2019-I-11

Since the remainder is $50$ when $p(x)$ is divided by $x – 1$, then by remainder theorem, we have

$\begin{array}{rcl}
p(1) & = & 50 \\
[2(1)^2 + 9(1) + 14][a(1) + b] & = & 50 \\
25( a + b) & = & 50 \\
a + b & = & 2 \ \ldots \unicode{x2460}
\end{array}$

Since the remainder is $-52$ when $p(x)$ is divided by $x + 2$, then by the remainder theorem, we have

$\begin{array}{rcl}
h(-2) & = & -52 \\
[2(-2)^2 + 9(-2) + 14][a(-2) + b] & = & -52 \\
4(-2a + b) & = & -52 \\
2a – b & = & 13 \ \ldots \unicode{x2461}
\end{array}$

$\unicode{x2460} + \unicode{x2461}$, we have

$\begin{array}{rcl}
3a & = & 15 \\
a & = & 5
\end{array}$

Sub. $a + 5$ into $\unicode{x2460}$, we have

$\begin{array}{rcl}
5 + b & = & 2 \\
b & = -3
\end{array}$

Therefore, the required quotient is $5x – 3$.

$\begin{array}{rcl}
p(x) & = & 0 \\
(2x^2 + 9x + 14)(5x – 3) & = & 0 \\
\end{array}$

Therefore, $x = \dfrac{3}{5}$ or $2x^2 + 9x + 14 = 0$.

The discriminant of $2x^2 + 9x + 14 = 0$

$\begin{array}{cl}
= & 9^2 – 4 \times 2 \times 14 \\
= & -31 \\
< & 0 \end{array}$

Therefore, $2x^2 + 9x + 14 = 0$ has not real roots.

Hence, the equation $p(x) = 0$ has only one rational root $\dfrac{3}{5}$.