Question

(х-1)(х+2)(х+3-х(х+2)(х+8)=0

Answer 1

Ответ:

Здраствуйте, ответ указен на фотографии.

Удачии!✌

Answer 2

To solve the equation $(x-1)(x+2)(x+3-x(x+2)(x+8)) = 0$, we can first simplify the equation by expanding and combining like terms, and then find the values of $x$ that make the equation equal to zero. Let's break it down step by step:

1. Expand the equation:

$(x-1)(x+2)(x+3-x(x+2)(x+8)) = 0$

2. Expand the second term inside the third parenthesis:

$(x-1)(x+2)(x+3-x(x^2 + 10x + 16)) = 0$

3. Distribute $x$ in the third term:

$(x-1)(x+2)(x+3 - x^3 - 10x^2 - 16x) = 0$

4. Distribute $x$ in the last term:

$(x-1)(x+2)(-x^3 - 10x^2 - 16x + 3) = 0$

5. Now, we have a polynomial equation. To find the values of $x$ that make it equal to zero, we can use the zero-product property. This means that either one or more of the factors on the left-hand side must equal zero.

So, we set each factor equal to zero and solve for $x$:

a) $x - 1 = 0$ Solving for $x$: $x = 1$

b) $x + 2 = 0$ Solving for $x$: $x = -2$

c) $-x^3 - 10x^2 - 16x + 3 = 0$

Solving this cubic equation can be more complex, and it may require numerical methods or software to find approximate solutions. We can't easily find the exact solutions algebraically.

So, the solutions to the equation are $x = 1$ and $x = -2$, and there may be additional solutions for the cubic equation, but they would require numerical methods to find.