Theresa, you wrote the initial factored problem as P(x) = −1/3 (x+3)
(x+1) (x−1)
(x−2).
In order to "state the 'zeros' of the function," we have to see what values of x will make P(x) = 0.
Each of the elements are multiplied, so if any one element is equal to 0, then the entire set becomes equal to 0.
Step 1) 1/3 can't be changed, because it's not multiplied by x.
Step 2) (x+3) = 0 if x = 3
Step 3) (x+1) = 0 if x = 1
Step 4) (x1) = 0 if x = 1
Step 5) (x2) = 0 if x = 2
Therefore, your solution set is {3, 1, 1, 2}
May 10

Jane B.
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