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Jika suku banyak f(x)= 2x^4 + ax^3 - 3x^2 + 5x +b dibagi (x^2 -1) menghasilkan sisa (6x + 5). 

diketahui S= (a^1009 mod 43) + (6b! x 4!  mod 40) . maka berapakah S^2017 mod 5.

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On 1/30/2017 at 1:35 PM, No name said:

Jika suku banyak $f(x)=2x^4+ax^3-3x^2+5x+b$ dibagi $(x^2-1)$ menghasilkan sisa $(6x+5)$ dan diketahui $S=(a^{1009}\bmod{43})+(6b!\times 4!\bmod{40})$, maka berapakah $S^{2017}\bmod 5$?


Dari persamaan umum suku banyak, $$f(x)=p(x)\cdot h(x)+s(x),$$ maka didapat persamaan \begin{align*}2x^4+ax^3-3x^2+5x+b&=(x^2-1)\cdot h(x)+(6x+5)\\&=(x-1)(x+1)\cdot h(x)+(6x+5).\end{align*}

Substitusi $x=1$ dan $x=-1$ didapat persamaan

$\begin{cases} 2+a-3+5+b=11\implies a+b=7\\2-a-3-5+b=-1\implies -a+b=5.\end{cases}$

Dengan menyelesaikan sistem persamaan di atas, diperoleh $a=1$ dan $b=6$.

\begin{align*}S&=(a^{1009}\bmod{43})+(6b!\times 4!\bmod{40})\\&=(1^{1009}\bmod{43})+(6\cdot 6!\times 4!\bmod{40})\\&=1+0\\&=1\end{align*}

Maka $S^{2017}\bmod{5}=1\bmod{5}=\boxed{1}.$

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