What are the real zeros of f(x) = x^3 +2x^2-5x-6

First of all Horner
First we try with -3
1. 2. -5. -6.
-3. 3. 6
1. -1. -2. 0
To verify: -3^3+2*-3^2+15-6=0

Now we have this equation
(X+3)*(x^2 - x -2)=0

I dont know If you know sum and product but that is the easiest way so sum= 1 product = 2 so 2 and -1 are the answers.
You can do this too by this way : b^2-4ac= D
X= (-b(+or-)d^1/2)/2a

So all the zeros
2, -1 and -3
If you want more explanation, send me a comment x

Here u go!

Step-by-step explanation:

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