Mathematics

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

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

##### 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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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