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)

$3+\frac{49}{2(x-4)}-\frac {13}{2(x-2)}$

Answer 2

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, $\frac{18x -23}{\left(x-2\right)\left(x-4\right)} = \frac{A}{x-2} + \frac{B}{x-4}$
$\Rightarrow \; 18 x - 23 = A (x - 4) + B (x - 2)$
$\Rightarrow \; 18 x - 23 = (A + B) x - 4 A - 2B$
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 E (i), we get
$\frac{3x^{2}+1}{x^{2} -6x +8} = 3- \frac{13}{2\left(x-2\right)} + \frac{49}{2\left(x-4\right)}$