$a(x)= x^4 - b^2x^3 - bx^2 -2x$ 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?
 A: "$x^4 - b^2x^3 - bx^2 -2x$ 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 $(x-2)$, $\qquad b=-\frac32$
A: As for how I might have solved it differently, I would've literally just done long division and tried to figure out which $b$ might work out - or at least that's my gut instinct. But just the thought of that now seems like a chore at best, and your method proves superior and is totally valid.
I guess if I really had to nitpick, I wouldn't have divided by $8$ in solving the quadratic, since the quadratic formula allows for your quadratic to have a leading coefficient which is not $1$ - I would've just jumped right into using it. But that's a personal thing; if you prefer to do the division, more power to you!
I would be interested in seeing other methods of solving the problem though.

Anyhow, to see the validity of these answers, simply plug each in for $b$ and check if $a(2) = 0$. If so, then $x-2$ divides the expression for $a(x)$ generated by that substitution of $b$, verifying the calculation.
$$\begin{align}
b=-1.5= -\frac 32 &\implies a(x) = x^4 - \frac 9 4 x^3 + \frac 3 2 x^2 - 2x \\
&\implies a(2) = 16 - \frac 9 4 \cdot 8 + \frac 3 2 \cdot 4 - 2 \cdot 2 \\
&\implies a(2) = 0 \\
b = 1 &\implies a(x) = x^4 - x^3 - x^2 - 2x \\
&\implies a(2) = 16 - 8 - 4 - 4 \\
&\implies a(2) = 0
\end{align}$$ 
Both solutions are thus valid. That $a(2)=0$ gives you a quadratic in $b$ for your original expression also ensures you have at most two possible values for $b$, ensuring that these two are the only solutions.
