Finding eigenvalues from simple characteristic polynomial gone wrong - Solution explanation

Question

Now I thought I had this understood, but apparently not. So I did some reduction and got a characteristic polynomial of the form:

$$-x^{3} + 2x^{2} + x - 2 = 0$$

Where I'm just using $x$ in place of $\lambda$ for typing purposes.

In my first attempt at solving it I went through this process:

$$-x(x^{2}-2x-1) = 2 \\ x = -2\ \text{or} \ (x^{2}-2x-1) = 2$$ From which I could rewrite (so I thought) as

$$x^{2}-2x-3 = 0 $$

From which we could factor to get

$$(x-3)(x+1)$$

Fortunately there is a solution I saw which gave the following factorization:

$$(x-1)(x-2)(x+1) = 0$$

Now I re worked this and the only way I could get this solution was by "guessing a root (using rational zeros)", then dividing my cubic polynomial, and finally factoring the quadratic that remains. Which for example would be of the form:

$$(x-1)(x^{2}-x-2) = 0$$

Now I know I should know the reason why but can't remember, but why didn't my first approach work? And what is a more efficient way of getting this factorization besides "guessing roots".

When a product of two expressions in an integral domain is $0$, then at least one of them must be $0$, but you can't replace $0$ with $2$ in that statement; cf. this answer — J. W. Tanner, Dec 01 '20 at 02:41
If you're looking for rational roots, the rational root theorem limits the possibilities — J. W. Tanner, Dec 01 '20 at 02:50
Yes @J.W.Tanner, I should've probably mentioned the rational root theorem, but since I am solving for roots in $\mathbb{R}$ I left it out. — D.C. the III, Dec 01 '20 at 02:57
The beauty of the expression being in a textbook and not what I'd probably encounter in the wild. — D.C. the III, Dec 01 '20 at 03:01
The sum of coefficient is 0, therefore one root of polynomial is $1$. So other factor comes out if you divide it by $(x-1)$. — sirous, Dec 01 '20 at 04:45

score 1 · Accepted Answer · answered Dec 01 '20 at 02:58

1

What went wrong with your first attempt is that $a\times b=2$ does not imply that $a=2$ or $b=2$, the way $a\times b=0$ implies $a=0$ or $b=0$ in an integral domain (such as a field). For further explication of that important point, see this answer.

Regarding how to find rational roots of a polynomial, see the rational root theorem, which says they will be factors of the constant term divided by factors of the leading coefficient; in your case the candidates would be $\pm1$ or $\pm2$.

answered Dec 01 '20 at 02:58

J. W. Tanner

60,406

Also thanks for the assistance. – D.C. the III Dec 01 '20 at 03:09
Actually to build on the rational roots idea, this means the only other way I could solve for the roots would have been to apply the equivalent of the quadratic formula, but the one used for 3rd degree polynomials. Does it even have a name? – D.C. the III Dec 01 '20 at 03:11
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https://en.wikipedia.org/wiki/Cubic_equation#General_cubic_formula is what you're looking for I think – mi.f.zh Dec 01 '20 at 03:18

Finding eigenvalues from simple characteristic polynomial gone wrong - Solution explanation

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