# Where is I'm wrong on getting to the remainder?

Let $f(x)=11x^3+15x^2+9x-2$. I want to calculate the remainder of $f(97)/11$. I know that $97\pmod{11}=9$. But it also doesn't help with calculation of remainder with the help of modular arithmetic. I also cant get to the answer with $f(9)$ with mod of $11$.

What should I do?

Since $11\equiv0\pmod{11}$, that $15\equiv4\pmod{11}$, and that $9\equiv-2\pmod{11}$, then for any integer $x$, $f(x)\equiv4x^2-2x-2\pmod{11}$. Since $97\equiv-2\pmod{11}$,\begin{align}f(97)&\equiv f(-2)\pmod{11}\\&\equiv4\times(-2)^2-2\times(-2)-2\pmod{11}\\&\equiv7\pmod{11}.\end{align}
Well, \begin{align}11(97)^3+15(97)^2+9(97)-2 & =4(-2)^2-2(-2)-2 \\ &= 5+2\pmod{11} \\ &=7\pmod{11}\end{align} because $97=-2\pmod{11}$.