Solve the following equation:

$$
6 x^{5}-x^{4}-43 x^{3}+43 x^{2}+x-6=0 .
$$
I. solution. Pair the symmetrically placed terms on the left side, whose coefficients are the negatives of each other. Then $x-1$ can be factored out:

\[
\begin{aligned}
& 6\left(x^{5}-1\right)-x\left(x^{3}-1\right)-43 x^{2}(x-1)=(x-1)\left[6 x^{4}+6 x^{3}+6 x^{2}+6 x+6-\right. \\
& \left.\quad-x\left(x^{2}+x+1\right)-43 x^{2}\right]=(x-1)\left(6 x^{4}+5 x^{3}-38 x^{2}+5 x+6\right)
\end{aligned}
\]

Therefore, the equation is satisfied for $x_{1}=1$, and the other roots are the values of $x$ for which

\[
6 x^{4}+5 x^{3}-38 x^{2}+5 x+6=0
\]

This is a symmetric equation, and according to the remark in problem 1113, the reciprocal of every root is also a root. Therefore, it is useful to divide the equation by $x^{2}$ and introduce the sum of $x$ and its reciprocal as a new variable. This way, we obtain a quadratic equation in the new variable:

\[
6 x^{2}+5 x-38+\frac{5}{x}+\frac{6}{x^{2}}=6\left(x^{2}+\frac{1}{x^{2}}\right)+5\left(x+\frac{1}{x}\right)-38=0
\]

If we let

\[
x+\frac{1}{x}=y, \quad \text{then} \quad x^{2}+\frac{1}{x^{2}}=\left(x+\frac{1}{x}\right)^{2}-2=y^{2}-2
\]

thus

\[
6\left(y^{2}-2\right)+5 y-38=6 y^{2}+5 y-50=0
\]

From this, $y_{1}=-10 / 3$ and $y_{2}=5 / 2$, so from the equations

\[
x+\frac{1}{x}=-\frac{10}{3} \quad \text{and} \quad x+\frac{1}{x}=\frac{5}{2}
\]

we get:

\[
x_{2}=-3, \quad x_{3}=-\frac{1}{3}, \quad \text{and} \quad x_{4}=2, \quad x_{5}=\frac{1}{2}
\]

The members of the two pairs of roots are indeed reciprocals of each other. Substitution shows that $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ all satisfy (1).

Since a fifth-degree equation can have at most 5 distinct roots, the solution is complete.

Ámon Magdolna (Győr, Zrínyi I. lg. III. o. t.)

II. solution. It is sufficient to deal with equation (2) from the first solution. Since all the coefficients of this equation are integers, we can find its rational roots, if any, by a finite number of substitution trials. It is well known that a rational root of an equation with integer coefficients, in its simplest form, must have a numerator that is a divisor of the constant term and a denominator that is a divisor of the coefficient of the highest power of the variable. In (2), both this coefficient and the constant term are 6, so only $1, 2, 3$, and 6, as well as their negatives, can be the numerator and denominator, respectively. Therefore, (2) can only be satisfied by the following rational numbers:

\[
\pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{1}{6}
\]

Among these, the value of the left side is 0 for $x_{2}=2$ and $x_{3}=-3$, so the remaining two roots are their reciprocals: $x_{4}=1 / 2$, $x_{5}=-1 / 3$.

Farkas Zoltán (Hódmezővásárhely, Bethlen G. g. IV. o. t.)

Remark. If we do not know the property of reciprocal equations used here, we can find the remaining two roots either by further trial and error or by determining them as the roots of the equation that results from dividing (2) by the product of the factors corresponding to the roots we have found, $(x-2)(x+3)$.[^0]

[^0]:    ${ }^{1}$ See K. M. L. 24 (1962/4) 152. p.

    ${ }^{2}$ Mathematics, gimn. IV. o. textbook, 194. p.