15 (1143). Find the integer roots of the equation:

a) $2 x^{3}-3 x^{2}-11 x+6=0$; b) $x^{4}+4 x^{3}-9 x^{2}-16 x+20=0$.
Solution. a) According to the statement proved in the previous problem, the integer roots of the given equation are among the divisors of the number 6, i.e., among the numbers $\pm 1, \pm 2, \pm 3, \pm 6$. By trial, we verify that $x=-2$ and $x=3$ are roots of the given equation and that it has no other integer roots.

3 Remark. After finding one integer root of the equation by substitution (replacing $x$ with the divisors of the constant term in the left side of the equation), the other roots (not only integer ones) can be found by reducing the degree of the equation by dividing its left side by $x-x_{1}$, where $x_{1}$ is a root of the equation. Thus, dividing the polynomial $2 x^{3}-3 x^{2}-11 x+6$ by $x-3$, where 3 is a root of the given equation, we get $2 x^{3}-3 x^{2}-11 x+6=(x-3)\left(2 x^{2}-3 x-2\right)=0$, from which $x_{2}=-2, x_{3}=\frac{1}{2}$.

b) Among the divisors of the number $20( \pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20)$, by trial, we find the roots of the given equation: $x_{1}=1, x_{2}=2$, $x_{3}=-2, x_{4}=-5$

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As in solving equation a), after finding one root $x_{1}$, the degree of the given equation can be reduced by dividing its left side by $x-x_{1}$ (for example, by $x-1$) and then looking for integer roots among the divisors of the constant term of the resulting cubic equation.

Alternatively, the left side of the equation can be divided by $(x-x_{1})(x-x_{2})$ (for example, by $(x-2)(x+2)=x^{2}-4$) and finding the roots of the resulting quadratic equation ($x^{2}+4 x-5=0$).

Answer: a) $-2 ; 3 ;$ b) $1 ; 2 ;-2 ;-5$.