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

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(C) Perform polynomial long division of $2x^4-x^3+3x^2-x+4$ by $x^2-3x+2$: The quotient is $f(x) = 2x^2+5x+14$. The remainder is $31x-28$. Thus,$\frac

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$f(x) = 2x^2+5x+14, A+B = 31$

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(C) Perform polynomial long division of $2x^4-x^3+3x^2-x+4$ by $x^2-3x+2$: The quotient is $f(x) = 2x^2+5x+14$. The remainder is $31x-28$. Thus,$\frac{2x^4-x^3+3x^2-x+4}{x^2-3x+2} = 2x^2+5x+14 + \frac{31x-28}{(x-1)(x-2)}$. Using partial fractions: $\frac{31x-28}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}$. $31x-28 = A(x-2) + B(x-1)$. For $x=1$: $31(1)-28 = A(1-2)$ $\Rightarrow 3 = -A$ $\Rightarrow A = -3$. For $x=2$: $31(2)-28 = B(2-1)$ $\Rightarrow 62-28 = B$ $\Rightarrow B = 34$. $A+B = -3+34 = 31$. Therefore,$f(x) = 2x^2+5x+14$ and $A+B = 31$.

If $\frac{2x^4-x^3+3x^2-x+4}{x^2-3x+2} = f(x) + \frac{A}{x-1} + \frac{B}{x-2}$,then:

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