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Question 274359: Solving Polynomial and Rational Inequalities?
I have did a little work with this but I got stuck on this one
x^3-4x^2+x-4<0
I do know i need to do somthing like this.
1 -4 1 -4
4* ____4__0___4__
1 0 1 0
I know that I need to bring down my 1 and * it with the 4 then add that to the -4 to get 0 and then * with the 4 to get 0 and add it to the 1 to get 1 and then I * it with the 4 to get 4 and add it to the -4 to get 0. then I will have 1x^2+1 then I will need to set it up like I just did.
1 1
1 -1
_______
1 0
is this right or do I need to go another way? can I get help thanks.
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website! Solving Polynomial and Rational Inequalities?
I have did a little work with this but I got stuck on this one
x^3-4x^2+x-4<0
I do know i need to do somthing like this.
1 -4 1 -4
4* ____4__0___4__
1 0 1 0
I know that I need to bring down my 1 and * it with the 4 then add that to the -4 to get 0 and then * with the 4 to get 0 and add it to the 1 to get 1 and then I * it with the 4 to get 4 and add it to the -4 to get 0. then I will have 1x^2+1 then I will need to set it up like I just did.
1 1
1 -1
_______
1 0
is this right or do I need to go another way? can I get help thanks.
Yep, you are doing this correctly. What you have determined is that the factors of  are  and  .
Now you should be able to recognize that is not factorable over the real numbers, so now we know that of the three roots of the original function, only one of them is real, namely  . So the graph of the boundary of your inequality only crosses the  -axis in one place, namely at the point (4,0). We are interested in determining over what interval the value of \ =\ x^3\ -\ 4x^2\ +\ x\ -\ 4) is negative, or less than zero.
Test a value larger than 4, and test a value smaller than 4. In other words, evaluate ) and ) . There are two ways to do this.
The first way is to simply substitute the selected values into the function and do the arithmetic. ^3\ -\ 4(3)^2\ +\ (3)\ -\ 4) for example.
Or you can use either 3 or 5 as a trial divisor in another iteration of synthetic division and examine the sign on the remainder. Whichever one of them is negative identifies the correct interval where the function is below the -axis. It turns out when you use synthetic division of ) using  as a trial divisor, the remainder is equal to ) .
John
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