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polynomial help

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2 Answers

 
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) (x-1) = 0 if x = 1
Step 5) (x-2) = 0 if x = 2
 
Therefore, your solution set is {-3, -1, 1, 2}
 
 
 

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No need for one tutor to overpost another, Jane.  I was trying to get the student to do some of the work, rather than handing her the answer.   There are plenty of unanswered questions to answer.

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Use the "zero factor property" - find the value of x that sets each factor to zero.
 
P(x) = −1/3 (x+3)(x+1)(x−1)(x−2)
 
P(x) will = 0 when (x+3)=0 or (x+1)=0 or (x-1)=0 or (x-2)=0
 
The first factor is (x+3).  The value of x = -3 will make it equal zero, so x = -3 is one the points.  Can you find the other values for the other factors?
 
The y-intercept occurs when x = 0.  Set x = 0 in P(x) to find the y-intercept.

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Yes, the y-intercept is (0,-2).  You'll note that it's the constant term in the polynomial; that is, the one term not multiplied by x.
 
If you look at the factored form, it's four factors multiplied together.  Any number multiplied by zero is zero.  So if any one of the factors equals zero, the whole polynomial equals 0 (P = 0).  So if (x+1)=0, then P=0.  (x+1)=0 when x=-1.  So (-1,0) is one of the P=0 points.  There are 3 others, one for each of the remaining three factors.

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