Need help in solving for x in polynomials. Don't want the answer, just pointed in the right direction.
How do I start breaking this fella down?

graph f(x)= 3x^3-29x+68x+26

Question

Need help in solving for x in polynomials.  Don't want the answer, just pointed in the right direction.
How do I start breaking this fella down?

graph f(x)= 3x^3-29x+68x+26

tkhunny · Answer

The very fist thing to do is make sure there IS a zero in the Real Numbers.  As this is of ODD degree, there is guaranteed to be ONE.  Anything else is just good fortune.  In this case we know there is one negative Real zero.  Do you know how to determine this?

The second thing to do would be to hope beyond normal hope that there is a Rational zero.  We should try:

Factors of 26

Divided by

Factors of 3

26 = 2*13

Try -1/3, -2/3, -1, -2, -13/3, -26/3, -13, -26  You should either find a Rational Zero or find that there isn't one.

Truthfully, after you determine that f(-1/3) > 0 and f(-2/3) < 0, you can quit.  There is NO Rational Zero.  You have only your favorite numerical method to find the Real Zero.  (Well, there is a way to find the exact value, but you do NOT want to know it!)

How's that for starters?

anonymous · Answer

Lookin for irrational zero then, huh?  Between f(-1/3) and f(-2/3)....am I getting into square root of -1 ("i") territory?

tkhunny · Answer

No, no.  There is a REAL zero.  Once you find that, you can reduce the polynomial and find the to Complex zeros with relative ease.

anonymous · Answer

Not giving up then...grr!  Thanks for help, tkhunny!

tkhunny · Answer

Did you figure out that x = -1/3 is REALLY CLOSE to a zero?  As in, it IS!?

I must have typed something funny or missed your x^2 typo, or something.