Solve the equation

$$
x^{4}+A x^{3}+29.25 x^{2}-40.28 x+B=0
$$

if we know that two of its roots are: $x_{1}=1.1, x_{2}=2.4$.
$x_{1}$ and $x_{2}$ satisfy the given equation, from which we get two linear equations for the unknown coefficients $A, B$. With $x_{1}$

$$
1.4641+1.331 A+35.3925-44.308+B=0,
$$

that is,

$$
1.331 A+B=7.4514
$$

similarly with $x_{2}$

$$
13.824 A+B=-104.9856
$$

From this system, $A=-9, B=19.4304$ and (1) becomes:

$$
x^{4}-9 x^{3}+29.25 x^{2}-40.28 x+19.4304=0
$$

On the other hand, the left side of (2) is divisible by the root factors $x-x_{1}=x-1.1$ and $x-x_{2}=x-2.4$, and thus by their product $x^{2}-3.5 x+2.64$ as well, because these root factors have no common divisor. The quotient will be the polynomial whose value is 0 at the other roots. From this, we get an equation for the other roots. By the known method

$$
\begin{aligned}
& \left(x^{4}-9 x^{3}+29.25 x^{2}-40.28 x+19.4304\right):\left(x^{2}-3.5 x+2.64\right)= \\
& \frac{-x^{4} \mp 3.5 x^{3} \pm 2.64 x^{2}}{-5.5 x^{3}+26.61 x^{2}} \\
& \frac{\mp 5.5 x^{3} \pm 19.25 x^{2} \mp 14.52 x}{7.36 x^{2}-25.76 x} \\
& -7.36 x^{2} \mp 25.76 x \pm 19.4304
\end{aligned}
$$

0

and from the resulting

$$
x^{2}-5.5 x+7.36=0
$$

we get $x_{3}=2.3, x_{4}=3.2$

Károly Szegö (Budapest, J. Apáczai Csere Practical Grammar School III. class)

Remarks. 1. We can more simply arrive at (3) instead of the division process if we take the coefficient of $x^{4}$ in the factorized form of the quartic equation to be 1, expand the parentheses:

$$
\begin{aligned}
& \left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right)=x^{4}-\left(x_{1}+x_{2}+x_{3}+x_{4}\right) x^{3}+ \\
& +\left(x_{1} x_{2}+x_{1} x_{3}+x_{1} x_{4}+x_{2} x_{3}+x_{2} x_{4}+x_{3} x_{4}\right) x^{2}-\left(x_{1} x_{2} x_{3}+x_{1} x_{2} x_{4}+\right. \\
& \left.+x_{1} x_{3} x_{4}+x_{2} x_{3} x_{4}\right) x+x_{1} x_{2} x_{3} x_{4}=0
\end{aligned}
$$

and note that the coefficient of $x^{3}$ is the negative of the sum of the roots, and the constant term is the product of the roots. Thus,

$$
1.1+2.4+x_{3}+x_{4}=9 \quad \text { and } \quad 1.1 \cdot 2.4 x_{3} x_{4}=19.4304
$$

from which

$$
x_{3}+x_{4}=5.5 \quad \text { and } \quad x_{3} x_{4}=7.36
$$

and (3) can be written from these.

Henrik Farkas (Eger, I. Dobó Grammar School IV. class)

2. Essentially, (1) can be solved without calculating the values of $A$ and $B$ by using the factorized form. Let the left side of (1) be denoted by $F(x)$, the quadratic polynomial that takes the value 0 at the known $x_{1}$ and $x_{2}$ by $G(x)$, that is,

$$
G(x)=\left(x-x_{1}\right)\left(x-x_{2}\right)=(x-1.1)(x-2.4)=x^{2}-3.5 x+2.64
$$

and the quadratic polynomial whose two 0-places are the remaining two roots, let it be the following:

$$
H(x)=x^{2}-s x+p
$$

Then the product $G(x) H(x)$ is identical to $F(x)$. Now

$$
\begin{gathered}
G(x) \cdot H(x) \equiv\left(x^{2}-3.5 x+2.64\right)\left(x^{2}-s x+p\right)=x^{4}-(3.5+s) x^{3}+ \\
+(p+3.5 s+2.64) x^{2}-(3.5 p+2.64 s) x+2.64 p
\end{gathered}
$$

It follows from the identity that the coefficients of the same powers of $x$ on the left and right are equal. Thus, from the coefficients of $x^{2}$ and $x$,

$$
\begin{aligned}
& p+3.5 s+2.64=29.25 \\
& 3.5 p+2.64 s \quad=40.28,
\end{aligned}
$$

we get a system of linear equations for the unknown coefficients of $H(x)$. From this, $p=7.36, s=5.5$, which means that $H(x)$ is identical to the left side of (3).

László Butor (Budapest, I. István Grammar School III. class)