Solve a cubic polynomial (given one root is four times a second root)?

Question

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.

Answer 1

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.

Answer 2

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 $P(x)=10(x-\frac{1}{5})(x+\frac{1}{2})(x+2)$.

Answer 3

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 $\left\{ -\dfrac 12, -2, \dfrac 15 \right\}$