Let be coprime such that , that is to say, can be represented rationally. Thus
so is even by definition. But if ,, then
ah! This cannot be, since we assumed were coprime! Thus we have a contradiction, and there is no rational representation of .
Problem 8.2. Prove that if , then (given that ).
Show solution
Let and , with and . Then
and since , their products are in , so .
Remark. Similar to complex conjugate notation (), we have radical conjugate notation: . (This is non-standard, however.)
Problem 8.4. Let . Show that there are no rational roots and that there exists at least one irrational root.
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By the Rational Root Theorem, if rational roots exist then .
and , so there exists a real root there. But as we just showed, this root cannot be rational.
Problem 8.5. Find a polynomial with rational coefficients that has and as roots.
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Clearly such a polynomial must be of degree at least :
The first pair gives
and the second pair gives . So
so .
Problem 8.6. Prove there exists such that for all and there is a root with .
Show solution
First try, with the quadratic formula:
There is a factor of to divide out, which leaves something of the form . Matching, , and (that is, ). This cannot be, since we need for all .
Instead, multiply out conjugates:
so
Its factors are , so .
Problem 8.7. Let . Find all roots and factor.
Show solution
By the Rational Root Theorem,
Since , we need , so . So the root cannot be in , and we now have
. An educated guess, since , is :
so there exists a root between and . Next, , so . seems possible:
so , but it must be irrational.
Let’s try , tracking signs in synthetic division:
So .
Now let . Then
(By the quadratic formula, , with .)
So
From ,; from ,.
Vieta’s Formulas
Problem 8.16. Let , and let be the roots of .
(a) Why must ?
Show solution
(a) Since , there are factors, so each of ,, divides with no remainder.
Expanding, , and , so
Problem 8.17.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 8.18. Let be the roots of . Find .
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Possible rational roots: , and then synthetic division, etc.
and , therefore
Problem 8.19. Let , and let the roots be .
Show solution
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 8.1.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.
Example 8.2. Let be the roots of .
(a) What is ?
(b) What is ?
(c) What is ?
Show solution
(a) .
(b) , so
(c)
Remark (Counting the terms). For each term of a polynomial:
Compare and .
Filling slots from :,,. Filling slots from :,,,,,.
Alongside: , and .
Problem 8.21. is a polynomial with root , and .
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The other root must be , since it must follow the property of polynomials. Then for the sum of the roots, since ,,,:
Now
We know , so , and .
So , so , so ,.
(Margin note: should be 1, not 9!)
Problem 8.22. Evaluate if are the roots of .
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Is that not just ?
,,.
.
Problem 8.23. has roots such that and . Find and .
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(The first attempt here is crossed out in the notes; the second attempt is below.)
Problem 8.24. 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 ?
Theorem 8.3 (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!
Problem 8.21 (Try again!).Show solution
. So
and
So .
Problem 8.22 (Try again!).Show solution
With :OR
so , so
Problem 8.23 (Second attempt).Show solution
,,.
,,,, so .
, and , so
So
therefore , so and . Let , so .
Now, with terms,
and so .
Problem 8.24 (Second attempt).Show solution
Since ,, so
and so . Next, :
so
giving or . And , so
Exercise 8.3.1.Show solution
Exercise 8.3.2.. Find .
Show solution
Exercise 8.3.3. 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 8.3.4., and is , with . Solve for ,.
Show solution
So
so ,,.
Exercise 8.3.5.Show solution
, so , so
Exercise 8.3.6. 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 :
Example 8.4 (Quartic with nonreal roots)., with and for all . Also and . What is ?
Show solution
Grouping :
which together make ; the first is .
, so
That is,
Naming roots and their conjugates: .
Exercise 8.3.6 (Second attempt).Show solution
. 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!
Using Roots to Make Equations
Problem 8.25. Let be a root of . Find .
Show solution
Let .. Also and , from :
oops lol.
So
Problem 8.26. Find the sum of the 20th powers of the roots of .
Show solution
Let be a root. So
Then (on a later page):
and because , so the sum is
Example 8.5 (Quartic with nonreal roots, second attempt).Show solution
and .
Now since and , for each there is . So since is complex, , and since ,. So let and , equivalently ,. So and
If , as the problem posits, then must be such that their imaginary parts are equal. shows this clearly. Using :
where is real. Then, since ,
(the middle terms being ), so
Other routes tried: let , so
and likewise, with ,.
Or , giving
Side work: ;
?;
,,,; then
Also:
so . And , so , and , so .
Problem 8.27. Let be the roots of (they are real). Prove that .
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? We have , so
Or:
Instead:
So since are roots, , so . Then
, so , therefore .
Exercise 8.4.1. Let be a root of . Evaluate .
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, so
Exercise 8.4.2. Let be a root of .
Show solution
, so .
(a) , so
So .
(b) , so . So , so .
(c)
Exercise 8.4.3. Find the sum of the 20th powers of the roots of .
Show solution
, so
Example 8.6. has roots , and has roots . Find .
Show solution
:
✗
✗
: ✗
Example 8.7..
Show solution
so and , and
Example 8.8. Let . What is ?
Show solution
, and .
Example 8.9. If are complex roots of the equation , where and , find in terms of and .
Show solution
The roots are , and , and
Expanding,
Also
So
Example 8.10. A cubic has leading coefficient , all its coefficients are real, and . If , find .
Show solution
Write , with roots , and . By Vieta,
Therefore
Test, dividing by :
leaves , whose roots are
Example 8.11 (Diagonal of a box). has roots . Find the length of the interior diagonal of a box with sides .
Show solution
By Vieta,
So the volume is , and is the surface area.
Expanding gives plus ,,, that is, plus . So
Let . The diagonal of the face has , and then the interior diagonal has . So
Example 8.12. Let with . What is the value of ?
Show solution
, so . Trying :. Candidates: .(unfinished)
Problem 8.56. Consider all lines that meet the graph of in four distinct points ,. Show that
is independent of the line and find its value.
Show solution
Let’s look at the line that intersects at . For that line, the are the roots of , so
And at height :