By performing the square root operation, determine the value of $x$ for which $x^{4}+6 x^{3}+11 x^{2}+3 x+31$ becomes a perfect square.
| $\sqrt{\left(x^{4}+6 x^{3}+11 x^{2}\right.}$  $-x^{4}$ | $+3 x+31)$ | $=x^{2}+3 x+1$ |
| :---: | :---: | :---: |
| $6 x^{3}+11 x^{2}$  $-6 x^{3} \pm 9 x^{2}$ |  | $:\left(2 x^{2}+3 x\right) 3 x$ |
|  | $2 x^{2}+3 x+31$  $-2 x^{2} \pm 6 x \pm 1$ | $:\left(2 x^{2}+6 x+1\right) \cdot 1$ |

The expression under the square root will be a perfect square for the value of $x$ that makes the remainder $-3 x+30=0$, which is when $x=10$. In this case, the value of the polynomial is: $10000+6000+1100+30+31=17161$.

Indeed:

$$
17161=131^{2}
$$

Tibor Banki (Szent László g. V. o. Bp. X.)