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 youit 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.
THEOREM
the Fundamental Theorem of Algebra
Every non-constant polynomial with complex coefficients has at least one zero in $\,\Bbb C\,$.
Comments on the Fundamental Theorem of Algebra:
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A ‘constant polynomial’ is so-named because its output never changesit is constant.
For example, the function $\,f\,$ defined by $\,f(x) = 5\,$ is a constant polynomialno matter what the input is, the output is always $\,5\,$.
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Most constant polynomials have no zeros.
For example, consider $\,f(x) = 5\,$.
Is $\,f(x)\,$ ever equal to zero? Noit's always $\,5\,$.
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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 $\,a\in\Bbb C\,$.
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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.
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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.
COROLLARY
to the Fundamental Theorem of Algebra
Let $\,n\,$ be a positive integer.
Every polynomial of degree $\,n\,$ with complex coefficients has exactly
$\,n\,$ zeros in $\,\Bbb C\,$,
counting multiplicities.
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Recall that a corollary is an interesting consequence of a theoremusually
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: $\,1\,$, $\,1\,$, $\,5\,$, $\,5\,$, $\,5\,$, $\,5\,$, $\,-9\,$
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‘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:
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Start with a polynomial with complex coefficients that has degree $\,n\,$.
For convenience, call this polynomial $\,P\,$ and let $\,z\,$ represent a typical input.
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By the Fundamental Theorem of Algebra, $\,P\,$ has a zero in $\,\Bbb C\,$.
Call this zero $\,z_1\,$.
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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 $\,n-1\,$.
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By the Fundamental Theorem of Algebra, $\,P_1\,$ has a zero in $\,\Bbb C\,$.
Call this zero $\,z_2\,$.
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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, $\,P(z) = (z - z_1)(z - z_2)P_2(z)\,$.
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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.
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Thus,
$\,P(z) = K(z - z_1)(z - z_2)\cdots(z - z_n)\,$ for some complex constant $\,K\,$.
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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!