Let f(x) = x^3+7x^2+9x+3
First notice that the sum of coefficients of the terms of odd powers of x is equal to the sum of the coefficients of the terms of even powers of x. That is 1 + 9 = 10 = 7 + 3.
As a result, x=-1 is a root of f(x) = 0
f(-1) = -1+7-9+3 = 0
So (x+1) is a factor of f(x)
x^3+7x^2+9x+3 = (x+1)(x^2+6x+3)
Next x^2+6x+3 is of the form ax^2+bx+c, with a=1, b=6 and c=3. This has discriminant Delta given by the formula:
Delta = b^2-4ac = 6^2-(4xx1xx3) = 36-12 = 24
Since this is positive, but not a perfect square x^2+6x+3=0 has irrational roots given by the quadratic formula:
x = (-b+-sqrt(b^2-4ac))/(2a) = (-b+-sqrt(Delta))/(2a)
=(-6+-sqrt(24))/2 = -3+-sqrt(6)
Hence x^2+6x+3 = (x+3+sqrt(6))(x+3-sqrt(6))
Putting it all together:
x^3+7x^2+9x+3 = (x+1)(x+3+sqrt(6))(x+3-sqrt(6))
