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