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 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 _i_symmetric_/i_ 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.
Set \(y=x+x^{-1}\). Then \[\frac{f(x)}{x^3}=g(y)=y^3-2y^2-2y+2.\]
Observe that \(x\) is of modulus 1 if and only if \(x=\cos t+i\sin t\)
for some \(t\), in which case \(y=2\cos t\); conversely, \(y=2\cos t\)
implies that \(x=\cos t\pm i\sin t\). In other words, \(|x|=1\) if and
only if \(y\) is real with \(-2\leq y\leq 2\), where to each such \(y\)
correspond two values of \(x\) if \(y\neq\pm2\). Therefore it remains
to show that \(g(y)\) has exactly two real roots in the interval
\((-2,2)\). To see this, it is enough to note that \(g(-2)=-10\),
\(g(0)=2\), \(g(2)=-2\), and that therefore \(g\) has a zero in each of
the intervals \((-2,0)\), \((0,2)\) and \((2,\infty)\).
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 (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 \(k=1,2,\dots,n\).
We will use induction on \(n\). The case \(n=1\) is trivial.
Assume that \(n > 1\) and write \(P(x)=(x-x_n)Q(x)\), where \(Q(x)=
(x-x_1)\cdots(x-x_{n-1})\). Let us compute the coefficient \(a_k\) of
\(P(x)\) at \(x^k\). Since the coefficients of \(Q(x)\) at \(x^{k-1}\)
and \(x^k\) are \(a_{k-1}^{\prime}=(-1)^{k-1}\sigma_{k-1}(x_1,\dots,x_{n-1})\)
and \(a_k^{\prime}=(-1)^k\sigma_k(x_1,\dots,x_{n-1})\) respectively, we have
\[a_k=-x_na_{k-1}^{\prime}+a_k^{\prime}=\sigma_k(x_1,\dots,x_n). \]
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.
Suppose that all its zeros \(x_1,x_2,\dots,x_n\) are real. They
satisfy \[\sum_i x_i=-2n,\quad\sum_{i < j} x_ix_j=2n^2.\] However,
by the mean inequality we have \[\sum_{i < j}x_ix_j=\frac12
\left(\sum_i x_i\right)^2-\frac12\sum_i x_i^2\leq\frac{n-1}{2n}
\left(\sum_i x_i\right)^2=2n(n-1),\] a contradiction.
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.
Let \(x_1,\dots,x_n\) be the roots of the given polynomial. Then
\[\begin{array}{c}x_1^2+x_2^2+\cdots+x_n^2=(\sum_i x_i)^2-
2(\sum_{i < j} x_ix_j)=a_{n-1}^2-2a_{n-2}\leq3;\newline x_1^2x_2^2\cdots
x_n^2=1.\end{array}\] By the mean inequality, the second equality
implies \(x_1^2+\cdots+x_n^2\geq n\); hence \(n\leq3\). The case \(n=3\)
is only possible if \(x_1,x_2,x_3=\pm1\). Now we can easily find all
solutions: \(x\pm1\), \(x^2\pm x-1\), \(x^3-x\pm(x^2-1)\).
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.
Since \(f(-\infty)=f(\infty)=
+\infty\), \(f(1) < 0\), \(f(3) > 0\) and \(f(5) < 0\), polynomial \(f\) has a
real zero in each of the intervals \((-\infty,1)\), \((1,3)\),
\((3,5)\), \((5,\infty)\), that is four in total.
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.
Write \(P(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_0\).
Suppose that \(P(0)=a_0\neq0\). For each \(r > 0\), let \(C_r\) be the
circle in the complex plane with the center at point 0 and radius
\(r\). Consider the continuous curve \(\gamma_r=P(C_r)=\{P(x)\mid
|x|=r\}\). The curve described by the monomial \(x^n\), i.e. \(\{x^n\mid
x\in C_r\}\) rounds point 0 \(n\) times. If \(r\) is large enough, for
example \(r > 1+|a_0|+\cdots+|a_{n-1}|\), we have \(|x^n| > |a_{n-1}x^
{n-1}+\cdots+a_0|=|P(x)-x^n|\), which means that the rest \(P(x)-x^n\)
in the expression of \(P(x)\) can not ``reach_quot_ point 0. Thus for such
\(r\) the curve \(\gamma_r\) also rounds point 0 \(n\) times; hence, it
contains point 0 in its interior.
For very small \(r\) the curve \(\gamma_r\) is close to point \(P(0)=
a_0\) and leaves point 0 in its exterior. Thus, there exists a
minimum \(r=r_0\) for which point 0 is not in the exterior of
\(\gamma_r\). Since the curve \(\gamma_r\) changes continuously as a
function of \(r\), it cannot jump over the point 0, so point
0 must lie on the curve \(\gamma_{r_0}\). Therefore, there is a zero
of polynomial \(P(x)\) of modulus \(r_0\).