# The number of roots for polynomials

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?

• The accepted terminology is degree and not grade. Commented Oct 13, 2016 at 15:31
• That is the only real root, though there are four complex roots. There isn't a general method for finding roots, sadly. It may help to remark that a multiple root is also a root of the derivative, and it's a lot easier to find common roots of two polynomials of close degree.
– lulu
Commented Oct 13, 2016 at 15:31
• Are you familiar with derivatives? If so, I can write out an approach for detecting multiple roots.
– lulu
Commented Oct 13, 2016 at 15:32
• Here is a picture which shows the location of all of the complex roots of your polynomial: i.sstatic.net/6EUZ8.png The roots are (approximately) {-0.320745 - 0.331599i, -0.320745 + 0.331599i, 1.08328 - 3.84111i, 1.08328 + 3.84111i, 1.47494} Commented Oct 13, 2016 at 15:51
• @AhmedAmir, you may be interested to know that there is likely no way to solve this "by hand" (meaning without a computer). Beyond 4th degree polynomials there is no general formula to write down their roots. This is known as the Abel-Ruffini theorem. Commented Oct 13, 2016 at 16:00

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?

Here it is a simple algorithm for generating a lot of polynomials with degree $5$ and a single real root:

1. Take a second-degree polynomial $p(x)$
2. Consider the fourth-degree polynomial $q(x)=p(x)^2$
3. 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.

• Well, this is interesting.. I will definitely look at this again after getting into calculus. Commented Oct 13, 2016 at 16:17
• May I ask the reason behind the downvote? Commented Oct 13, 2016 at 17:16

Fundamental Theorem of Algebra

Every non-zero, single-variable, degree $n$ polynomial with complex coefficients has, counted with multiplicity, exactly $n$ complex roots.

• This does not answer the question. The comments clarify that the OP is curious about unique zeroes. Commented Oct 13, 2016 at 15:22
• You might have made less mystifying for the OP if you had said "exactly $n$ complex roots". Commented Oct 13, 2016 at 15:23
• @Sloan My answer was posted before the comments and I explained that you need to look at multiplicity. If there are further questions motivated by the answer, anyone can ask in the comments...
– Edu
Commented Oct 13, 2016 at 15:28

Yes. The polynomial

(x - r)^5


has one root r with multiplicity 5.

In general one can construct a polynomial ( function ) with desired properities