The partial fractions of frac{x^2 - 5}{x^2 - | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given expression: $f(x) = \frac{x^2 - 5}{x^2 - 3x + 2}$.Since the degree of the numerator is equal to the degree of the denominator,we perform lon

Vedclass · Accepted Answer

$1 + \frac{4}{x - 1} - \frac{1}{x - 2}$

Vedclass · Answer

(D) Given expression: $f(x) = \frac{x^2 - 5}{x^2 - 3x + 2}$.Since the degree of the numerator is equal to the degree of the denominator,we perform long division:$\frac{x^2 - 5}{x^2 - 3x + 2} = 1 + \frac{3x - 7}{x^2 - 3x + 2}$.Now,factor the denominator: $x^2 - 3x + 2 = (x - 1)(x - 2)$.We express the remainder as partial fractions: $\frac{3x - 7}{(x - 1)(x - 2)} = \frac{A}{x - 1} + \frac{B}{x - 2}$.$3x - 7 = A(x - 2) + B(x - 1)$.Setting $x = 1$: $3(1) - 7 = A(1 - 2)$ $\Rightarrow -4 = -A$ $\Rightarrow A = 4$.Setting $x = 2$: $3(2) - 7 = B(2 - 1) \Rightarrow -1 = B$.Thus,$\frac{x^2 - 5}{x^2 - 3x + 2} = 1 + \frac{4}{x - 1} - \frac{1}{x - 2}$.

The partial fractions of $\frac{x^2 - 5}{x^2 - 3x + 2}$ are

Similar Questions

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

If $\frac{2x^4-3x^2+4}{(x^2+1)(x^2+2)} = a + \frac{px+q}{x^2+1} + \frac{mx+n}{x^2+2}$,then $\frac{n}{q} =$

If $\frac{x^2}{2 x^4+7 x^2+6}=\frac{A x+B}{x^2+a}+\frac{C x+D}{a x^2+3}$, then find the value of $A+B+C-2 D$. (in $a$)

If $\frac{9x-7}{(x+3)(x^2+1)} = \frac{A}{x+3} + \frac{Bx+C}{x^2+1}$, where $A, B, C \in \mathbb{R}$, then $A+B+C = $

If $\frac{3x+2}{(x+1)(2x^2+3)} = \frac{A}{x+1} + \frac{Bx+C}{2x^2+3}$,then $A-B+C=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module