(x² + x)² - 8(x² + x) + 12 = 0.​

Ответ:

Объяснение:

y^2-8y+12

=> y^2-6y-2y+12

=> y (y-6)-2 (y-6)

=> (y-6)(y-2)

Putting the value of y

=> (x^2-x-6)(x^2-x-2)

=> (x^2-3x+2x-6)(x^2-2x+x-2)

=> [x (x-3)+2 (x-3)][x (x-2)+1 (x-2)]

=> (x-3)(x+2)(x-2)(x+1)