What are the zeroes of the polynomial function f(x)=x^3+2x^2-x-2

[tex]f(x)=x^3+2x^2-x-2\\\\f(x)=0\iff x^3+2x^2-x-2=0\\\\x^2(x+2)-1(x+2)=0\\\\(x+2)(x^2-1)=0\iff x+2=0\ \vee\ x^2-1=0\\\\x+2=0\qquad\text{subtract 2 from both sides}\\\\\boxed{x=-2}\\\\x^2-1=0\qquad\text{add 1 to both sides}\\\\x^2=1\to x=\pm\sqrt1\\\\\boxed{x=-1\ \vee\ x=1}\\\\Answer:\ -2,\ -1,\ 1[/tex]
Answer: -2, -1, 1
Step-by-step explanation:
f(x) = x³ + 2x² - x - 2
0 = x³ + 2x² - x - 2
= x²(x + 2) -1(x + 2)
= (x² - 1) (x + 2)
= (x - 1) (x + 1) (x + 2)
0 = x - 1 0 = x + 1 0 = x + 2
1 = x -1 = x -2 = x