# Zeroes of Polynomials

In the first section we described some basic properties of polynomials. In this section we describe some further properties and at the end we prove that every complex polynomial actually has a root.

As we pointed out, in some cases the zeros of a given polynomial can be exactly determined. The case of polynomials of degree 2 has been known since the old age. The well-known formula gives the solutions of a quadratic equation $$ax^2+bx+c=0$$ ($$a\neq0$$) in the form $x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\:.$

When $$f$$ has degree 3 or 4, the (fairly impractical) formulas describing the solutions were given by the Italian mathematicians Tartaglia and Ferrari in the 16-th century. We show Tartaglia’s method of solving a cubic equation.

At first, substituting $$x=y-a/3$$ reduces the cubic equation $$x^3+ax^2+bx+c=0$$ with real coefficients to $y^3+py+q=0, \quad\mbox{where}\quad p=b-\frac{a^2}3,\quad q=c-\frac{ab}3+ \frac{2a^3}{27}.$ Putting $$y=u+v$$ transforms this equation into $$u^3+v^3+(3uv+p)y+q=0$$. But, since $$u$$ and $$v$$ are variable, we are allowed to bind them by the condition $$3uv+p=0$$. Thus the above equation becomes the system $uv=-\frac p3,\quad u^3+v^3=-q$ which is easily solved: $$u^3$$ and $$v^3$$ are the solutions of the quadratic equation $$t^2+qt-\frac{p^3}{27}=0$$ and $$uv=-p/3$$ must be real. Thus we come to the solutions:

Theorem 2.1 (Cardano’s formula)

The solutions of the equation $$y^3+py+q=0$$ with $$p,q\in\mathbb{R}$$ are $y_i=\epsilon^j\sqrt[3]{-\frac q2+\sqrt{\frac{q^2}4+ \frac{p^3}{27}}}+\epsilon^{-j}\sqrt[3]{-\frac q2-\sqrt{\frac{q^2}4 +\frac{p^3}{27}}},\quad j=0,1,2,$ where $$\epsilon$$ is a primitive cubic root of unity.

A polynomial $$f(x)=a_nx^n+\cdots+a_1x+a_0$$ is symmetric if $$a_{n-i}=a_i$$ for all $$i$$. If $$\deg f=n$$ is odd, then $$-1$$ is a zero of $$f$$ and the polynomial $$f(x)/(x+1)$$ is symmetric. If $$n=2k$$ is even, then $f(x)/x^k=a_0(x^k+x^{-k})+\cdots+a_{k-1}(x+x^{-1}) +a_k$ is a polynomial in $$y=x+x^{-1}$$, for so is each of the expressions $$x^i+x^{-i}$$ (see problem 3 in section 7). In particular, $$x^2+x^{-2}=y^2-2$$, $$x^3+x^{-3}=y^3-3y$$, etc. This reduces the equation $$f(x)=0$$ to an equation of degree $$n/2$$.

Problem 2

Show that the polynomial $$f(x)=x^6-2x^5+x^4-2x^3+x^2-2x+1$$ has exactly four zeros of modulus 1.

How are the roots of a polynomial related to its coefficients?

Consider a monic polynomial $P(x)=x^n+a_1x^{n-1}+\cdots+a_{n-1}x+ a_n=(x-x_1)(x-x_2)\cdots(x-x_n)$ of degree $$n> 0$$. For example, comparing coefficients at $$x^{n-1}$$ on both sides gives us $$x_1+x_2+\cdots+x_n=-a_1$$. Similarly, comparing the constant terms gives us $$x_1x_2\cdots x_n=(-1)^na_n$$. The general relations are given by the Vieta formulas below.

Definition

Elementary symmetric polynomials in $$x_1,\dots,x_n$$ are the polynomials $$\sigma_1,\sigma_2,\dots,\sigma_n$$, where $\sigma_k= \sigma_k(x_1,x_2,\dots,x_n)=\sum x_{i_1}x_{i_2}\dots x_{i_k},$ the sum being over all $$k$$-element subsets $$\{i_1,\dots,i_k\}$$ of $$\{1,2,\dots,n\}$$.

In particular, $$\sigma_1=x_1+x_2+\cdots+x_n$$ and $$\sigma_n= x_1x_2\cdots x_n$$. Also, we usually set $$\sigma_0=1$$ and $$\sigma_k=0$$ for $$k> n$$.

Theorem 2.2 (Vieta’s formulas)

If $$\alpha_1,\alpha_2, \dots,\alpha_n$$ are the zeros of polynomial $$P(x)=x^n+a_1x^{n-1}+ a_2x^{n-2}+\dots+a_n$$, then $$a_k=(-1)^k\sigma_k(\alpha_1,\dots, \alpha_n)$$ for $$k=1,2,\dots,n$$.

Example

The roots $$x_1,x_2,x_3$$ of polynomial $$P(x)=x^3- ax^2+bx-c$$ satisfy $$a=x_1+x_2+x_3$$, $$b=x_1x_2+x_2x_3+x_3x_1$$ and $$c=x_1x_2x_3$$.

Problem 3

Prove that not all zeros of a polynomial of the form $$x^n+2nx^{n-1}+2n^2x^{n-2}+\cdots$$ can be real.

Problem 4

Find all polynomials of the form $$a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x +a_0$$ with $$a_j\in\{-1,1\}$$ ($$j=0,1,\dots,n$$), whose all roots are real.

One contradiction is enough to show that not all zeros of a given polynomial are real. On the other hand, if the task is to show that all zeros of a polynomial are real, but not all are computable, the situation often gets more complicated.

Problem 5

Show that all zeros of a polynomial $$f(x)=x(x-2)(x-4)(x-6)+(x-1) (x-3)(x-5)(x-7)$$ are real.

We now give the announced proof of the fact that every polynomial has a complex root. This fundamental theorem has many different proofs. The proof we present is, although more difficult than all the previous ones, still next to elementary. All imperfections in the proof are made on purpose.

Theorem 2.3 (Fundamental Theorem of Algebra)

Every nonconstant complex polynomial $$P(x)$$ has a complex zero.

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