I have an exam in a couple of days :smack: . Some of the problems include finding the roots of a cubic equation . I’ll have like around 2 -3 minutes to find the real root and then solve for the remaining roots , which would become pretty simple once I know the real root. The equations will not have any roots which can be found out by observation.
So what I’m looking for is a quick solution for finding one of the roots of the equation.
Thanx
Seems to me like if this is an exam, you must have already learned something that you are forgetting. Just review all of your notes about roots, etc.
In the notes that I have they have magically found out the root to the third decimal by trial and error!
The factor theorem: x-a is a root of f(x) iff f(a)=0.
For example, take
f(x) = x[sup]3[/sup] -3x[sup]2[/sup] +3x -2.
Then f(2) = 0, so x-2 is a factor of f and by division we find
f(x) = (x-2)(x[sup]2[/sup]-x +1)
Numerical Analysis/Trial and Error: The simplest version is
If f(a)>0 and f(b)<0 [or vice versa] and f is continuous on [a,b] then f has a root in [a,b].
Use:
Synthetic Division
Rational Roots Theorem
Descartes Law of Signs
Finding the root that way (only as an approximation) might not get you enough information as to be able to solve for the other two roots. What did your notes do after finding it the first root magically? Other posters have suggested other methods–do they look familiar?
What sort of class is this for?
OK. Here’s an example of the Rational Roots Theorem:
3X^4 - 11X^3 + 10X - 4 = 0
p is the absolute value of the last monomial. Here, it is 4.
q is the absolute value of the first coefficient. Here, it is 3.
Now factor each of these (remember, the factors can be positive or negative) so ± will mean that the number can be positive or negative and thus you will later have to check both:
p: ± 1, ± 2, ± 4
q: ± 1, ± 3
p/q: ±1, ±1/3, ± 2, ± 2/3, ±4, ±4/3
if the solution has any rational roots they must be one of these (the set of numbers for p/q).
Now test each of them through synthetic division (I am not going to go through all of them, but you may have to go through at least some that don’t work until you find the ones that do):
-1l 3 -11 0 10 -4
-3 14 -14 4
3 -14 14 -4 0
Since the last number is a zero, you’ve got a factor.
Factoring out:
(X + 1) (3X^3 - 14X^2 + 14X - 4)
Synthetic division again to find the factors of
3X^3 - 14X^2 + 14X - 4:
2/3l 3 -14 14 -4
2 -8 4
3 -12 6 0
You know have: (X + 1)(X - 2/3)(3X^2 - 12X + 6)
Use the quadratic formula to find the roots of 3X^2 - 12X + 6 (I’m not even going to ry to represent that here)
and you get the roots: {2 + [sq.root of 2], 2- [sq.root of 2]}
So your roots for the original equation are: { -1, 2/3, 2 ± [sq.root of 2]}
And don’t forget about the Bounds Theorem during synthetic division (if there are no negative coefficients it is an upper bound; if coefficients alternate between non-negative and non-positive, then it is a lower bound).
sigh I wish there would be more math symbols that I could include, all the underlining was killing me. Hope it looks all right.
Shit. It didn’t quite work out the way I planned.
For this:
-1l 3 -11 0 10 -4
-3 14 -14 4
3 -14 14 -4 0
just move all the numbers one space (one number space) over.
That is, the 3 gets brought down. Multiply the 3 by -1 to get -3. Add -3 to -11 to get -14. Multiply -14 by -1 to get 14. Add 14 to 0 to get 14. Multiply 14 by -1 to get -14. Add -14 to 10 to get -4. Multiply -4 by -1 to get 4. Add 4 to -4 to get 0.
Do the same for:
2/3l 3 -14 14 -4
2 -8 4
3 -12 6 0
Also, use the rational zero test if necessary:
A cubic equation would be in the form: ax^3+bx^2+cx+d
All rational zeros (this doesn’t work for imaginary ones) will be in the form p/q, where p and q have no common factors other than 1, p is a factor of the constant d, and q is a factor of the coefficient a.
Only works for integer coefficients.
Example:
3x^3+2x^2-5x-2
Therefore the zeros can only be +2/3, -2/3, +1, -1, +1/3, -1/3
Damn, sleeping already wrote that. Don’t I feel stupid 
To use synthetic division we need one root of the equation . Finding that one root is why I posted this thread
I really won’t have the time to do that and root obtained is surely almost going to be upto 2 decimals everytime as 2.34 or 3.57.
And the equations itself are gonna in decimal form (So no Rational zero test)
The approximated root was the exact root upto the 2 decimal , and then using synthetic division the solved for the other roots.
Anyways thanx for all your suggetions guys. If I can’t find an easy solution I guess I’ll just have to skip the problem if it comes!