1. How do you solve third-grade equations algebraicly?
For example y=(ax^3)+(bx^2)+d

2.

3. Cardano’s method

The working can be very complicated and tedious – best if you have a computer program for it.

PS: Such an equation is called a third-degree equation, not third-grade equation.

4. Dang. I was still working on addition and subtraction when I was in the third grade.

5. Originally Posted by Harold14370
Dang. I was still working on addition and subtraction when I was in the third grade.
Heh!

And I thought the answer would involve Fermat's Last Theorem! Or is that for equations of degree 4 and greater?

6. Originally Posted by sunshinewarrior
And I thought the answer would involve Fermat's Last Theorem! Or is that for equations of degree 4 and greater?
If you're just interested in finding roots and not integer roots, then FLT won't help you at all. Furthermore, FLT only talks about a very specific set of equations.

For equations of degree less than or equal to 4, you can always find the roots via arithmetic operations and root extractions (i.e., square root, cube root, etc.). However, beginning with degree 5 and continuing on forever, there exist polynomials whose roots cannot be expressed as such. For example, the family of polynomials x<sup>n</sup>-x-1, n ≥ 5, are such polynomials. However, this only says that there are no elementary algebraic methods of solving these polynomials. So, for example, there are analytic methods of finding roots, so not all hope is lost!

7. Originally Posted by serpicojr
Originally Posted by sunshinewarrior
And I thought the answer would involve Fermat's Last Theorem! Or is that for equations of degree 4 and greater?
If you're just interested in finding roots and not integer roots, then FLT won't help you at all. Furthermore, FLT only talks about a very specific set of equations.

For equations of degree less than or equal to 4, you can always find the roots via arithmetic operations and root extractions (i.e., square root, cube root, etc.). However, beginning with degree 5 and continuing on forever, there exist polynomials whose roots cannot be expressed as such. For example, the family of polynomials x<sup>n</sup>-x-1, n ≥ 5, are such polynomials. However, this only says that there are no elementary algebraic methods of solving these polynomials. So, for example, there are analytic methods of finding roots, so not all hope is lost!
Had forgotten about integer roots, though my post was, in any case, made more light-heartedly. Thanks for that.

8. Originally Posted by sunshinewarrior
Had forgotten about integer roots, though my post was, in any case, made more light-heartedly. Thanks for that.
My light-heartedness detectors turn off when math is involved. It's just a mathematician's instinct!

9. Originally Posted by serpicojr
Originally Posted by sunshinewarrior
Had forgotten about integer roots, though my post was, in any case, made more light-heartedly. Thanks for that.
My light-heartedness detectors turn off when math is involved. It's just a mathematician's instinct!
I could tell SunshineWarrior was being light-hearted; I just couldn’t think of a light-hearted reply to make. :P

I know, it can be difficult to be light-hearted when trying to explain certain things in maths.

10. Originally Posted by serpicojr
Originally Posted by sunshinewarrior
Had forgotten about integer roots, though my post was, in any case, made more light-heartedly. Thanks for that.
My light-heartedness detectors turn off when math is involved. It's just a mathematician's instinct!
Don't worry about it - we love you just the way you are. And it's just wonderful to have maths-expert guys like you and Lizzie's elder sister and Ratty here!

11. couldn't you just plug the equation into your graphing calculator and just find when where the graph intercepts the x axis?

12. If you're going to use a calculator, you may as well use it the right way. Most graphing calculators come with a polynomial solver. Use this instead of the graphing function.

13. Thansk for all the posts but could anybody please tell me, if I have the function y=ax<sup>3</sup>+bx<sup>2</sup>+cx+d what solutions does it have? I know it can be solved algebraicly and this I want to learn. Please don't answer that you can factorize it in (x-a)(x-b)(x-c) where a,b & c are roots.

14. I don't think any of us here know it off the top of our heads. For serious, look it up on Wikipedia:

http://en.wikipedia.org/wiki/Cubic_e...cubic_function

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