will substitute a specific value of x into this equation. Although any value can be substituted for x we will choose a value which simplifies the equation as much as possible. We note that substituting x = —2 will simplify the r.h.s. of the equation since the term A(x + 2) will then be zero. Similarly, substituting in x = —\ will simplify the r.h.s. because the term 5(x + 1) will then be zero. So x = — 1 and x = —2 are two convenient values to substitute into Equation (1.9). We substitute each in turn.

Evaluating Equation (1.9) with x = — 1 gives

Evaluating Equation (1.9) with x = —2 gives

Substituting A = 2, B = 4 into Equation (1.8) yields

Thus the required partial fractions are \frac{2}{x+1} \text { and } \frac{4}{x+2}.

The constants A and B could have been found by equating coefficients.

Equating coefficients

Equation (1.9) may be written as

Equating the coefficients of x on both sides gives

Equating the constant terms on both sides gives

Thus we have two simultaneous equations in A and B, which may be solved to give A = 2 and B = 4 as before.