THE FUNDAMENTAL THEOREM OF ALGEBRA

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Does a polynomial equation always have a solution?

Short Answer: It depends! What kind of solutions are you looking for?

If you're looking specifically for real number solutions,
then the answer to the question ‘Does a polynomial equation always have a solution?’ is NO.

For example, the polynomial equation $\,x^2 = -1\,$ doesn't have any real number solutions.
Why not? Because every real number, when squared, is greater than or equal to zero (hence can't possibly equal $\,-1\,$).

If you're looking for complex number solutions (which include any real number solutions), then you're in luck.
Indeed, the Fundamental Theorem of Algebra tells us that there always exists a solution,
as long as you look in the set of complex numbers.

(By the way, both $\,i\,$ and $\,-i\,$ (where $\,i = \sqrt{-1}\,$) are complex number solutions of the equation $\,x^2 = -1\,$.)

The Fundamental Theorem of Algebra

The statement of the Fundamental Theorem of Algebra is short and simple.
Don't let its simplicity fool you—it is a very powerful result.

Recall that the symbol $\Bbb C\,$ (blackboard bold C) represents the set of complex numbers,
and the symbol $\Bbb R\,$ (blackboard bold R) represents the set of real numbers.

Comments on the Fundamental Theorem of Algebra:

• A ‘constant polynomial’ is so-named because its output never changes—it is constant.
  For example, the function $\,f\,$ defined by $\,f(x) = 5\,$ is a constant polynomial—no matter what the input is, the output is always $\,5\,$.
• Most constant polynomials have no zeros.
  For example, consider $\,f(x) = 5\,$.
  Is $\,f(x)\,$ ever equal to zero? No—it's always $\,5\,$.
• On the other hand, the constant function $\,g\,$ defined by $\,g(x) = 0\,$ (the zero function) is always zero.
  It has infinitely many zeros.
• For these reasons, the Fundamental Theorem of Algebra only deals with non-constant polynomials.
• Remember that the set of complex numbers includes the real numbers.
  In other words, if $\,x\in\Bbb R\,$, then $\,x\in\Bbb C\,$.
  That is, every real number is also a complex number.
  Indeed, the real numbers are the $\,x\,$-axis (the real axis) in the complex plane.
• There are loads of complex numbers that are not real numbers.
  Any number that doesn't lie on the real axis in the complex plane isn't a real number.
• A ‘polynomial with complex coefficients’ allows terms of the form $\,ax^n\,$, where $\,a\in\Bbb C\,$.
• Is a polynomial like $\,P(x) = 5x^3 - 2x + 1\,$ a ‘polynomial with complex coefficients’?
  YES! Every real number is a complex number.
  The Fundamental Theorem of Algebra covers (in particular) all polynomials with real coefficients.
• Recall that ‘at least one’ means greater than or equal to one.
• The Fundamental Theorem of Algebra is an existence theorem.
  It guarantees that there exists a complex number $\,c\,$ for which $P(c) = 0$,
  as long as the function $\,P\,$ is a non-constant polynomial with complex coefficients.

• Recall that a corollary is an interesting consequence of a theorem—usually something that is easy to prove.
• The positive integers are: $1, 2, 3, \ldots$
• Consider the polynomial $\,P\,$ defined by $\,P(x) = (x-1)^2(x-5)^4(x+9)\,$.
  It has degree $\,7\,$—if it were multiplied out, the highest power on $\,x\,$ would be $\,7\,$.
  How many zeros does it have?
  This question is a bit ambiguous, so needs some clarification, as follows:
• The polynomial $\,P\,$ has three distinct (that is, different) zeros: $\,1\,$, $\,5\,$, and $\,-9\,$.
  If you look at the graph of $\,P\,$, it crosses the $\,x\,$-axis at exactly these three places.
• The polynomial $\,P\,$ has seven zeros, counting multiplicities:
  • the zero $\,1\,$ has multiplicity $\,2\,$ (from the two factors of $\,(x-1)\,$, written as $\,(x-1)^2\,\ $)
  • the zero $\,5\,$ has multiplicity $\,4\,$ (from the four factors of $\,x-5\,$, written as $\,(x-5)^4\,\ $)
  • the zero $\,-9\,$ is a simple zero (multiplicity $\,1\,$) (from the factor $\,(x+9)\,\ $)
  Counting multiplicities, the seven zeros of $\,P\,$ are: $\,1\,$, $\,1\,$, $\,5\,$, $\,5\,$, $\,5\,$, $\,5\,$, $\,-9\,$
• ‘Counting multiplicities’ means that if $\,c\,$ is a zero of multiplicity $\,k\,$, then $\,c\,$ is counted as a zero $\,k\,$ times.

How is this corollary an easy consequence of the Fundamental Theorem of Algebra?

Here's the idea:

• Start with a polynomial with complex coefficients that has degree $\,n\,$.
  For convenience, call this polynomial $\,P\,$ and let $\,z\,$ represent a typical input.
• By the Fundamental Theorem of Algebra, $\,P\,$ has a zero in $\,\Bbb C\,$.
  Call this zero $\,z_1\,$.
• Since $\,z_1\,$ is a zero of $\,P\,$,   $\,z - z_1\,$ is a factor of $\,P(z)\,$.
  Thus, $\,P(z) = (z-z_1)P_1(z)\,$, where $\,P_1\,$ is a polynomial with complex coefficients that has degree $\,n-1\,$.
• By the Fundamental Theorem of Algebra, $\,P_1\,$ has a zero in $\,\Bbb C\,$.
  Call this zero $\,z_2\,$.
• Since $\,z_2\,$ is a zero of $\,P_1\,$,   $\,z - z_2\,$ is a factor of $\,P_1(z)\,$.
  Thus, $\,P_1(z) = (z-z_2)P_2(z)\,$, where $\,P_2\,$ is a polynomial with complex coefficients that has degree $\,n-2\,$.
  Combining results thus far, $\,P(z) = (z - z_1)(z - z_2)P_2(z)\,$.
• Repeat as needed.
  Eventually, we have $\,P(z) = (z - z_1)(z - z_2)\cdots(z - z_n)P_n(z)\,$,
  where $\,P_n(z)\,$ is a polynomial with complex coefficients that has degree $\,n - n = 0\,$.
• Since $\,P_n(z)\,$ has degree zero, it is a constant; call it $K$ for simplicity.
• Thus, $\,P(z) = K(z - z_1)(z - z_2)\cdots(z - z_n)\,$ for some complex constant $\,K\,$.
• Thus, $\,P\,$ has the $\,n\,$ zeros $\,z_1\,,\,z_2\,,\,\ldots \,,\,z_n\,$.
  Some of the zeros $\,z_i\,$ may be identical, which is why we need to count multiplicities.
• Voila!

Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
polynomials with real number coefficients:
non-real zeros must occur in complex conjugate pairs