Without using a calculator, determine the number of real zeros of the function:
f(x)= x^3+ 4x^2+x-6

*the clue on pluto, synthetic division finds that there are three real zeros. as suggested by the highest degree in the function*

1 , -3, -2

It factors into:

(x-1) (x+3) (x+2)

Step-by-step explanation:

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

it's both, now can you answer my most recent question?