Question

X^3 + 2x^2 - x - 2 =0 решить уравнение​

Answer 1

$x^3+2x^2-x-2=0;\\x^2(x+2)-(x+2)=0;\\(x+2)(x^2-1)=0;\\(x+2)(x+1)(x-1)=0;\\x_1=-2;x_2=-1;x_3=1.$

Ответ: x ∈ {-2; -1; 1}.

Answer 2

To solve the cubic equation $x^3 + 2x^2 - x - 2 = 0$, we can use various methods. One common method is to use numerical techniques or software, but I'll show you how to solve it using a simplified version of the method called the "Rational Root Theorem."

The Rational Root Theorem states that if a polynomial equation has any rational roots, they will be of the form $p/q$, where $p$ is a factor of the constant term, and $q$ is a factor of the leading coefficient.

For the given equation $x^3 + 2x^2 - x - 2 = 0$, the leading coefficient is 1 (coefficient of $x^3$), and the constant term is -2.

Now, we need to find the factors of -2 and 1.

The factors of -2 are: ±1, ±2 The factors of 1 are: ±1

Therefore, the possible rational roots of the equation are: ±1, ±2

Now, we can use these possible roots and apply the synthetic division or long division method to find the actual roots. Fortunately, in this case, one of the rational roots is apparent:

If we substitute $x = 1$ into the equation, we get: $1^3 + 2(1)^2 - 1 - 2 = 1 + 2 - 1 - 2 = 0$

So, $x = 1$ is one of the roots of the equation.

Now, to find the other roots, we need to divide the original cubic equation by $x - 1$ using long division or synthetic division:

Dividing $x^3 + 2x^2 - x - 2$ by $x - 1$:

$$\begin{array}{r|rrrr} 1 & 1 & 2 & -1 & -2 \\ \hline & & 1 & 3 & 2 \\ \end{array}$$

The result of the division is $x^2 + 3x + 2$.

Now, we have a quadratic equation $x^2 + 3x + 2 = 0$, which can be factored as $(x + 1)(x + 2) = 0$.

Setting each factor to zero and solving for $x$: $x + 1 = 0 \Rightarrow x = -1$ $x + 2 = 0 \Rightarrow x = -2$

So, the solutions to the cubic equation are: $x = 1$ $x = -1$ $x = -2$