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

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Question

(A) Divide the numerator by the denominator: $\frac{6x^4 + 5x^3 + x^2 + 5x + 2}{6x^2 + 5x + 1} = x^2 + \frac{5x + 2}{6x^2 + 5x + 1}$.Factor the denomi

Vedclass · Accepted Answer

$x^2 + \frac{1}{1 + 2x} + \frac{1}{1 + 3x}$

Vedclass · Answer

(A) Divide the numerator by the denominator: $\frac{6x^4 + 5x^3 + x^2 + 5x + 2}{6x^2 + 5x + 1} = x^2 + \frac{5x + 2}{6x^2 + 5x + 1}$.Factor the denominator: $6x^2 + 5x + 1 = (2x + 1)(3x + 1)$.Now,express $\frac{5x + 2}{(2x + 1)(3x + 1)}$ as partial fractions: $\frac{5x + 2}{(2x + 1)(3x + 1)} = \frac{A}{2x + 1} + \frac{B}{3x + 1}$.$5x + 2 = A(3x + 1) + B(2x + 1)$.For $x = -\frac{1}{2}$,$5(-\frac{1}{2}) + 2 = A(3(-\frac{1}{2}) + 1) \implies -\frac{1}{2} = A(-\frac{1}{2}) \implies A = 1$.For $x = -\frac{1}{3}$,$5(-\frac{1}{3}) + 2 = B(2(-\frac{1}{3}) + 1) \implies \frac{1}{3} = B(\frac{1}{3}) \implies B = 1$.Thus,the expression is $x^2 + \frac{1}{2x + 1} + \frac{1}{3x + 1}$.

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

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