If frac{2x^3+1}{2x^2-x-6} = ax+b+frac{A}{px-2 | vedclass

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(A) Given the expression $\frac{2x^3+1}{2x^2-x-6} = ax+b+\frac{A}{px-2}+\frac{B}{2x+q}$.Performing polynomial long division of $2x^3+1$ by $2x^2-x-6$:

Vedclass · Accepted Answer

$23$

Vedclass · Answer

(A) Given the expression $\frac{2x^3+1}{2x^2-x-6} = ax+b+\frac{A}{px-2}+\frac{B}{2x+q}$.Performing polynomial long division of $2x^3+1$ by $2x^2-x-6$:$\frac{2x^3+1}{2x^2-x-6} = (x + \frac{1}{2}) + \frac{\frac{17}{2}x+4}{2x^2-x-6}$.Factoring the denominator: $2x^2-x-6 = (x-2)(2x+3)$.Using partial fractions for $\frac{\frac{17}{2}x+4}{(x-2)(2x+3)} = \frac{A}{x-2} + \frac{B}{2x+3}$.Solving for $A$ and $B$: $A = \frac{17}{7}$ and $B = \frac{23}{14}$.Comparing with the given form,we identify $a=1, b=\frac{1}{2}, p=1, q=3, A=\frac{17}{7}, B=\frac{23}{14}$.Calculating $51apB = 51 	imes 1 	imes 1 	imes \frac{23}{14} = \frac{1173}{14}$.Calculating $23bqA = 23 	imes \frac{1}{2} 	imes 3 	imes \frac{17}{7} = \frac{1173}{14}$.Thus,$51apB = 23bqA$.

If $\frac{2x^3+1}{2x^2-x-6} = ax+b+\frac{A}{px-2}+\frac{B}{2x+q}$,then $51apB=$ (in $bqA$)

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