ANSWER
x = -1, x = -1, x = 1;
-1 has a multiciplicity of 2
+1 has a multiciplicity of 1
EXPLANATION
We have the function:
[tex]f(x)=x^3+x^2-\text{ x - 1}[/tex]
To find the zeros of the function, we have to make the function = 0.
That is:
[tex]x^3+x^2\text{ - x - 1 = 0}[/tex]
Now, factorise the function:
[tex]\begin{gathered} \Rightarrow x^2(x\text{ + 1) - 1(x + 1) = 0} \\ \Rightarrow(x^2\text{ - 1)(x + 1) = 0} \\ Factorise(x^2\text{ - 1):} \\ (x\text{ + 1)(x - 1)}(x\text{ + 1) = 0} \\ \Rightarrow\text{ x = -1, x = 1 and x = -1} \end{gathered}[/tex]
Those are the zeros of the function.
The multiciplicities of this function simply refer to the number of times a certain root (or zero) of the function appears in the function.
Therefore:
-1 has a multiciplicity of 2 (it appears twice)
+1 has a multiciplicity of 1
