Let's determine the real roots of the following equation:

$$
x^{4}-16 x^{3}+91 x^{2}-216 x+180=0
$$
I. Solution: First of all, it is obvious that the real roots can only be positive, because in the case of a negative $x$, every term on the left side is positive and thus cannot be 0. Let's examine whether the equation has a rational root?

In this case, the coefficient of the highest degree term is 1, so the rational roots can only be integers. Therefore, taking into account the aforementioned observation, the rational roots can only be chosen from the 180 positive divisors. By substituting the values $1,2,3,4,5,6,9,10$ etc. in sequence, we can verify that $2,3,5$ and 6 satisfy our equation, and since there cannot be more than four roots, we do not need to try further values; we have solved the problem.

Parlagh Gyula (Kecskemét, Katona J. g. III. o. t.)

II. Solution: The necessary and sufficient condition derived in problem 656 (K. M. L. XI. 2. sz., 1955. Oct. 57. p.) for a quartic equation to be reducible to a quadratic equation is satisfied in this case. Indeed,

$$
a^{3}-4 a b+8 c=-16^{3}+4 \cdot 16 \cdot 91-8 \cdot 216=16(-256+364-108)=0
$$

Using the transformation given there

$$
x=z-\frac{a}{4}=z+4
$$

we arrive at the equation

$$
z^{4}-5 z^{2}+4=0
$$

From this, $z_{1,2}= \pm 1, z_{3,4}= \pm 2$, and thus

$$
x_{1}=5, \quad x_{2}=3, \quad x_{3}=6, \quad x_{4}=2
$$

Grell Mihály (Bp. XVI., Corvin Mátyás g. III. o. t.)

III. Solution: Complete the first two terms on the left side of (1) to a perfect square:

$$
x^{4}-16 x^{3}+64 x^{2}+27 x^{2}-216 x+180=\left(x^{2}-8 x\right)^{2}+27\left(x^{2}-8 x\right)+180=0
$$

from which

$$
\left(x^{2}-8 x\right)=-15, \text { or }-12
$$

From

$$
x^{2}-8 x+15=0, \text { or } x^{2}-8 x+12=0
$$

the values obtained in the previous solutions are derived.

Tusnády Gábor (Sátoraljaújhely, Kossuth g. I. o. t.)

Note: The equations under (2) can also be obtained by trying to bring the left side of (1) into the form $\left(x^{2}+a x+b\right)^{2}-c$.