3rd order something-or-other

[alkatran]

Sorry I don't know the term, but I take math in french.
What we use would translate into "3rd degree equation"
it's

x^3+2x^2+5x+6=0

It's a "level above" quadratic equations correct? So how would I solve this??


extra: My math teacher says x^3 cuts the axis 3 times, that can't be true right??? it'd have to be x^2+x^3-.01???

[bugzpodder]

cubic, and assuming you are talking about the x-axis, if it has 3 real roots then yes.

[bugzpodder]

like the quadratic formula, there is also a cubic formula. but i won't recommend it for your level. search for it if you are interested. usually you guesstimate what the roots are and sub it into f(X) if f(a)=0 then it is a root, or (x-a) is a factor of the equation.

[alkatran]

Hmm I still don`t understand, and we are covering it in math now.
My math teacher does something similar to what you siad (f(x) = x^3 + x^2 ....)

I understand up to assuming x-a, but what do I do after that?

[bugzpodder]

long division or synthetic division. if you know synthetic division, use it, otherwise you have to use long division

[Guv]

y = x³ crosses the X-Axis only once at (0, 0), but a more general 3rd order polynomial will cut it once or three times.

The analytic method of solving for roots of a third order polynomial is so messy that nobody uses it anymore, unless they are taking certain mathematics courses. Numerical solutions like the Newton method are far faster and easier to program for a computer.

The general idea behind the analytic method is as follows.

First, divide by the leading coefficient (if necessary) to get the following form.

P(x) = x³ + bx² + cx + d

Then substitute x = (u - b / 3). After some algebraic manipulation, you get the following form.

P(u) = u³ + pu + q, where p & q can be expressed in terms of b, c, & d.

Then substitute u = (v - p / 3v) I am not 100% sure of this substitution, but it is close to correct. After some algebraic manipulation, you get something like the following.

P(v) = v⁶ + mv³ + n, which can be solved for v³ using the formula for a 2nd order (quadratic) polynomial.

Find three cube roots and reverse the substitutions. There are actually 6 cube roots, considering the positive and negative solutions to the 2nd order polynomial. I am not sure of the details. I suppose that there are some equal values, resulting in only 3 distinct cube roots.

There is always a real root of a 3rd order polynomial. I could find it with the Newton method & then the other roots via synthetic division. If I used pencil and paper, I could probably find the roots faster than I could using the analytic approach. If I used a hand calculator, I know I could do it faster.

[alkatran]

Thanks guys!
On long division, I wasn't taught that way of diving in school
From what I understand...

1321
5/5875
08
17
15
0

??? Right? It's very confusing when you have to learn a whole method of dividing in under 45 minutes

x^2
x-2 / x^3+X^2-x+5 Am I starting correctly?
x^3

[bugzpodder]

1321
5/5875
08
17
15
0

i don't know what the heck these numbers are but the x^3 thing looks ok

[TheAlchemist]

hey man,
if you're just lbegginging to learn about cubic functions then i suggest that you try and understand this basic form of solving the functions so that you can see more clearly what the other guys have been trying to say.
so far we know that the general cubic function has 3 roots(solutions). lets call these p,q and r. now we can use these in to get actual values: for the cubic function then;

(x-p)(x-q)(x-r) = f(x)= ax^3 + bx^2 + cx + d.

multiplying out the factors we get:

x^3 - (p+q+r)x^2 + (rp + qr + pq)x + pqr

now comparing the coeficients of different powers of x with the general equation we can see that

p+q+r = b --(i) , rp+qr+pq = c --(ii) and pqr = d --(iii)

we can then solve these linear equations simultaeneously to get the values of p,q and r

i hope that helped!!

[alkatran]

My teacher taught us how to estimate a factor, then use synthetique/long division.

Only P I have now is if factor isn`t a whole number O_O

[Guv]

TheAlchemist: Your simultaneous are not linear. If you try to solve them, you will end up with a cubic or worse if you make a mistake in algebraic manipulation. This is not the way to find an analytical solution.

See my previous post to this thread, which describes the general concept of the Tartaglia/Cardano analytical solution to cubics.

In the computer age, solutions to real problems always rely on numerical methods using successive approximations for finding roots of polynomials. Other methods are used in the academic community for purposes of teaching mathematical insight, understanding of basic concepts, and some mental discipline. The academic methods are not used for the sole purpose of finding the roots.

BTW: It has been proven that there can be no analytical method for finding roots of 5th order or higher polynomials. My intuition tells me that such a proof cannot exist, one more example of my intuition being incorrect.

[TheAlchemist]

hey Guv,
Thanks for correcting the mistake,you are right there will be a quadratic involved in solving the equations.
We're going to start on numerical methods from next week at school so perhaps then i may be able to understand your thread better.
Just two questions: what are u and v in yuor substitutions and why do we substitute these for x?
thanks again

[Guv]

Alchemist: The first substitution of x = u - b/3 results in a cubic equation with no second order term, which is easier to deal with than the general cubic. It looks like the following.

u³ + pu + q, where p and q can be expressed terms of the original coefficients.

If you can solve that equation for u, you can then find x, which is (u - b/3).

The next substitution results in a quadratic in v³, which looks like the following.

v⁶ + mv³ + n, where m and n can be expressed in terms of p and q.

The ordinary quadratic formula can be used to determine v³, which can be used to find three values for v. These can be used to find values for u, which are needed to find values for x.

To give an idea of a successive approximations method, consider solving x² - 100 = 0, which is really finding the simple square root of 100. Solving this special case polynomial should give some insight into successive approximations methods.

The algorithm is easy. Guess a value. Divide that value into 100. Average the quotient and the guessed value to find the next approximate value. Keep this up. Suppose you did not know that ten is the square root of 100 and that you are a lousy guesser. You might start with a first guess of 20.

First guess 20

100/20 = 5 ---- Average (5 and 20) = 25/2, or 12.5

100/12.5 = 8 ----- (8 + 12.5)/2 = 10.25 — We are closer now. The following are the values, showing no arithmetic.

20.000 000 000
12.500 000 000
10.250 000 000
10.003 048 751
10.000 000 465
10.000 000 000