Mathematics

Resolve $$\displaystyle \frac{2x^2-11x+5}{(x-3)(x^2+2x+5)}$$ into partial fractions.


ANSWER

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


SOLUTION
Let            $$\displaystyle \frac{2x^2-11x+5}{(x-3)(x^2+2x+5)} =\frac{A}{x-3}+\frac{Bx+C}{x^2+2x+5}$$                                 ...(i)

$$\Longrightarrow                     2x^2-11+5=A(x^2+2x+5)+(Bx+C)(x-3)$$                      ...(ii)

Putting $$x-3=0$$ or $$x=3$$ in (i), we obtain

                    $$2(3)^2-11(3)+5=A(3^2+2.3+5)+0$$

                                $$18-33+5=20A$$

$$\therefore                                                                     A= \displaystyle \dfrac{-1}{2}$$

equation the coefficient of $$x^2$$ and $$x$$ in (ii), we have

$$\Longrightarrow                                          2=A+B$$

$$\therefore                                                B=2-A=2-(-\dfrac{1}{2})$$

$$\therefore                                                B=\dfrac 52$$

and                   $$-11=2A-3B+C$$

$$\Longrightarrow                           -11=-1-\dfrac{15}{2}+C$$

$$\Longrightarrow                             C = -11 + 1+\dfrac{15}{2}$$

                            $$= -10+\dfrac{15}{2}$$

                            $$= -\dfrac{5}{2}$$

Substituting the value of A, B and C in (i), we have

            $$\displaystyle \frac{2x^2-11x+5}{(x-3)(x^2+2x+5)} = -\frac{1}{2(x-3)} + \frac{(5x-5)}{2(x^2+2x+5)}$$

which are the required partial fractions.

Hence, option 'B' is correct.
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Single Correct Medium Published on 17th 09, 2020
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