Question

(x+2)(x-1)(x-3)(x-6)=16​

Answer 1

$(x + 2)(x - 1)(x - 3)(x - 6) = 16 \\ ( x + 2)(x - 6)(x - 1)(x - 3) = 16 \\ ( {x}^{2} - 4x - 12)( {x}^{2} - 4x + 3) = 16 \\ (t - 12)(t + 3) = 16 \\ {t}^{2} - 9t - 36 = 16 \\ {t}^{2} - 9t - 52 = 0 \\ d = 81 + 4 \times 52= 81 + 208 = 289 = {17}^{2} \\ t1 = \frac{9 + 17}{2} = \frac{26}{2} = 13 \\ t2 = \frac{9 - 17}{2} = - \frac{8}{2} = - 4$

${x}^{2} - 4x = 13 \\ {x}^{2} - 4x - 13 = 0 \\ d = 16 + 4 \times 13 = 16 + 52 = 68 = {(2 \sqrt{17)} }^{2} \\ x1 = \frac{4 + 2 \sqrt{17} }{2} = \frac{2(2 + \sqrt{17}) }{2} = 2 + \sqrt{17} \\ x2= \frac{4 - 2 \sqrt{17} }{2} = \frac{2(2 - \sqrt{17}) }{2} = 2 - \sqrt{17}$

${x}^{2} - 4x = - 4 \\ {x}^{2} - 4x + 4 = 0 \\ {(x - 2)}^{2} = 0 \\ x3 = 2$

Ответ: 2+√17, 2-√17, 2

Answer 2

It seems you have an equation involving a quartic polynomial. The equation is given by:

$(x+2)(x-1)(x-3)(x-6) = 16$.

To solve for the values of $x$, you'll need to expand the left side of the equation and then work through the algebraic steps to isolate $x$. However, this equation might not have straightforward analytical solutions, so numerical methods or graphing might be necessary to find the approximate solutions.

Let's start by expanding the left side of the equation:

$(x+2)(x-1)(x-3)(x-6) = (x^2 + x - 2)(x^2 - 3x - 6)\\ = x^4 - 3x^3 - 6x^2 + x^3 - 3x^2 - 6x + 2x^2 + 6x + 12\\ = x^4 - 2x^3 - 6x^2 + 12$.

So, the equation becomes:

$x^4 - 2x^3 - 6x^2 + 12 = 16$.

Now we can rearrange the equation to solve for $x$:

$x^4 - 2x^3 - 6x^2 - 4 = 0$.

At this point, it might be challenging to find exact solutions algebraically. You can attempt to use numerical methods like the Newton-Raphson method, or you can plot the function $f(x) = x^4 - 2x^3 - 6x^2 - 4$ and observe where it intersects the $y = 0$ line. Graphing software or calculators can be useful for this purpose.

Keep in mind that quartic equations can have multiple solutions, including real and complex solutions, so the exact nature of the solutions will depend on the coefficients of the equation.