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Question 362638: list all rational zeros
use synthetic division to test the possible rational roots and find an actual root
x^3-2x^2-7x-4=0
Answer by shorty4aboo(8)   (Show Source):
You can put this solution on YOUR website! Ok this is going to be a hard one to explain and i apologize now that i probably won't explain it good enough. Ok anyways......
Your problem is {{f(x)=x^3-2x^2-7x-4=0}}}
so with that we will list the possible zeros first.
In order to list these it will look like this.
P=+1,-1,+2,-2,+4,-4 [these are your factors of 4 the last # in your equ.]
Q=+1,-1 [these are your factors of 1 the first # in your equ.]
Then you put  to make your list so lets do that:
 , , This list converts to
 , , These are the only three numbers you will use within your synthetic division to find your zeros. Now when you go to synthetically divide if it is a true zero, the prob will end in zero.
so lets look at this:When you synthetically divide by
1 you get No zero because the remainder is -12
2 you get  No zero because the remainder is 18
4 you get  This is a zero because the remainder is 0
Lets rewrite the equ now with this zero in and we should be able to regulary factor out the last zero(s)[Don't forget to change your synthetic #'s sign].
Let us now factor  ...... This factors to 
So the problem now looks like this:
Now that it is completely factored, lets find those zeros:
 would turn into  then this would turn into 
 would turn into  then this would turn into 
So your Zeros are : 4 and 1 (w/ multiplicty of 2)
Your # of zeros will always be either the same or lower than the power of your leading term. Out leading term in this equ. was  which means we could have 3 or less zeros and we have 3. Good Job!
Again I apologize because this is even complicated for me to understand and i wrote it. Yet try with what you know to make these numbers show up on your paper with pencil. If I just confused you so much more I sincerely apologize.
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