how to get the real solution x^3 + 4x^2 + 25x + 100 = 0?

Answer 1

factor it -- notice that the relation of the x³ coefficient to the x² coefficient is the same as the relation of the x coefficient to the constant (1:25), therefore try x+4 as your first root and use long division to divide it evenly into your equation.

x³ + 4x² + 25x + 100 =
(x+4)(x²+25)

So your real root is x = -4

(You have two other complex roots of x = -5i, x = 5i)

Answer 2

Factor by grouping.
Split the equation in half, and see what you can factor out of both terms.

In:
x^3+4x^2
you can take out an x^2, leaving:
x^2(x+4)

and in the next part:
take out a 25
25(x+4)

Only if the two parts have the number in parenthesis equal to each other, you can factor by grouping. In this case there is (x+4) in both parts, so you can.

Your solution is:

(x^2 + 25) (x+4) = 0

Answer:
-4

The other is imaginary, since you cannot square root a negative number.

Answer 3

x^3+4x^2+25x+100=0
x^2(x+4)+25(x+4)=0
(x^2+25)(x+4)=0
x^2+25=0 or x+4=0
so you get the real solution only from x+4=0,x=-4

Answer 4

x^3 + 4x^2 + 25x + 100
(x^3 + 4x^2) + (25x + 100)
x^2(x + 4) + 25(x + 4)
(x^2 + 25)(x + 4)

to take this further, it would be

(x + 5i)(x - 5i)(x + 4)

so as you can see

x = -4