$a(x)= x^4 - b^2x^3 - bx^2 -2x$ is divisible by $x-2$. Find $b$. $a(x)= x^4 - b^2x^3 - bx^2 -2x$ is divisible by $x-2$ for certain value $b$. Calculate all possible values $b$ can assume.
According to the factor theorem $x=2$ is a root when $a(x)=0$
$a(x)= x^4 - b^2x^3 - bx^2 -2x=0$
$a(2)= 2^4 - b^22^3 - b2^2 -2*2=0$
$8b^2 + 4b- 12=0$
$b^2 + 0.5b- 1.5=0$
After solving the quadratic equation, I get $b_1=-1,5$ and $b_2=1$
Is this right? Would you have solved it differently?

"$x^4 - b^2x^3 - bx^2 -2x$ divisible by $(x-2)$" is equivalent to 
"$x^3 - b^2x^2 - bx -2$ divisible by $(x-2)$".
\begin{align}
&\quad\; x^3 - b^2x^2 - bx -2 \\
&=(x^3-8) - b^2x^2 - bx +6 \\
&=(x^3-8) - (bx-2)(bx+3)
\end{align}
$(x^3-8)$ is divisible by $(x-2)$ apparently.
For $(bx-2)(bx+3)$ divisible by $(x-2)$, there are two possibilities:
$\qquad (bx-2)$ divisible by $(x-2)$, $\qquad b=1$
$\qquad (bx+3)$ divisible by $(x-2)$, $\qquad b=-\frac32$
