How can I factor a 3rd order polynomial using the cubic formula?

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To factor a 3rd order polynomial like x^3 + 8x^2 - 21x + 10, it's suggested to start by identifying potential rational roots, which are typically factors of the constant term. Once a root is found, polynomial division can be used to simplify the polynomial into a linear factor and a quadratic factor. The quadratic can then be solved using the quadratic formula. While the cubic formula is available for finding roots, it is considered tedious, and numerical methods or online tools may offer easier alternatives. The discussion highlights the challenge of dealing with irrational roots in cubic polynomials.
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Homework Statement


What are the steps to factoring 3rd order polynomials like x^3+8x^2-21x+10?
It's to find eigenvalues of a matrix in linear algebra, I completely forgot how to factor and it's killing me.

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The Attempt at a Solution


None, unless its a polynomial that I can factor by grouping, I have no clue how to begin.
 
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EDIT: Nevermind the below, that factor trick (the 'rational root test') doesn't work in this case. The discriminant is positive, but it turns out that all three roots are irrational.

For a cubic polynomial with integer coefficients, which is what you have, I believe one of the roots is typically a factor of the constant term. So, try the factors of 10 and see if one of them is a root. Once you've figured out one of the roots, you can factor an (x-root) term out by polynomial division, leaving you with (x-root)*quadratic, and the quadratic you can factor with the quadratic formula.
 
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:( so the pq thing with synthetic and what-the-other-one's-name division?
I don't know why i didnt think of it (but that sounds really tedious).
Thanks for your help! :)
 
Got it, so i happened to pick the only example in the book of a matrix with irrational eigenvalues. I'm still not so sure how to find the irrational roots of a 3rd order polynomial... i should change the name of this thread haha.
 
There's always the cubic formula, but using it is a bit... tedious. :eek:
 
eumyang said:
There's always the cubic formula, but using it is a bit... tedious. :eek:

To add to what "eumyang"" said, using the Cubic Formula to get an exact algebraic result is, indeed, a bit tedious.

On the other hand, if you are fine with numerical results, there are several free online utilities available; some of them are listed at the bottom of that Wikipedia page.

In addition, I refer you to an old post I made at the bottom of the following thread:

https://www.physicsforums.com/showthread.php?t=580075&page=2

A search of these forums would probably turn up much more information about factoring a cubic polynomial.
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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