Ans: (a) $5x+2$ (b) (i) $a=2$, $b=-1$ (ii) $-3$, $1$ or $\dfrac{-2}{5}$
- By long division, we have
$\require{enclose}\begin{array}{rl}
& \ \ 5x + 2 \\
x^2 + 2x -3 & \enclose{longdiv}{5x^3 + 12x^2 -9x -7} \\
& \ \ \underline{5x^3 + 10x^2 -15x \phantom{00}\ \ } \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2x^2 + 6x – 7 \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline{2x^2 +6x – 7\ } \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +2x – 1 \\
\end{array}$Therefore, the quotient is $5x+2$.
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- $a=2$, $b=-1$.
- By the result of (a), we have $g(x)=(x^2+2x-3)(5x+2)$. Then
$\begin{array}{rcl}
g(x) & = & 0 \\
(x^2+2x-3)(5x+2) & = & 0 \\
(x+3)(x-1)(5x+2) & = & 0
\end{array}$$\therefore x=-3$ or $x=1$ or $x=\dfrac{-2}{5}$.
