1. ## Solving cubic equation

Hi Guys,

Need a little help with solving this cubic equation.

I did the rational root test and figured out that it has either 1 or 0 rational roots. Further by the descartes method, assuming the rational root would be of either of, -1, 1, -1/2, 1/2, -2, 2, or -4, or 4.

f(2) is negative and f(4) is positive so I figured the root has to be near 3. And this further leads me to think there is no rational root.

I have run out of ideas here. The method I have been following so far have been to try to get a rational root by trial and error/via synthetic division using the assumed roots in the descartes method, then you end up with a quadratic that is easy to solve.

Without the first root, its a little hard to proceed.

I looked around on the web and found the "Cubic Formula". This thing is enormous! Can you guys guide if I missed any simpler method to do this. Do I memorize this?!

2. Originally Posted by mathguy80
Hi Guys,

Need a little help with solving this cubic equation.

I did the rational root test and figured out that it has either 1 or 0 rational roots. Further by the descartes method, assuming the rational root would be of either of, -1, 1, -1/2, 1/2, -2, 2, or -4, or 4.

f(2) is negative and f(4) is positive so I figured the root has to be near 3. And this further leads me to think there is no rational root.

I have run out of ideas here. The method I have been following so far have been to try to get a rational root by trial and error/via synthetic division using the assumed roots in the descartes method, then you end up with a quadratic that is easy to solve.

Without the first root, its a little hard to proceed.

I looked around on the web and found the "Cubic Formula". This thing is enormous! Can you guys guide if I missed any simpler method to do this. Do I memorize this?!

I suspect that one of the minuses should be a + so that m = -1 will be a factor. Go back and check the equation, ask your teacher if there's a typo. You would not be expected to solve the cubic you posted by hand. Where has it come from?

3. If last term is changed to +4 : m^3 - m^2 - m + 4 = 0
then m = 2, m = -2 and m = 1 are the solutions.

4. Originally Posted by Zap
Check your last bracket of the second line. It would give you a +4 instead of a -4.

5. Ah yes, should have known not to try and think without my cup of coffee...going back to bed

6. Originally Posted by Zap
Ah yes, should have known not to try and think without my cup of coffee...going back to bed
Zap!!

7. @Mr Fantastic, I check the problem its not a typo. From the responses so far I am guessing that it's a problem with a greater difficulty level that I am currently at. I guess I will have to come back to it in the future then!

8. Originally Posted by mathguy80
@Mr Fantastic, I check the problem its not a typo. From the responses so far I am guessing that it's a problem with a greater difficulty level that I am currently at. I guess I will have to come back to it in the future then!
I'm curious. Did this come from a text where you are given the answer? I'd like to see what the supposed answers are.

I wouldn't give this problem to anyone who wasn't one of the better students in my College Algebra class trying for extra credit. Solving general cubic equations is a pain in the...umm..patootie. Wherever you got the problem from, I'm sure there is a typo in the original question.

-Dan

9. Originally Posted by mathguy80
@Mr Fantastic, I check the problem its not a typo. From the responses so far I am guessing that it's a problem with a greater difficulty level that I am currently at. I guess I will have to come back to it in the future then!
I am convinced that there is a typo in the question.
As written is has only one real root that is very complicated.
See the solition.

10. Originally Posted by topsquark
I'm curious. Did this come from a text where you are given the answer? I'd like to see what the supposed answers are.
-Dan
It didn't come from my textbook. I was having some difficulty in solving cubic equations. I can solve them but it just takes time. So I thought it would be a good idea to try to practice solving problems from elsewhere. Searching around the web I found this book with a chapter on Solving Cubic Polynomials. The problem no. is Page 54, 5.5.1, exercise 3.

It doesn't have solutions unfortunately. That's the extent of my adventure so far!

11. Originally Posted by mathguy80
The problem no. is Page 54, 5.5.1, exercise 3.
These 3 equations appear:
x^3 + x^2 - 5x + 3 = 0 ; solution: (x + 3)(x - 1)(x - 1) = 0
y^3 - 3y^2 - 16y - 12 = 0 ; solution: (y + 2)(y + 1)(y - 6) = 0
m^3 - m^2 - 4m - 4 = 0 ; solution: as complicated as a mother-in-law!
IF -4 changed to +4:
m^3 - m^2 - 4m + 4 = 0 ; solution: (m + 2)(m - 2)(m - 1) = 0

So isn't it obvious to you that the book has a typo?

Btw, no need to look all over the joint to find exercises; make 'em up yourself:
like write down something like (x + 2)(x - 5)(x - 1) = 0 ; expand : x^3 - 4x^2 - 7x + 10 = 0
Now pretend it's an exercise you found!

12. @Wilmer, yes it is probably a typo. Nevertheless this typo got me aware of possible solutions, so its all good. Good tip on making up examples. Thanks.

13. Originally Posted by mathguy80
It didn't come from my textbook. I was having some difficulty in solving cubic equations. I can solve them but it just takes time. So I thought it would be a good idea to try to practice solving problems from elsewhere. Searching around the web I found this book with a chapter on Solving Cubic Polynomials. The problem no. is Page 54, 5.5.1, exercise 3.

It doesn't have solutions unfortunately. That's the extent of my adventure so far!
Yes, it does have solutions as Plato showed. They are just very complicated. At first I thought something titled "Solving Cubic Polynomials" should have a section on the "Cubic Formula" but if this is intended for Grade 12 probably it doesn't.