Ok not sure if it's just me with a block
solve the following equation in the real number system.
x^4 +3x^3 + 3x^2 + 21x - 28 = 0
Since the general method to solve a quartic is very complicated, you have to assume that there will be one or two easy-to-find roots.
Try x = 1. y=1+3+3+21-28 = 0
so, y = (x-1)(x^3 + 4x^2 + 7x + 28)
Now we see there will be a negative root, since all the coefficients are positive.
Working with the cubic now,
Try x = -1. y=32
x = -2. y=22
x = -3 y=16
x = -4 y=0
y = (x-1)(x+4)(x^2 + 5x + 7)
The quadratic has no real roots.