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.
 A: If $a$, $4a$ are the two roots in question and $b$ is the third, then you know by Vieta that $a+4a+b=-\frac{23}{10}$ and $ab+4ab+4a^2=\frac5{10}$ and $4a^2b=\frac2{10}$.
Alternatively (inspired by Taladris' approach, but without the need to know how to solve quadratics! As a penalty, you need to work with a few bigger numbers, though)
With $P(x)=10x^3+23x^2+5x-2$, let $a$ be the root such that $4a$ is also a root. Then $a$ is a root of $P(x)$ and of $P(4x)=640x^3+368x^2+20x-2$.
But $a$ is also a root of any linear combinations of these polynomials (adjusted to eliminate the highest powers of $x$), especially of (essentially we are computing the $\gcd$)
$$ Q(x):=\frac16(P(4x)-64P(x))=-184x^2-50x+21$$
then of
$$R(x):=92P(x)+5xQ(x) = 1866x^2+565x-184$$
and finally of $$933Q(x)+92R(x)=5530x+2665,$$
hence we must have $a=\frac12$. The rest is easy.
A: Another (more brutal) method: 
Let $x$ a root of $P(x)=10x^3+23x^2+5x-2$ such that $4x$ is also a root of $P$. Then, substituting $x$ by $4x$ in $P(x)$ implies that $x$ is a root of $640x^3+368x^2+20x-2$. Substracting $P$, we obtain that $x$ is a root of $630x^3+345x^2+15x=15x(42x^2+23x+1)$. Since $x$ is not zero, $x$ is a root of $42x^2+23x+1$. The discriminant is $19^2$, so $x$ is $\frac{-1}{2}$ or $\frac{-1}{21}$ (and $4x$ is $-2$ or $\frac{-4}{21}$). We check easily that $-2$ is a root of $P$, so we can factor $P$ as $P(x)=10(x-\frac{1}{5})(x+\frac{1}{2})(x+2)$.
A: 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 $\left\{ -\dfrac 12, -2, \dfrac 15 \right\}$
