Factoring a third degree polynomial as part of a limits problem

AI Thread Summary
The discussion centers on factoring third-degree polynomials to solve a limit problem as x approaches 3. The user initially attempts polynomial long division but encounters errors, particularly with sign mistakes during subtraction. A suggestion is made to use the rational root theorem, specifically testing x = 3 as a factor. The user ultimately acknowledges their mistakes and expresses gratitude for the guidance received. Correctly applying polynomial division is essential for reaching the right solution in limit problems involving cubic polynomials.
EcKoh
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Homework Statement



I have a limit problem, however I do know how to work limits, I guess what I need is more of a refresher on how to work third degree polynomials. The polynomial(s) I am trying to work with are the following:

x3-2x2+2x-15
-and-
x3-5x2+10x-12

The limit is a limit where x approaches 3, so I believe that it is correct to assume that 3 is the rational root that we use in this case. However, my attempts haven't come up with a correct solution. I am just wondering where I have gone wrong and what I need to do to reach the correct answer.

2. The attempt at a solution

x3-2x2+2x-15
x3-2x2+2x-15 ÷ x+3

First, I divide to get x2, then multiply x-3 by this to reach x3-3x2
Then I bring down 2x to get x2+2s
Next I divide again to get x, then multiply x-3 again to reach x2-3x
Finally I bring down -15 to get x-15. However if I divide again I get 1, but 1•-3 does not equal -15 so I get a remainder of 12.

What am I doing wrong here?
 
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EcKoh said:

Homework Statement



I have a limit problem, however I do know how to work limits, I guess what I need is more of a refresher on how to work third degree polynomials. The polynomial(s) I am trying to work with are the following:

x3-2x2+2x-15
-and-
x3-5x2+10x-12

The limit is a limit where x approaches 3, so I believe that it is correct to assume that 3 is the rational root that we use in this case. However, my attempts haven't come up with a correct solution. I am just wondering where I have gone wrong and what I need to do to reach the correct answer.

2. The attempt at a solution

x3-2x2+2x-15
x3-2x2+2x-15 ÷ x+3

First, I divide to get x2, then multiply x-3 by this to reach x3-3x2
Then I bring down 2x to get x2+2s
Next I divide again to get x, then multiply x-3 again to reach x2-3x
Finally I bring down -15 to get x-15. However if I divide again I get 1, but 1•-3 does not equal -15 so I get a remainder of 12.

What am I doing wrong here?

A root x = 3 corresponds to a factor of (x-3). Try that.
 
Last edited:
EcKoh said:
First, I divide to get x2, then multiply x-3 by this to reach x3-3x2
Then I bring down 2x to get x2+2s
Next I divide again to get x, then multiply x-3 again to reach x2-3x
Finally I bring down -15 to get x-15. However if I divide again I get 1, but 1•-3 does not equal -15 so I get a remainder of 12.

What am I doing wrong here?
When you subtracted x^2-3x from x^2+2x, you made a sign error.
 
If you can't get a hold on what <polynomial long division> means, then try this thing

x^3 - 2x^2 + 2x -15 = (x-3) (Ax^2 + Bx + C)

No matter what value of x. Then try x=0, x=1, x=-1...
 
edit - never mind
 
Thanks to all of you. I realize now what my errors were.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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