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Question 149768: Graph the polynomial function  to approximately find the function's zeros, then use synthetic division and the remainder theorem to exactly find its zeros.
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! First, let's graph the function to get
From the graph, we can see that the graph has the approximate zeros  and 
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Now let's use the Rational Root Theorem to list all of the possible rational roots
Rational Root Theorem:
 where p and q are the factors of the last and first coefficients
So let's list the factors of -2 (the last coefficient):
Now let's list the factors of 1 (the first coefficient):
Now let's divide each factor of the last coefficient by each factor of the first coefficient
Now simplify
These are all the distinct rational zeros of the function that could occur
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Now let's use synthetic division to test each possible zero
Let's see if the possible zero  is really a root for the function 
So let's make the synthetic division table for the function given the possible zero :
| 1 | | | 1 | 1 | -3 | -5 | -2 | | | | | 1 | 2 | -1 | -6 | | | 1 | 2 | -1 | -6 | -8 |
Since the remainder  (the right most entry in the last row) is not equal to zero, this means that is not a zero of
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Let's see if the possible zero  is really a root for the function
So let's make the synthetic division table for the function given the possible zero :
Since the remainder  (the right most entry in the last row) is equal to zero, this means that is a zero of
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Let's see if the possible zero  is really a root for the function
So let's make the synthetic division table for the function given the possible zero :
| -1 | | | 1 | 1 | -3 | -5 | -2 | | | | | -1 | 0 | 3 | 2 | | | 1 | 0 | -3 | -2 | 0 |
Since the remainder (the right most entry in the last row) is equal to zero, this means that is a zero of
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Let's see if the possible zero  is really a root for the function
So let's make the synthetic division table for the function given the possible zero :
| -2 | | | 1 | 1 | -3 | -5 | -2 | | | | | -2 | 2 | 2 | 6 | | | 1 | -1 | -1 | -3 | 4 |
Since the remainder  (the right most entry in the last row) is not equal to zero, this means that is not a zero of
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Summary:
So only and are actually rational roots.
Now looking back at the table for the test zero , we see
| -1 | | | 1 | 1 | -3 | -5 | -2 | | | | | -1 | 0 | 3 | 2 | | | 1 | 0 | -3 | -2 | 0 |
The bottom row of coefficients (minus the last one) form the quotient
Now let's perform synthetic division using the other zero on the function
Since the remainder (the right most entry in the last row) is equal to zero, this means that is a zero of
Once again the first three coefficients in the bottom row form the quotient
Let's use the quadratic formula to find the zeros of
 Start with the quadratic formula
 Plug in  ,  , and 
 Square to get .
 Multiply  to get
 Subtract from to get
 Multiply and to get .
 Take the square root of to get .
 or  Break up the expression.
 or  Combine like terms.
 or Simplify.
So the zeros of are or or just with a multiplicity of 2
Now there are 3 instances where we get a zero of . So this tells us that the zero has a multiplicity of 3
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Answer:
So the zeros of are (with a multiplicity of 3) and
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