146(1174). A motorcyclist left point $A$ for point $B$ and at the same time a cyclist left point $B$ for point $A$. The motorcyclist arrived at point $B$ 2 hours after meeting the cyclist, while the cyclist arrived at point $A$ 4.5 hours after meeting the motorcyclist. How many hours were the motorcyclist and the cyclist on the road
Solution. Let the motorcyclist and the cyclist have been on the road for $t$ hours before meeting. If $v_{1}$ kilometers per hour is the speed of the motorcyclist, and $v_{2}$ kilometers per hour is the speed of the cyclist, then we have the system of equations:

$$
\left\{\begin{array}{l}
v_{1}=\frac{v_{2} t}{2} \\
v_{2}=\frac{v_{1} t}{4.5}
\end{array}\right.
$$

from which $t^{2}=9, t=3$.

Then the time of the motorcyclist's movement is 5 hours $(3+2=5)$, and the time of the cyclist's movement is 7.5 hours $(3+4.5=7.5)$.

Answer: 5 hours; 7.5 hours.

## § 9. PROBLEMS OF INCREASED DIFFICULTY IN THE ALGEBRA COURSE FOR THE 9TH GRADE

$\triangle 1(1129)^{\prime}$. Find the roots of the polynomial $2 x^{5}+x^{4}-10 x^{3}-$ $-5 x^{2}+8 x+4$. (A root of a polynomial in one variable is a value of the variable for which the value of the polynomial is zero.)

Solution. Factorize the given polynomial:

$$
2 x^{5}+x^{4}-10 x^{3}-5 x^{2}+8 x+4=
$$

$$
=x^{4}(2 x+1)-5 x^{2}(2 x+1)+4(2 x+1)=(2 x+1)\left(x^{4}-5 x^{2}+4\right) .
$$

Set each factor equal to zero. We have $2 x+1=0$, from which $x_{1}=-\frac{1}{2}$, or $x^{4}-5 x^{2}+4=0$, from which $x^{2}=1, x^{2}=4$, i.e., $x_{2}=1, x_{3}=-1, x_{4}=2, x_{5}=-2$.

Answer: $-\frac{1}{2} ;-1 ;-2 ; 1 ; 2$.