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?!
Thanks for your help!
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.
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!
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!