Expression Involving Roots of a Polynomial

Question

I'm working on part of a problem, which reads as follows:

Let $\theta$ be a root of the equation $x^{3}+x^{2}+x+2=0$. Express $(\theta^{2}+\theta+1)(\theta^{2}+\theta)$ and $(\theta-1)^{-1}$ in the form $a\theta^{2}+b\theta+c$, where $a,b,c\in\mathbb{Q}$.

My attempt at a solution: I let $f(x)=x^{3}+x^{2}+x+2$; then I was able to show that $f$ is irreducible over $\mathbb{Q}$. Using the division algorithm, we have $f(x)=(x-1)(x^{2}+2x+3)+5$. Since $f(\theta)=0$ by assumption, we evaluate at $x=\theta$ and get $$\begin{align}0=f(\theta)=(\theta-1)(\theta^{2}+2\theta+3)+5&\Longrightarrow \frac{-5}{\theta-1}=\theta^{2}+2\theta+3\\&\Longrightarrow(\theta-1)^{-1}=-\frac{1}{5}\theta^{2}-\frac{2}{5}\theta-\frac{3}{5}.\end{align}$$ My trouble is with finding $(\theta^{2}+\theta+1)(\theta^{2}+\theta)$. I think this is what you do, but I'm not sure: since $0=\theta^{3}+\theta^{2}+\theta+2$, we have $\theta^{3}=-(\theta^{2}+\theta+2)$. Hence, using the fact that $\theta^{4}=\theta\cdot\theta^{3}$, we have $$\begin{align}(\theta^{2}+\theta+1)(\theta^{2}+\theta)&=\theta^{4}+2\theta^{3}+2\theta^{2}+\theta\\ &=-\theta(\theta^{2}+\theta+2)-2(\theta^{2}+\theta+2)+2\theta^{2}+\theta\\ &=-\theta^{3}-\theta^{2}-3\theta-4\\ &=(\theta^{2}+\theta+2)-\theta^{2}-3\theta-4\\ &=-2\theta-2. \end{align}$$

Does my solution look correct? Thank you in advance for any feedback.

Answer 1

Yes, it is correct. But you could do it a little quicker:

$$(\theta^{2}+\theta+1)(\theta^{2}+\theta) = \underbrace{(\theta^{3}+\theta^2+\theta)}_{=-2}(\theta+1) =-2\theta-2$$

Answer 2

The systematic way for the first item is to follow what you did for $\theta-1$: compute the remainder of $(x^2+x+1)(x^2+x)$ when divided by $x^3+x^2+x+2$: $$ (x^2+x+1)(x^2+x) = (x + 1)(x^3+x^2+x+2)+(-2 x - 2) $$ Therefore, $$ (\theta^{2}+\theta+1)(\theta^{2}+\theta) = -2\theta-2 $$