If frac{6 x^3+7 x^2-14 x+11}{6 x^3+x^2-10 x+3 | vedclass

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Question

(A) Perform polynomial division: $\frac{6 x^3+7 x^2-14 x+11}{6 x^3+x^2-10 x+3} = 1 + \frac{6 x^2-4 x+8}{6 x^3+x^2-10 x+3}$.Factor the denominator: $6

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$2$

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(A) Perform polynomial division: $\frac{6 x^3+7 x^2-14 x+11}{6 x^3+x^2-10 x+3} = 1 + \frac{6 x^2-4 x+8}{6 x^3+x^2-10 x+3}$.Factor the denominator: $6 x^3+x^2-10 x+3 = (x-1)(3 x-1)(2 x+3)$.Using partial fractions: $\frac{6 x^2-4 x+8}{(x-1)(3 x-1)(2 x+3)} = \frac{A}{x-1} + \frac{B}{3 x-1} + \frac{C}{2 x+3}$.Solving for constants: $A=1, B=-3, C=2$.Thus,the expression is $1 + \frac{1}{x-1} - \frac{3}{3 x-1} + \frac{2}{2 x+3}$.Comparing with $a+\frac{b}{x+p}+\frac{c}{q x+3}+\frac{d}{3 x+p}$,we identify $a=1, b=1, p=-1, c=2, q=2, d=-3$.Finally,$\frac{a+b}{p+q} = \frac{1+1}{-1+2} = \frac{2}{1} = 2$.

If $\frac{6 x^3+7 x^2-14 x+11}{6 x^3+x^2-10 x+3}=a+\frac{b}{x+p}+\frac{c}{q x+3}+\frac{d}{3 x+p}$ then $\frac{a+b}{p+q}=$

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