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Question:
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Let $H(x)=3 x^4+6 x^3-2 x^2+1$ and $g(x)$ be a polynomial of degree one. If
$\begin{aligned}
& \frac{H(x)}{(x-1)(x+1)(x-2)}=f(x)+\frac{g(x)}{(x-1)(x+1)(x-2)} \\
& \text { then } H(-1)+2 H(2)-3 H(1)=
\end{aligned}$
Options:
$\begin{aligned}
& \frac{H(x)}{(x-1)(x+1)(x-2)}=f(x)+\frac{g(x)}{(x-1)(x+1)(x-2)} \\
& \text { then } H(-1)+2 H(2)-3 H(1)=
\end{aligned}$
Solution:
1716 Upvotes
Verified Answer
The correct answer is:
$g(-1)+2 g(2)-3 g(1)$
We have,
$\frac{H(x)}{(x-1)(x+1)(x-2)}=f(x)+\frac{g(x)}{(x-1)(x+1)(x-2)}$
$\begin{aligned} & \Rightarrow \quad H(x)=(x-1)(x+1)(x-2) f(x)+g(x) \\ & \Rightarrow \quad H(-1)=0+g(-1)=g(-1) \\ & H(2)=0+g(2)=g(2) \\ & H(1)=0+g(1)=g(1) \\ & \therefore H(-1)+2 H(2)-3 H(1)=g(-1)+2 g(2)-3 g(1) \\ & \end{aligned}$
$\frac{H(x)}{(x-1)(x+1)(x-2)}=f(x)+\frac{g(x)}{(x-1)(x+1)(x-2)}$
$\begin{aligned} & \Rightarrow \quad H(x)=(x-1)(x+1)(x-2) f(x)+g(x) \\ & \Rightarrow \quad H(-1)=0+g(-1)=g(-1) \\ & H(2)=0+g(2)=g(2) \\ & H(1)=0+g(1)=g(1) \\ & \therefore H(-1)+2 H(2)-3 H(1)=g(-1)+2 g(2)-3 g(1) \\ & \end{aligned}$
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