<s>[INST] The sum of the coefficients of $P(x)$ of 4th degree given that $P(x)\equiv x-1 \pmod{2x^2+4}\equiv x-1 \pmod{x^2-x-2}\equiv 141 \pmod{x+2}$ Let there be a polynomial $P(x)$ of 4th degree. The remainder of $P(x)$ divided by both $2x^2+4$ and $x^2-x-2$ is equal to $x-1$. So:
$$P(x)\equiv x-1 \pmod{2x^2+4}\equiv x-1 \pmod{x^2-x-2}$$
Given that $P(x)\equiv 141 \pmod{x+2}$, I'm asked to calculate the sum of the coefficients of $P(x)$. 
I don't have any idea how I should go about solving the question, any help would be appreciated. Thanks.
 [/INST] Hint 1:$$P\left( x \right) =a\left( 2{ x }^{ 2 }+4 \right) \left( { x }^{ 2 }-x-2 \right) +x-1\\ P(x)\equiv 141{ \quad \left( mod\quad x+2 \right) \Rightarrow  }P\left( -2 \right) =141\\ P\left( -2 \right) =48a-3=141\\ 48a=144\\ a=3\\ P\left( x \right) =3\left( 2{ x }^{ 2 }+4 \right) \left( { x }^{ 2 }-x-2 \right) +x-1\\ $$
Hint 2
$P(1)=$sum of the coefficients of $P(x)$.
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