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

The Correct Option is 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 \; 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)} $

Answer 2

We have Problem Equation \( \frac{3x^2 + 1}{x^2 - 6x + 8} \).
After  factorizing  the denominator we get  \( x^2 - 6x + 8 = (x - 4)(x - 2) \). Now, we rewrite the expression:
\[ \frac{3x^2 + 1}{x^2 - 6x + 8} = \frac{3x^2 + 1}{(x - 4)(x - 2)} \]
The partial fraction decomposition is:
\[ \frac{3x^2 + 1}{(x - 4)(x - 2)} = \frac{A}{x - 4} + \frac{B}{x - 2} \]
Multiplying both sides by \( (x - 4)(x - 2) \), we get:
\[ 3x^2 + 1 = A(x - 2) + B(x - 4) \]
Expanding and comparing coefficients, we find:
\[ 3x^2 + 1 = Ax - 2A + Bx - 4B \]
Matching coefficients of \( x^2 \), \( x \), and constants, we get the following system of equations:
\[ A + B = 0 \]
\[ -2A - 4B = 1 \]
From the first equation, we get \( A = -B \). Substituting into the second equation, we find:
\[ -2(-B) - 4B = 1 \]
\[ 2B - 4B = 1 \]
\[ -2B = 1 \]
\[ B = -\frac{1}{2} \]
Therefore, \( A = \frac{1}{2} \).
Substituting back into the partial fraction decomposition, we get:
\[ \frac{3x^2 + 1}{x^2 - 6x + 8} = \frac{\frac{1}{2}}{x - 4} - \frac{\frac{1}{2}}{x - 2} \]
Simplifying, we have:
\[ \frac{3x^2 + 1}{x^2 - 6x + 8} = \frac{1}{2(x - 4)} - \frac{1}{2(x - 2)} = 3 + \frac{1}{2(x - 4)} - \frac{1}{2(x - 2)} \]
So the correct  answer is Option(B): \( 3 + \frac{1}{2(x - 4)} - \frac{1}{2(x - 2)} \).