If frac{x^4-6 x^3+9 x^2+5 x-20}{x^2-x-2}=f(x) | vedclass

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(D) Given: $\frac{x^4-6 x^3+9 x^2+5 x-20}{x^2-x-2}=f(x)+\frac{a}{x+p}+\frac{b}{x+q}$ ... $(i)$Performing polynomial long division of the numerator by

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$f(4)$

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(D) Given: $\frac{x^4-6 x^3+9 x^2+5 x-20}{x^2-x-2}=f(x)+\frac{a}{x+p}+\frac{b}{x+q}$ ... $(i)$Performing polynomial long division of the numerator by the denominator:$\frac{x^4-6 x^3+9 x^2+5 x-20}{x^2-x-2} = (x^2-5x+6) + \frac{x-8}{x^2-x-2}$Since $x^2-x-2 = (x+1)(x-2)$,we use partial fractions for $\frac{x-8}{(x+1)(x-2)}$:$\frac{x-8}{(x+1)(x-2)} = \frac{A}{x+1} + \frac{B}{x-2}$$x-8 = A(x-2) + B(x+1)$For $x=-1$,$-9 = A(-3) \Rightarrow A=3$.For $x=2$,$-6 = B(3) \Rightarrow B=-2$.So,$\frac{x^4-6 x^3+9 x^2+5 x-20}{x^2-x-2} = (x^2-5x+6) + \frac{3}{x+1} - \frac{2}{x-2}$ ... (ii)Comparing $(i)$ and (ii),$f(x) = x^2-5x+6$,$a=3$,$b=-2$.Then $2(a+b) = 2(3-2) = 2(1) = 2$.Evaluating the options:$f(7) = 49-35+6 = 20$$f(6) = 36-30+6 = 12$$f(5) = 25-25+6 = 6$$f(4) = 16-20+6 = 2$Thus,$2(a+b) = f(4)$.

If $\frac{x^4-6 x^3+9 x^2+5 x-20}{x^2-x-2}=f(x)+\frac{a}{x+p}+\frac{b}{x+q}$,then $2(a+b)=$

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