Question

$\frac{3x^2+1}{x^2-6x+8}$ is equal to

• A. $\frac{49}{2(x-4)}-\frac{13}{2(x-2)}$
• B. $3+\frac{49}{2(x-4)}-\frac{13}{2(x-2)}$
• C. $\frac{49}{2(x-4)}+\frac{13}{2(x-2)}$
• D. $\frac{-49}{2(x-4)}+\frac{13}{2(x-2)}$

Answer 1

Answer: (B)

Given, $\frac{3x^2+1}{x^2-6x+8}$
On dividing, we get
$\frac{3x^2+1}{x^2-6x+8}=3+\frac{18x-23}{x^2-6x+8}.....\left(i\right)$
Now, $\frac{18x-23}{\left(x-2\right)\left(x-4\right)}=\frac{A}{x-2}+\frac{B}{x-4}$
$\Rightarrow 18x-23=A(x-4)+B(x-2)$
$\Rightarrow 18x-23=(A+B)x-4A-2B$
Equating the coefficient of x and constant
term, we get
A + B = 18
- 4 A - 2 B = -23
On solving these equations, we get
$A=-\frac{13}{2},B=\frac{49}{2}$
$\therefore \frac{18x-23}{\left(x-2\right)\left(x-4\right)}=-\frac{13}{2\left(x-2\right)}+\frac{49}{2\left(x-4\right)}$
Then, from Eq. (i), we get
$\frac{3x^2+1}{x^2-6x+8}=3-\frac{13}{2\left(x-2\right)}+\frac{49}{2\left(x-4\right)}$