(х-2)(х-3)(х+4)(х+5)=1320

Предыдущий
вопрос Следующий
вопрос

To solve this equation, we need to expand the left side of the equation by multiplying the factors:

(x-2)(x-3)(x+4)(x+5) = 1320
(x^2 - 3x - 2x + 6)(x^2 + 5x + 4x + 20) = 1320
(x^2 - 5x + 6)(x^2 + 9x + 20) = 1320
x^4 + 9x^3 + 20x^2 - 5x^3 - 45x^2 - 100x + 6x^2 + 54x + 120 = 1320
x^4 + 4x^3 - 19x^2 + 74x + 120 = 1320
x^4 + 4x^3 - 19x^2 + 74x - 120 = 0

Now, we need to find the roots of this polynomial equation. We can use numerical methods or factorize the equation using rational roots theorem. Let's factorize this equation using rational roots theorem:

The possible rational roots are factors of the constant term, 120, divided by factors of the leading coefficient, 1.
±1, ±2, ±3, ±4, ±5, ±6, ±8, ±10, ±12, ±15, ±20, ±24, ±30, ±40, ±60, ±120

By trying out these values, we find that x = 5 is a root of the equation:

(5)^4 + 4(5)^3 - 19(5)^2 + 74(5) - 120 = 0
625 + 500 - 475 + 370 - 120 = 0
125 + 70 + 370 - 120 = 0
195 + 370 - 120 = 0
565 - 120 = 0
445 = 0

Therefore, x = 5 is a root of the equation. To find the other roots, we can divide the polynomial by (x - 5) using long division or synthetic division to find the quadratic equation that makes the next factor.