How many real number solutions does the equation x3 + 5x2 + 3x – 9 = 0 have?

f(x) = x³ + 5x² + 3x - 9 = 0
There is 1 sign change in f(x), so there is 1 positive real root.

f(-x) = -x³ + 5x² - 3x - 9
There are 2 sign changes in f(-x), so there are either
(a) 2 negative real roots, or
(b) a pair of complex roots.

Use the Remainder Theorem to test for the positive real root.
f(1) = 1 + 5 + 3 - 9 = 0
Therefore (x-1) is a factor of f(x).

Perform synthetic division.
1 | 1  5  3  -9
        1  6   9
  ---------------
    1   6  9  0

f(x) = (x-1)(x² + 6x + 9) = (x-1)(x+3)²
The zeros or roots are 1, -3, -3.
Therefore there are 3 real solutions for f(x) = 0.

Answer: 3 real solutions