If $\;$ $2x^3 + ax^2 + bx - 2$ $\;$ when divided by $\;$ $2x - 3$ $\;$ and $\;$ $x + 3$ $\;$ leaves remainder $7$ and $-20$ respectively, find the values of $a$ and $b$.
Let $f \left(x\right) = 2x^3 + ax^2 + bx - 2$
Given: $\;$ $f \left(x\right)$ when divided by $2x - 3$ leaves remainder $7$
$\therefore \;$ By remainder theorem, $\;$ $f \left(\dfrac{3}{2}\right) = 7$
i.e. $\;$ $2 \times \left(\dfrac{3}{2}\right)^3 + a \times \left(\dfrac{3}{2}\right)^2 + b \times \dfrac{3}{2} - 2 = 7$
i.e. $\;$ $\dfrac{27}{4} + \dfrac{9a}{4} + \dfrac{3b}{2} = 9$
i.e. $\;$ $9a + 6b = 36 - 27 = 9$
i.e. $\;$ $3a + 2b = 3$ $\;\;\; \cdots \; (1)$
Given: $\;$ $f \left(x\right)$ when divided by $x + 3$ leaves remainder $-20$
$\therefore \;$ By remainder theorem, $\;$ $f \left(-3\right) = -20$
i.e. $\;$ $2 \times \left(-3\right)^3 + a \times \left(-3\right)^2 + b \times \left(-3\right) - 2 = -20$
i.e. $\;$ $9a - 3b = -18 + 54 = 36$
i.e. $\;$ $3a - b = 12$ $\;\;\; \cdots \; (2)$
Solving equations $(1)$ and $(2)$ simultaneously, we get,
$3b = -9$ $\implies$ $b = -3$
Substituting the value of $b$ in equation $(2)$, we get,
$3a + 3 = 12$
i.e. $\;$ $3a = 9$ $\implies$ $a = 3$
$\therefore \;$ $a = 3$, $\;$ $b = -3$