I1.1 If $A$ is the sum of the squares of the roots of $x^{4}+6 x^{3}+12 x^{2}+9 x+2$, find the value of $A$.
Let $f(x)=x^{4}+6 x^{3}+12 x^{2}+9 x+2$
By division,
$$
\begin{array}{l}
f(-1)=1-6+12-9+2=0 \\
f(-2)=16-48+48-18+2=0
\end{array}
$$
$\therefore$ By factor theorem, $\left(x^{2}+3 x+2\right)$ is a factor of $f(x)$. $x^{4}+6 x^{3}+12 x^{2}+9 x+2=(x+1)(x+2)\left(x^{2}+3 x+1\right)$
The roots are $-1,-2$ and $\alpha, \beta$; where $\alpha+\beta=-3, \alpha \beta=1$
$$
\begin{aligned}
A & =(-1)^{2}+(-2)^{2}+\alpha^{2}+\beta^{2}=5+(\alpha+\beta)^{2}-2 \alpha \beta \\
& =5+9-2 \\
A & =12
\end{aligned}
$$

Method 2 By the change of subject, let $y=x^{2}$, then the equation becomes
$$
\begin{array}{l}
x^{4}+12 x^{2}+2=-x\left(6 x^{2}+9\right) \Rightarrow y^{2}+12 y+2=\mp \sqrt{y}(6 y+9) \\
\left(y^{2}+12 y+2\right)^{2}-y(6 y+9)^{2}=0
\end{array}
$$

Coefficient of $y^{4}=1$, coefficient of $y^{2}=24-36=-12$
If $\alpha, \beta, \delta$ and $\gamma$ are the roots of $x$, then $\alpha^{2}, \beta^{2}, \delta^{2}$ and $\gamma^{2}$ are the roots of $y$
$$
\alpha^{2}+\beta^{2}+\delta^{2}+\gamma^{2}=-\frac{\text { coefficient of } y^{3}}{\text { coefficient of } y^{4}}=12
$$

Method 3
Let $\alpha, \beta, \delta$ and $\gamma$ are the roots of $x$, then by the relation between roots and coefficients,
$$
\begin{array}{l} 
\alpha+\beta+\delta+\gamma=-6 \cdots \cdots(1) \\
\alpha \beta+\alpha \delta+\alpha \gamma+\beta \delta+\beta \gamma+\delta \gamma=12 \cdots \cdots(2) \\
\alpha^{2}+\beta^{2}+\delta^{2}+\gamma^{2}=(\alpha+\beta+\delta+\gamma)^{2}-2(\alpha \beta+\alpha \delta+\alpha \gamma+\beta \delta+\beta \gamma+\delta \gamma) \\
=(-6)^{2}-2(12)=12
\end{array}
$$