10.12 The Fundamental Theorem of Algebra and the Identity Theorem
A polynomial of degree has at most roots. This page states the Fundamental Theorem of Algebra and uses that count to show when a polynomial must be zero and when two polynomials must be equal.
Theorem 10.12.1 (Fundamental Theorem of Algebra). Every one-variable polynomial of degree has at most roots in and exactly roots in , counted with multiplicity.
Problem 10.12.2. Let be a polynomial of degree . Show that the graphs of and cannot intersect at more than points.
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Given that the polynomial is of degree , we can factor it into total zeros.
For , either ,, or .
If , then for to become less than at some point, it must cross . a point such that . So , and since is constant, .
Algebraic applications of the Fundamental Theorem
Problem 10.12.3. Suppose is a polynomial such that and there are at least different values of for which .
(a) If , then what does the Fundamental Theorem of Algebra (FTA) tell us about the number of different values of for which ?
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(a) FTA tells us there exists a max of roots for a polynomial of degree , but this contradicts our premise. Thus our premise cannot be.
(b) If there are different values for which , then sends all to zero, so the function is .
Note (Trick!!). If is a polynomial of degree at most and there are different values for which , then is the zero polynomial, i.e.
Problem 10.12.4. Let be of degree with , and .
(a) Let . What is the max degree of ?
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(a) The max is , if has .
(b) We have
So we have values of for which , but has max degree , so must be the zero function. Thus , so .
Problem 10.12.5. Let and be in with degree at most . Show that if for distinct values of , then for all .
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Consider . Each has degree at most , so is of degree at most . So has a max of roots. So if for distinct values of , then for distinct values of we have . But , which can be at most times, which is a contradiction. So must be the zero function. If is the zero function then for all , so for all ,
Theorem 10.12.6 (The Identity Theorem). If and are polynomials of at most degree and for distinct values of , then for all .
Problem 10.12.7.(a) Consider the system of equations:
These equations determine . Show that
Additionally, prove that and .
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Let , where
We show is the zero function. See that for the values , respectively.
Note that is at most degree , since . Then there are at most values for which . Ah! But consider . (EQ1), (EQ2), (EQ3), (EQ4) give these values for , but we just saw that also equals for . Thus the only explanation is that is the zero function, since ,, and . Thus for all , so
For the second part, we let , so , so , so . Then
Scratch: other values of give other relations. With :
and reads .
Take 2.
WTS: AND .
What we know:
Expanding the left side,
Comparing the terms, , so .
For the second identity: .
Problem 10.12.8. There exists a unique with , where , such that ,,,. What is ?
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Consider the function that results from the difference between and . The roots of such a function are clearly through . So for we have
where . Given that , we see that must be . So our polynomial is
For it is immediately clear that will be the only surviving term, so
Note (Trick!!). If is an unknown polynomial of degree and you know values that can be modeled as a function (e.g. ), then you can construct a new function , where can be manipulated to produce a polynomial that can give you .
Problem 10.12.9. Let . Earlier we showed that is not necessarily a factor of if every root of is a root of (we did this by counterexample: but ). What if was restricted to not have any multiple roots? Must then be a factor of ?
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First attempt. We can represent this with a function (we essentially ask if the image of is necessarily in for some function ). Let . If , it’s trivial. If , then , since every root of is a root of . So, given that has no multiple roots, we have
Given that is a polynomial, it’s obviously bijective.
Scratch: . Trying examples:
Second attempt. If, in , has no multiple roots, then there exists a one-to-one correspondence between each root of and a root of , so evenly divides , so is a factor of .
Third attempt. Write with or .
Let , so that it has roots. Let those roots be given by . Since for each with , we have
But , so . So for all , which means has at least roots, since each is distinct. Ah! But if , so must be zero for all , so
Exercises
Exercise 10.12.10. (a)
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(a) Clearly a root is , and another is , since . So
Only factors are .
Exercise 10.12.11.(a) Show that if is of degree with and , then must be .