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

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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:$H(x)=(x-1)(x+1)(x-2)f(x)+g(x)$Now, we evaluate $H(x)$ at $x=-1, 2, 1$:For $x=-1$: $H(-1)=( -1-1)( -1+1)( -1-2)f(-1)+g(-1) = 0+g(-1) = g(-1)$For $x=2$: $H(2)=(2-1)(2+1)(2-2)f(2)+g(2) = 0+g(2) = g(2)$For $x=1$: $H(1)=(1-1)(1+1)(1-2)f(1)+g(1) = 0+g(1) = g(1)$Substituting these into the expression $H(-1)+2H(2)-3H(1)$:$H(-1)+2H(2)-3H(1) = g(-1)+2g(2)-3g(1)$

Let $H(x)=3x^4+6x^3-2x^2+1$ and $g(x)$ be a polynomial of degree one. 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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