I don't know the steps, pls help me

Thanks.

2. Note that $\displaystyle \frac{\alpha}{2}$ is a root of $\displaystyle 2x^2+x+1$. This means:
$\displaystyle 2\left(\frac{\alpha}{2}\right)^2+\frac{\alpha}{2}+ 1=0$
$\displaystyle \frac{\alpha^2}{2}+\frac{\alpha}{2}+1=0$
$\displaystyle \frac{\alpha^2}{2}=-\frac{\alpha}{2}-1$
$\displaystyle \alpha^2=-\alpha-2$

Use this, along with $\displaystyle \alpha^4=(\alpha^2)^2$, to get your result.

--Kevin C.

3. Originally Posted by TwistedOne151
Note that $\displaystyle \frac{\alpha}{2}$ is a root of $\displaystyle 2x^2+x+1$. This means:
$\displaystyle 2\left(\frac{\alpha}{2}\right)^2+\frac{\alpha}{2}+ 1=0$
$\displaystyle \frac{\alpha^2}{2}+\frac{\alpha}{2}+1=0$
$\displaystyle \frac{\alpha^2}{2}=-\frac{\alpha}{2}-1$
$\displaystyle \alpha^2=-\alpha-2$

Use this, along with $\displaystyle \alpha^4=(\alpha^2)^2$, to get your result.

--Kevin C.
I've tried that before,
$\displaystyle \begin{array}{l} \alpha ^2 = - 2 - \alpha \\ \alpha ^4 = \left( { - 2 - \alpha } \right)^2 = \alpha ^2 + 4\alpha + 4 \\ \alpha ^2 + 4\alpha + 4 = p\alpha + q \\ ????? \\ \end{array}$

4. Good job getting that far.
You still have an $\displaystyle \alpha^2$. Use again to get the answer

You still have an $\displaystyle \alpha^2$. Use again to get the answer
oh ya! stupid me, I didn't realize I can still sub $\displaystyle \alpha^2$ again!