Question

X^{4} -2 x^{3} -2x+3=

Answer 1

$x^4-2x^3-2x+3=(x-1)(x^3-x^2-x-1)$

Заметим, что при x=1

$x^4-2x^3-2x+3=1-2-2+3=0$

,значит (x-1) - множитель. Делим столбиком (картинка), получаем разложение.

Answer 2

To solve the equation $X^{4} - 2x^{3} - 2x + 3 = 0$, we need to find the values of $x$ that satisfy the equation. Unfortunately, there is no general algebraic method to find the exact solutions for quartic equations (equations with a degree of 4). However, we can use numerical methods or graphing to approximate the solutions.

One common numerical method is the Newton-Raphson method. Let's try to find one of the solutions of the equation using this method. We will start with an initial guess and iteratively improve it until we get a good approximation.

Let's use $x_0 = 1$ as the initial guess:

1. Calculate $f(x_0) = x_0^4 - 2x_0^3 - 2x_0 + 3$.
2. Calculate the derivative of $f(x)$ with respect to $x$, denoted as $f'(x)$: $f'(x) = 4x^3 - 6x^2 - 2$.
3. Update the guess using the Newton-Raphson formula: $x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}$.
4. Repeat steps 1 to 3 until the value of $x$ stops changing significantly.

Let's perform the calculations:

1. $f(1) = 1^4 - 2(1)^3 - 2(1) + 3 = 0$.
2. $f'(1) = 4(1)^3 - 6(1)^2 - 2 = -4$.
3. $x_1 = 1 - \frac{0}{-4} = 1$.

Since the value of $x$ has not changed significantly, it appears that the solution is $x = 1$ (with an approximation of $x \approx 1$).

However, keep in mind that there may be other real or complex solutions to the equation. To find them all, you'll need to use more advanced numerical methods or graph the equation to visualize the roots.