Multiplying out and comparing coefficients ties the coefficients of a polynomial to sums and products of its roots.
Problem 10.14.1. Let , and let be the roots of .
(a) Why must ?
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(a) Since , there are factors, so each of ,, divides with no remainder.
Expanding, , and , so
Problem 10.14.2.Show solution
, with roots :
So for ,
Therefore, since and , we have , so . Therefore the sum of the roots for is .
We consider factored as :
Multiplying the cubic by one more factor :
The numbers of terms in the coefficients are ,,,,.
Now: , so . Finally, , so .
Problem 10.14.3. Let be the roots of . Find .
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Possible rational roots: , and then synthetic division, etc.
and , therefore
Problem 10.14.4. Let , and let the roots be .
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So , which begins
where (a sum of terms), and next comes .
So, next: gives the term. So apply the same to :
so it’s ? Then , and so on, up to .
In general (with , and in ):
where the inner sum runs over the ways of choosing of the roots.
Check: ✓, and
Definition 10.14.5.Elementary symmetric polynomials, or symmetric sums, are the sums of all products, represented by , for a set of variables , where the sum is of all products of the taken at a time:
for :,;
for :,,;
and so on.
Theorem 10.14.6 (Vieta’s Formulas). For of degree with roots :
where each is the symmetric sum of the roots of taken at a time.
Proof. For every choice of of the , taking from the other binomial factors of the polynomial, we form , where is the sum of the roots taken at a time, which has terms. Expanded:
(Technically the first term is , but .) Comparing coefficients gives the formulas. ∎
Remark (Why “symmetric”?). Why do we call multivariate symmetric sums symmetric? Well:
so we can swap their positions. So if is any permutation of (with giving ), then
The elementary symmetric polynomials build (full) symmetric polynomials!
Remark (Counting the terms). For each term of a polynomial:
Compare and .
Filling slots from :,,. Filling slots from :,,,,,.
Alongside: , and .
Example 10.14.7. Let be the roots of .
(a) What is ?
(b) What is ?
(c) What is ?
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(a) .
(b) , so
(c)
Problem 10.14.8. Evaluate if are the roots of .
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Is that not just ?
,,.
.
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With :OR
so , so
Problem 10.14.9. has roots such that and . Find and .
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,,.
,,,, so .
, and , so
So
therefore , so and . Let , so .
Now, with terms,
and so .
Problem 10.14.10. The roots of satisfy . Find all possible .
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So , and , so , so . Then
so one option… ✗ no wait.
so ,, and , so . Then
so ?
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Since ,, so
and so . Next, :
so
giving or . And , so
Exercises
Exercise 10.14.11.Show solution
Exercise 10.14.12.. Find .
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Exercise 10.14.13. Two factors of are and . Find all roots.
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There are many ways to do this:
and . Now and , so , so , so ,, and . So ,, and ✓. So is a double root. So we have
Exercise 10.14.14.Show solution
, so , so
Exercise 10.14.15. Find the two values of for which has a double root.
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Signs of the coefficients: . With roots :
So , and
so , that is,
Also, from :
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. Say there is a double root, i.e. . So
; or :
so . Then , and . So , and .
So or gives a double root.
Let’s check! (Just one :) ) :, so
giving , and ✓:
So , that is, , yay!
Exercise 10.14.16. Let be a quadratic polynomial with integer coefficients and integer roots.
(a) If , can be a root of ?
(b) If , can be a double root of ?
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(a) Yes, since .
(b) No. By Vieta’s (for monic ),, and if then , so this cannot be.
Furthermore, the roots are unique and there exist at most TWO roots in by the FTA, and we know both are in ; so since , we get , which pins down the polynomial.
If , however, would give a double root of . But here .