Question

(x^2+3x+1)(x^2+3x+3)=35

Answer 1

 Пусть t = x² + 3x + 1.
Тогда x² + 3x + 3 = t + 2
t(t + 2) = 35
t² + 2t = 35
t² + 2t - 35 = 0
t² + 2t + 1 - 36 = 0
(t + 1)² - 6² = 0
(t + 1 - 6)(t + 1 + 6) = 0
t = 5; -7
Обратная замена:
x² + 3x + 1 = 5
x² + 3x - 4 = 0
x² + 4x - x - 4 = 0
x(x + 4) - (x + 4) = 0
(x - 1)(x + 4) = 0
x = -4; 1
x² + 3x + 1 = -7
x² + 3x + 8 = 0
D = 9 - 4·8 < 0 ⇒ нет корней.
Ответ: x = -4; 1. 

Answer 2

X^4+6x^3+13x^2+12x+3=35
x^4-x^3+7x^2+20x^2-20x+32x-32=0
(x-1)*(x^3+7x^2+20x+32)=0
(x-1)(x+4)(x^2+3x+8)=0
x-1=0
x+4=0
x^2+3x+8=0
x1=-4
x2=1