Can the roots of $f(x)=x^4-x^3+2x^2-x-1$ be found algebraically?

Question

Can the roots of $f(x)=x^4-x^3+2x^2-x-1$ be found algebraically? Are there multiple methods for doing so?

What have you tried? Hint: start by searching for simple roots (e.g. integers). — lulu, Jul 08 '15 at 14:02
There are formulas but they're not pretty. See what WA gives. — lhf, Jul 08 '15 at 14:04
You can at least factor out $x - 1$, which will make the equation cubic. — Keith, Jul 08 '15 at 14:07

score 6 · Answer 1 · answered Jul 08 '15 at 14:04

6

Yes, polynomial equations of degree four (or less) have formulas that will let you compute the (possible) roots. For more on how to do this in this case, you can take a look at this Wikipedia article about solving the quartic equation.

Another way would be to guess a root. Try for example $x=0,-1, \color{red}{1}$. If you find a root $\alpha$, then you can divide the polynomial by $x-\alpha$ (using long division). That way to reduce the problem to finding roots of a lower degree polynomial.

answered Jul 08 '15 at 14:04

Thomas

43,555

Thanks! One thing -- is there a more methodical way of guessing roots based on the function besides starting at zero and working up/down through the integers? Also, I remember learning some different factoring techniques for solving quartic functions in high school; guess those can't really be applied here? – Rasputin Jul 08 '15 at 14:11
Never mind about the factoring part, wikipedia covers that. – Rasputin Jul 08 '15 at 14:12
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@Rasputin Yes, you can always produce a finite list of rational numbers guaranteed to include all rational roots---this is the Rational Root Theorem. For more, see this recent question: http://math.stackexchange.com/questions/1351111/guessing-one-root-of-a-cubic-equation-for-hit-and-trial/1351158#1351158 – Travis Willse Jul 08 '15 at 14:13
@Rasputin: It is in general hard to find roots. You can give a list of possible rational roots. You can use things like the the "Rational Roots Theorem" https://en.wikipedia.org/wiki/Rational_root_theorem – Thomas Jul 08 '15 at 14:13

score 1 · Answer 2 · answered Jul 08 '15 at 14:25

Yes, all perfect polynomials of degree $4$ or less can be solved algebraically, but perfect polynomials of degree $5$ or higher can't. By a "perfect" polynomial I mean a polynomial which has terms in all lower degrees, for example, your polynomial.

There are multiple methods of solving a polynomial. I think the easiest one is trial and error by checking for integer solutions, and then factoring. For example, in this case, $x=1$ is a root, so the polynomial takes form of $(x-1)*something$, and that turns out to be $(x-1)(x^3+2x+1)$, utilizing polynomial long division, which is also a good thing to know when solving tricky polynomials.

The solution for quadratic polynomial can be easily found by algebra, but cubics and quartics are trickier. However, every cubic $ax^3+bx^2+cx+d$ can be expressed as a depressed cubic $t^3+pt+q$ which can then in turn be actually solved using Vieta's substition and quadratic formula. Also, every quartic has a depressed form, but those can only be solved by Ferrari's method, if I'm not mistaken.

It is (or should be) well known that a parabola can be expressed geometrically by it's focus and directrix. Other geometrical interpretations of polynomials can also be made: for example, Vieté's trigonometric expression for the roots of a cubic.

So, to answer your question: useful methods for solving your polynomial are trial and error, factoring, polynomial long division, expressing polynomials in their depressed forms, and geometrical interpretations.

If you expand $(x-1)^5$, it seems to fit your definition of a "perfect" polynomial. Clearly we can solve it algebraically. Not all such polynomials are soluble but that is different from stating none are. — Macavity, Jul 08 '15 at 17:21

score 0 · Answer 3 · answered Jul 08 '15 at 14:06

Yes; 4 is the max grade where the roots can be found with a formula: https://en.wikipedia.org/wiki/Quartic_function#General_formula_for_roots It has been proved that for grade >=5, such a formula can't exist: https://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem

However, people don't use grade 3 and 4 formulas (unless they have a computer, at least). Try using Gauss method!

score 0 · Answer 4 · answered Jul 08 '15 at 15:04

If you read French, you can have a description of Ferrari's method, and links to a bunch of methods for algebraic equations (some exist in other languages).

In the present case, however, as $1$ is a root,we have a factorisation by $x-1$: $$ x^4-x^3+2x^2-x-1=(x-1)(x^3+2x+1). $$ So we have a cubic equation in standard form. Its discriminant is $\Delta=-4\cdot 2^3-27\cdot 1^2=-59<0$, hence it has only $1$ real root (we could as well check this point with the derivative), and we can use Cardano's method:

Set $x=u+v$. The equation rewrites as: $$u^3+v^3+(3uv+2)(u+v)=1=0.$$ As we've replaced $1$ unknown with $2$, we can impose a condition: $3uv+2=0$. The equation is now equivalent to the system: $$\begin{cases}uv=-\frac23\\u^3+v^3=-1\end{cases}\iff\begin{cases}u^3v^3=-\frac8{27}\\u^3+v^3=-1\end{cases} $$ Thus the problem boils down to the standard problem on quadratic equations: find two numbers, given their sum and their product. They are the roots of the equation: $$t^2+t-\frac8{27}=0.$$ The discriminant of this equation is $\;1+\dfrac{32}{27}=\dfrac{59}{27}\;$ so $$u^3,v^3=\frac{-1\pm\sqrt{\frac{59}{27}}}2$$ and finally: $$ x=u+v= \sqrt[3]{\frac{-1+\sqrt{\frac{59}{27}}}2} + \sqrt[3]{\frac{-1-\sqrt{\frac{59}{27}}}2} $$

Can the roots of $f(x)=x^4-x^3+2x^2-x-1$ be found algebraically?

4 Answers4