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

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(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^

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None of these

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(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:

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