<s>[INST] Solve a cubic polynomial (given one root is four times a second root)? So, I've been stuck on a question for a long time now:
"Solve the equation $10x^3 + 23x^2 + 5x - 2 = 0$ given that one root is four times a second root."
How would you go about solving this?
Any help would be greatly appreciated.
 [/INST] Solve $10x^3 + 23x^2 + 5x - 2 = (x-u)(x-4u)(x-v)$.
This can also be solved using synthetic division.
\begin{array}{r|rrrr}
    & 10 &    23 &             5 &             -2 \\
 u  &  0 &   10u &     23u+10u^2 & 5u+23u^2+10u^3 \\
\hline
   & 10 & 23+10u & 5+ 23u+ 10u^2 & \color{red}{-2+5u+23u^2+10u^3 = 0} \\
 4u &  0 &    40u &   92u+200u^2  \\
\hline
    & 10 & 23+50u & \color{red}{5+115u+210u^2=0}
\end{array}
We find $5+115u+210u^2=0 \implies u \in \{-\frac 12, -\frac{1}{21} \}$
We compute that 
$\left. -2+5u+23u^2+10u^3 \right|_{u=-\frac{1}{21}} \ne 0$
and
$\left. -2+5u+23u^2+10u^3 \right|_{u=-\frac 12} = 0$.
So $u = -\dfrac 12$ and $4u =-2$. Redoing the table above and continuing, we get
\begin{array}{r|rrrr}
           & 10 &  23 &  5 & -2 \\
 -\frac 12 &  0 &  -5 & -9 &  2 \\
\hline
           & 10 &  18 & -4 &  0 \\
        -2 &  0 & -20 &  4  \\
\hline
           & 10 &  -2 & 0 \\
  \frac 15 &  0 &   2 \\
\hline
           & 10 &   0
\end{array}
So the roots are $\left\{ -\dfrac 12, -2, \dfrac 15 \right\}$
</s>