Question

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

Answer 1

Смотри........................

Answer 2

Let's simplify the equation step by step:

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

Expanding both sets of parentheses:

(x^2 + 2x - x - 2)(x + 3) - x(x + 2)(x + 8) = 0

Simplifying further:

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

Expanding again:

(x^3 + 3x^2 + x^2 + 3x - 2x - 6) - (x^3 + 10x^2 + 16x) = 0

Combining like terms:

x^3 + 4x^2 + x^2 - 2x - 6 - x^3 - 10x^2 - 16x = 0

Simplifying further:

(x^2 - 17x - 6) = 0

Now, we have a quadratic equation. We can solve it by factoring, completing the square, or using the quadratic formula. Factoring may not be straightforward in this case, so let's use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

For our equation (x^2 - 17x - 6) = 0, the coefficients are: a = 1 b = -17 c = -6

Using the quadratic formula:

x = (-(-17) ± √((-17)^2 - 4(1)(-6))) / (2(1))

Simplifying:

x = (17 ± √(289 + 24)) / 2

x = (17 ± √(313)) / 2

Therefore, the solutions to the equation are:

x = (17 + √313) / 2

and

x = (17 - √313) / 2