<s>[INST] $2x^2 + 3x +4$is not divisible by $5$ I tried by $x^2 \equiv 0, 1, 4 \pmod 5$ but how can I deal with $3x$? 
I feel this method does not work here.
 [/INST] The method does work. How did you deal with $2x^2$ anyway? You know that $x^2 \equiv 0, 1, 4 \pmod 5$. So you just double those, like this: $0, 2, 8$, rewrite as $0, 2, 3$.
Likewise with $3x$, you just have to triple $0, 1, 2, 3, 4$ to get $0, 3, 6, 9, 12$ which we rewrite as $0, 3, 1, 4, 2$.
And lastly $4 \equiv 4 \pmod 5$, obviously.
So the possibilities are:


*

*If $x \equiv 0 \pmod 5$, then $2x^2 + 3x + 4 \equiv 0 + 0 + 4 = 4$.

*If $x \equiv 1 \pmod 5$, then $2x^2 + 3x + 4 \equiv 2 + 3 + 4 = 9 \equiv 4$.

*If $x \equiv 2 \pmod 5$, then $2x^2 + 3x + 4 \equiv 8 + 6 + 4 = 18 \equiv 3$.

*If $x \equiv 3 \pmod 5$, then $2x^2 + 3x + 4 \equiv 18 + 9 + 4 = 31 \equiv 1$.

*If $x \equiv 4 \pmod 5$, then $2x^2 + 3x + 4 \equiv 32 + 12 + 4 = 48 \equiv 3$.


As we failed to find a $0$ this way, we conclude that $2x^2 + 3x + 4$ is never divisible by $5$. Not as elegant as completing the square, but easy enough for a child to do.
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