Without using a calculator, determine the number of real zeros of the function
f(x) = x3 + 4x2 + x − 6.

First, we use the rational root theorem to determine any solutions of p(x). = x3 + 4x2 + x − 6

Factoring -6: 

1 
-1 
2 
-2 
3 
-3 
6 
-6 

x = 1 

p(1) = 1^3 + 4 * 1^2 + 1 - 6 = 6 - 6 = 0 
x = 1 is a solution. 

(x^3 + 4x^2 + x - 6) / (x - 1) = 

x^3 / x = x^2 
x^2 * (x - 1) = x^3 - x^2 
x^3 + 4x^2 - x^3 + x^2 = 5x^2 

5x^2 / x = 5x 
5x * (x - 1) = 5x^2 - 5x 
5x^2 + x - 5x^2 + 5x = 6x 

6x / x = 6 
6 * (x - 1) = 6x - 6 
6x - 6 - 6x + 6 = 0 

(x - 1) * (x^2 + 5x + 6) 

x^2 + 5x + 6 factors to (x + 3) * (x + 2) 

Factors: 

(x - 1) 
(x + 2) 
(x + 3) 

roots: 

x = 1 
x = -2 
x = -3