Can a $5$-th grade polynomial have only one solution? for example: $$x^5 - 3x^4 + 17x^3 - 12x^2 - 11x - 5 = 0$$
I mean that it's not necessary for every seventh grade polynomial to have seven solutions. There may be only one or three. The same for a sixth grade polynomial, there may be only two solutions.
If this is true, then how I can decide if a fifth grade polynomial has only one solution or three and not five solutions?
$\endgroup$ 154 Answers
$\begingroup$A 5th degree polynomial (with real coefficients) has at least 1 real root. A polynomial of odd degree has at least 1 real root.
The fundamental theorem of algebra says that a polynomial roots equal to its degree. However, they may be complex and they may be roots of multiplicity.
$x^2-1$ has 2 real roots. $x^2+1$ has 2 complex roots. $x^2-2x + 1$ has one root of multiplicity 2.
Does this help?
$\endgroup$ $\begingroup$Here it is a simple algorithm for generating a lot of polynomials with degree $5$ and a single real root:
- Take a second-degree polynomial $p(x)$
- Consider the fourth-degree polynomial $q(x)=p(x)^2$
- Take some $C\in\mathbb{R},D\in\mathbb{R}^+$ and define $Q(x)$ as $C+D\int_{0}^{x}q(t)\,dt.$
$Q(x)$ is a fifth-degree polynomial with a single real root, since it is a continuous, weakly increasing and unbounded (in both directions) function over $\mathbb{R}$. For instance, with the choices $p(x)=x^2+3$, $C=0,D=5$ we get $$ Q(x) = x^5+10x^3+45x $$ whose only real root lies at $x=0$.
The main algorithm for counting the number of real zeroes of a polynomial is given by Sturm's theorem, but yet the computation of the discriminant gives you some information.
$\endgroup$ 2 $\begingroup$Fundamental Theorem of Algebra
Every non-zero, single-variable, degree $n$ polynomial with complex coefficients has, counted with multiplicity, exactly $n$ complex roots.
$\endgroup$ 3 $\begingroup$Yes. The polynomial
(x - r)^5 has one root r with multiplicity 5.
In general one can construct a polynomial ( function ) with desired properities
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