The partial fraction of frac{6x^4 + 5x^3 + x^ | vedclass

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

Question

(D) First,perform polynomial long division of the numerator $6x^4 + 5x^3 + x^2 + 5x + 2$ by the denominator $6x^2 + 5x + 1$.Dividing $6x^4 + 5x^3 + x^

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) First,perform polynomial long division of the numerator $6x^4 + 5x^3 + x^2 + 5x + 2$ by the denominator $6x^2 + 5x + 1$.Dividing $6x^4 + 5x^3 + x^2 + 5x + 2$ by $6x^2 + 5x + 1$ gives the quotient $x^2$ and the remainder $4x + 2$.So,$\frac{6x^4 + 5x^3 + x^2 + 5x + 2}{6x^2 + 5x + 1} = x^2 + \frac{4x + 2}{6x^2 + 5x + 1}$.Now,factor the denominator: $6x^2 + 5x + 1 = (2x + 1)(3x + 1)$.Using partial fractions for $\frac{4x + 2}{(2x + 1)(3x + 1)} = \frac{A}{2x + 1} + \frac{B}{3x + 1}$.$4x + 2 = A(3x + 1) + B(2x + 1)$.Setting $x = -1/2$,we get $4(-1/2) + 2 = A(3(-1/2) + 1) \implies 0 = A(-1/2) \implies A = 0$.Setting $x = -1/3$,we get $4(-1/3) + 2 = B(2(-1/3) + 1) \implies 2/3 = B(1/3) \implies B = 2$.Thus,the expression is $x^2 + \frac{2}{3x + 1}$.Comparing this with the given options,none match the result.

The partial fraction of $\frac{6x^4 + 5x^3 + x^2 + 5x + 2}{1 + 5x + 6x^2}$ is:

Similar Questions

The Effective Atomic Number $(EAN)$ of $Co(CO)_4$ is $35$,hence it is less stable. It attains stability by:

The logical statement $(\sim(\sim p \vee q) \vee(p \wedge r)) \wedge(\sim q \wedge r)$ is equivalent to

If $t_n$ denotes the number of triangles formed with $n$ points in a plane,no three of which are collinear,and if $t_{n+1}-t_n=36$,then $n$ is equal to

Which of the following plants are used as green manure in crop fields and in sandy soils?

The value of the series $x \log _e a + \frac{x^3}{3!} (\log _e a)^3 + \frac{x^5}{5!} (\log _e a)^5 + \dots$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module