Question

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

• A. $\frac{1}{2(x-3)}-\frac{(5x-5)}{2(x^2+2x+5)}$
• B. $\frac{1}{2(x-3)}+\frac{(5x-5)}{2(x^2+2x+5)}$
• C. $\frac{1}{(x-3)}+\frac{(5x-5)}{(x^2+2x+5)}$
• D. $\frac{1}{2(x+3)}+\frac{(5x-5)}{2(x^2+2x+5)}$

Answer 1

Answer: (B)

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

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

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

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

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

$18-33+5=20A$

$\therefore A=\frac{-1}{2}$

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

$⟹2=A+B$

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

$\therefore B=\frac{5}{2}$

and $-11=2A-3B+C$

$⟹-11=-1-\frac{15}{2}+C$

$⟹C=-11+1+\frac{15}{2}$

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

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

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

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

which are the required partial fractions.

Hence, option 'B' is correct.