Question

(x-2)(x+1)(x+4)(x+7)=19​

Answer 1

Ответ:

(x-2)(x+1)=x^2-x-2

(x+4)(x+7)=x^2+11x+28

(x^2-x-2)(^2+11x+28)=x^4+11x^2+28x^2-x^3-11x^3-28x-2x^2-22x-56=19

x^4+37x^2-12x^3-50x=77

Answer 2

To solve the equation $(x-2)(x+1)(x+4)(x+7) = 19$, we can start by expanding the left side of the equation:

$(x-2)(x+1)(x+4)(x+7) = 19$ $(x^2 - x - 2)(x^2 + 5x + 4) = 19$

Expanding further:

$(x^2 - x - 2)(x^2 + 5x + 4) = 19$ $x^4 + 5x^3 + 4x^2 - x^3 - 5x^2 - 4x - 2x^2 - 10x - 8 = 19$ $x^4 + 4x^3 - 3x^2 - 14x - 8 = 19$ $x^4 + 4x^3 - 3x^2 - 14x - 27 = 0$

Now, we have a quartic equation. Unfortunately, there is no general algebraic formula to solve quartic equations like there is for quadratic equations. However, we can use numerical methods or approximation techniques to find the solutions.

One common numerical method is to use numerical solvers or graphing calculators that can find the roots of the equation. Another approach is to approximate the solutions by using techniques like the Newton-Raphson method or the bisection method.

Keep in mind that finding the exact solutions may not always be possible or practical for higher degree equations.