Xác định số hữu tỉ a sao cho x³ + ax² + 5x + 3 chia hết cho x² + 2x + 3.

Ta có:$\begin{array}{cccccccc}& x^3& +& {}^{}{}^{}{}^{}{}^{}a{x}^2& +& {}^{}{}^{}{}^{}{}^{}{}^{}5x& +& {}^{}{}^{}{}^{}3\\ \overline{}& x^3& +& {}^{}{}^{}{}^{}{}^{}2x^2& +& {}^{}{}^{}{}^{}{}^{}{}^{}3x& & \\ & & & \left(a-2\right)x^2& +& {}^{}{}^{}{}^{}{}^{}{}^{}2x& +& {}^{}{}^{}{}^{}3\\ & & \overline{}& \left(a-2\right)x^2& +& \left(2a-4\right)x& +& 3a-6\\ & & & & & \left(6-2\text{a}\right)x& +& 9-3a\end{array}\begin{array}{c}{x}^2+2x+3\\ x+\left(a-2\right)\\ \\ \\ \end{array}$

Để x³ + ax² + 5x + 3 ⋮ x² + 2x + 3

Vậy a = 3 thì x³ + ax² + 5x + 3 chia hết cho x² + 2x + 3.