Let H(x) = 3x^4 + 6x^3 - 2x^2 + 1 and g(x) be | vedclass

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(C) Given the equation: $\frac{H(x)}{(x-1)(x+1)(x-2)} = f(x) + \frac{g(x)}{(x-1)(x+1)(x-2)}$.Multiplying both sides by $(x-1)(x+1)(x-2)$,we get: $H(x)

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$g(-1) + 2g(2) - 3g(1)$

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(C) Given the equation: $\frac{H(x)}{(x-1)(x+1)(x-2)} = f(x) + \frac{g(x)}{(x-1)(x+1)(x-2)}$.Multiplying both sides by $(x-1)(x+1)(x-2)$,we get: $H(x) = f(x)(x-1)(x+1)(x-2) + g(x)$.Since $H(x)$ is a polynomial of degree $4$ and the divisor is a polynomial of degree $3$,$f(x)$ must be a linear polynomial of the form $ax+b$.Substituting the values $x = -1, 2, 1$ into the equation $H(x) = f(x)(x-1)(x+1)(x-2) + g(x)$,we note that the term $f(x)(x-1)(x+1)(x-2)$ becomes $0$ at these points.Thus,$H(-1) = g(-1)$,$H(2) = g(2)$,and $H(1) = g(1)$.Substituting these into the expression $H(-1) + 2H(2) - 3H(1)$,we get $g(-1) + 2g(2) - 3g(1)$.

Let $H(x) = 3x^4 + 6x^3 - 2x^2 + 1$ and $g(x)$ be a linear polynomial. If $\frac{H(x)}{(x-1)(x+1)(x-2)} = f(x) + \frac{g(x)}{(x-1)(x+1)(x-2)}$,then $H(-1) + 2H(2) - 3H(1) =$

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