By Mario123
Sample Lecture For Christmas Camp
Symmetric Polynomials
1. Introduction
This handout will cover some properties of symmetric polynomials. There will be a few definitions and
theorems, and then, there will be some problems.
2. An Introduction to Symmetric Polynomials
Definition: A symmetric polynomial is a polynomial where if you switch any pair of the variables, the
polynomial is still the same. For example, x + y + z is a symmetric polynomial, since switching x and y
yields y + x + z, which is the same as x + y + z, switching x and z yields z + y + x, which is the same as
x + y + z, and switching y and z yields x + z + y, which is the same as x + y + z.
Definition: The elmentary symmetric polynomials in the variables x1 , x2 , x3 , ..., xn are defined as follows:
S1 = x1 + x2 + x3 + · · · + xn
S2 = x1 x2 + +x1 x3 + x1 x4 + · · · + xn−1 xn−2
S3 = x1 x2 x3 + x1 x2 x4 + · · · + xn−2 xn−1 xn
·
·
·
Sn = x1 x2 x3 · · · xn
Don’t be intimidated by the notation. Elementary symmetric polynomials are actually very simple! For
example, for the variables x1 , x2 , and x3 , the elementary symmetric polynomials are
S1 = x1 + x2 + x3
S2 = x1 x2 + +x1 x3 + x2 x3
S3 = x1 x2 x3
The elmentary symmetric polynomials can be thought of as the building blocks of any symmetric polynomial
because any symmetric polynomial can be expressed in terms of elementary symmetric polynomials. For
example, we can express the symmetric polynomial x2 + y 2 + z 2 as an elementary symmetric polynomial by
letting a variable a = x2 , a variable b = y 2 , and a variable c = z 2 . Therefore, we now have the elementary
symmetric polynomial a + b + c. You may be asking why this is useful. One of the main reasons why is
because elementary symmetric polynomials are directly linked to the powerful Vieta’s Formulas.
3. Vieta’s Formulas
Theorem: Vieta’s Formulas are a set of equations relating the roots and the coefficients of polynomials.
Given a polynomial an xn + an−1 xn−1 + · · · + a1 x + a0 with roots r1 , r2 , · · ·, rn , the following equations hold
true:
r1 + r2 + r3 + · · · + rn =
1
−an−1
an
r1 r2 + r1 r3 + r1 r4 + · · · + rn−1 rn−2 =
r1 r2 r3 + r1 r2 r4 + · · ·rn−2 rn−1 rn =
an−2
an
−an−3
an
·
·
·
r1 r2 r3 r4 · · · rn = −(1)n aan0
The left sides of these equations are the elementary symmetric polynomials in the variables r1 , r2 , · · ·, rn .
Once again, don’t be intimidated by the notation. Vieta’s Formulas are actually quite simple! For example,
if the polynomial x3 + 3x2 + 2x + 1 has roots x1 , x2 , and x3 , the following equations hold:
x1 + x2 + x3 = −3
x1 x2 + +x1 x3 + x2 x3 = 2
x1 x2 x3 = −1
4. The General Method for Attacking Symmetric Polynomial Problems
So now what? We have explored elementary symmetric polynomials, and we have seen their application to
the powerful Vieta’s Formulas. Let’s try to develop a general method for attacking symmetric polynomials.
Earlier, I mentioned that the elmentary symmetric polynomials can be thought of as the building blocks
of any symmetric polynomial because any symmetric polynomial can be expressed in terms of elementary
symmetric polynomials. Our general method for attacking symmetric polynomials revolves around utilizing
that fact:
1. Given a symmetric polynomial, express it in terms of elementary symmetric polynomials.
2. Find the values of those elementary symmetric polynomials.
3. Construct a polynomial using the values of those elementary symmetric polynomials.
4. Find the roots of the newly constructed polynomial, and use this information to solve the problem. Note
that the roots of a polynomial f (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 are those values of x such that f (x)
= 0.
This general method won’t solve every question involving symmetric polynomials. However, you can use it
to solve many problems. Note that Vieta’s Formulas will probably be used in at least one of the four steps.
Also, Vieta’s Formulas are useful in many other problems besides those that involve symmetric polynomials.
I will demonstrate this general method with a couple example problems.
5. Some Example Problems
Problem: Given that the polynomial x3 + 3x2 + 4x + 9 has the roots r1 , r2 , and r3 , find the value of
r12 + r22 + r32 .
Solution: We want to express r12 + r22 + r32 in terms of elementary symmetric polynomials. We know the
elementary symmetric polynomials for the variables r1 , r2 , and r3 are r1 + r2 + r3 , r1 r2 + r1 r3 + r2 r3 , and
r1 r2 r3 . Our goal is to express the r12 + r22 + r32 in terms of these elementary symmetric polynomials. We note
that r12 + r22 + r32 has squared terms, and none of our elementary symmetric polynomials have squared terms,
so we probably have to square one of our elementary symmetric polynomials. So, let’s square r1 + r2 + r3 .
This yields r12 + r22 + r32 + 2r1 r2 + 2r1 r3 + 2r2 r3 . We see our desired expression, r12 + r22 + r32 , in this expansion
but we also have a few extra terms, 2r1 r2 + 2r1 r3 + 2r2 r3 . If we factor a 2 out of these extra terms, we have
2(r1 r2 + r1 r3 + r2 r3 ). Aha! We recognize the expression multiplied by 2 as one of our elementary symmetric
polynomials! So we now have r12 + r22 + r32 = (r1 + r2 + r3 )2 − 2(r1 r2 + r1 r3 + r2 r3 ). We have reduced the
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problem to finding the values of r1 + r2 + r3 and r1 r2 + r1 r3 + r2 r3 . However, we know these values! Vieta’s
Formulas applied to the polynomial x3 + 3x2 + 4x + 9 gives us r1 + r2 + r3 = −3 and r1 r2 + r1 r3 + r2 r3 = 4.
Therefore, r12 + r22 + r32 = (r1 + r2 + r3 )2 − 2(r1 r2 + r1 r3 + r2 r3 ) = (−3)2 − 2(4) = 9 − 8 = 1
Problem: Find x, y, and z if
x+y+z =6
x2 + y 2 + z 2 = 26
x3 + y 3 + z 3 = 90
and x ≥ y ≥ z
Solution: The first thing we notice is that the left hand sides of each of the three equations are symmetric
polynomials. We want to express these polynomials in terms of elementary symmetric polynomials, but
luckily for us, the problem already gives us one: x + y + z. Now, our goal is to find the two other symmetric
polynomials in the variables x, y, and z, which are xy + xz + yz and xyz. Based on our work in the previous
problem, we know that x2 + y 2 + z 2 = (x + y + z)2 − 2(xy + xz + yz). In this problem, we are given
the values of x2 + y 2 + z 2 and x + y + z, so we can find the value of xy + xz + yz. Plugging our values
of x2 + y 2 + z 2 and x + y + z into the equation x2 + y 2 + z 2 = (x + y + z)2 − 2(xy + xz + yz), we have
26 = 62 − 2(xy + xz + yz), so xy + xz + yz = 5. Now, we have to find the value of xyz. In order to
get this xyz expression, we probably have to manipulate at least one of the given equations to have this
xyz expression. Since xyz has degree 3, it’s reasonable to cube both sides of the equation x + y + z = 6.
This yields x3 + 3x2 y + 3xy 2 + y 3 + 3x2 z + 6xyz + 3y 2 z + 3xz 2 + 3yz 2 + z 3 = 216. This looks a little
intimidating, so let’s try to break it down. Firstly we recognize the expression 6xyz contains our desired
expression, xyz. Next, we recognize the expression x3 + y 3 + z 3 , and we know this equals 90. So we
have reduced the equation x3 + 3x2 y + 3xy 2 + y 3 + 3x2 z + 6xyz + 3y 2 z + 3xz 2 + 3yz 2 + z 3 = 216 to
3x2 y + 3xy 2 + 3x2 z + 6xyz + 3y 2 z + 3xz 2 + 3yz 2 = 126. We can divide both sides of the equation by 3, which
yields x2 y + xy 2 + x2 z + 2xyz + y 2 z + xz 2 + yz 2 = 42. We are trying to find the value of xyz, so we isolate
the expression 2xyz on the left handside, which yields 2xyz= 42 − (x2 y + xy 2 + x2 z + y 2 z + xz 2 + yz 2 ).
So essentially, our goal now is to find the value of x2 y + xy 2 + x2 z + y 2 z + xz 2 + yz 2 . We recognize
this expression as a symmetric polynomial, so now, our task is to express this symmetric polynomial in
terms of elementary symmetric polynomials. In order to find how the way to do this, let’s experiment
a bit. We can rewrite the expression as x(xy) + y(yx) + x(xz) + y(yz) + z(xz) + z(yz). Each term in
this expression is a term of the elementary symmetric polynomial x + y + z multiplied by a term of the
elementary symmetric polynomial xy + yz + xz. Perhaps, we can multiply x + y + z by xy + yz + xz to create
x2 y+xy 2 +x2 z+y 2 z+xz 2 +yz 2 . Let’s try it. (x+y+z)(xy+yz+xz) = x2 y+xy 2 +x2 z+3xyz+y 2 z+xz 2 +yz 2 .
Success! We see x2 y + xy 2 + x2 z + y 2 z + xz 2 + yz 2 within the expression. We also see the extra 3xyz, so
we need to deal with that. Firstly, we note that (x + y + z)(xy + yz + xz) = (6)(5) = 30. So 30 =
x2 y + xy 2 + x2 z + y 2 z + xz 2 + yz 2 + 3xyz. We isolate x2 y + xy 2 + x2 z + y 2 z + xz 2 + yz 2 because that is
the value we seek, so we have 30 − 3xyz = x2 y + xy 2 + x2 z + y 2 z + xz 2 + yz 2 . We substitute this expression
into the equation 2xyz= 42 − (x2 y + xy 2 + x2 z + y 2 z + xz 2 + yz 2 ), which yields 2xyz= 42 − (30 − 3xyz),
which yeilds 2xyz= 42 − 30 + 3xyz, which yields −xyz = 12, so xyz = −12. We now have the values of all
three symmetric polynomials in the variables x, y, and z:
x+y+z =6
xy + yz + xz = 5
xyz = −12
Now we can construct a polynomial using the values of these elementary symmetric polynomials. We can
achieve this by letting x, y, and z be the roots of a polynomial in the variable t, and by Vieta’s Formulas,
this polynomial is t3 − 6t2 + 5t + 12. Now, we have to find the roots of this polynomial, in other words, we
have to find the values of t such that t3 − 6t2 + 5t + 12 = 0. We could do this in several ways. Some of course
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are more efficient than others. For example if you know the Factor Theorem or Rational Root Theorem,
then you could easily solve this. However, for those of you who don’t have any experience with tools such
as the Factor Theorem or Rational Root Theorem, you could simply guess the roots by plugging in a few
values of t. If you use this method, then I suggest that you start with small values of t, such as −1, 0, or
1, and then start trying larger values. We find that the roots are 3, 4, and −1. The problem states that
x ≥ y ≥ z, so our solution is (x, y, z) = (4, 3, −1)
6. Problems
What follows are some problems that involve the material from this handout. Not all of these problems can
be solved by the general method that I have gone over. Be flexible. Remember that Vieta’s Formulas extend
beyond symmetric polynomials, so always keep those formulas in mind.
1. Given that the polynomial x3 + 3x2 + 4x + 9 has the roots r1 , r2 , and r3 , find the value of r13 + r23 + r33 .
2. Given that the polynomial x3 + 4x2 + 2x + 8 has the roots r1 , r2 , and r3 , find the value of
1
r1
+
1
r2
+
1
r3 .
3. Suppose the polynomial 5x3 + 4x2 − 8x + 6 has three real roots a, b, and c. Find the value of a(1 + b +
c) + b(1 + a + c) + c(1 + a + b).
4. Suppose that the sum of the squares of two complex numbers x and y is 7 and the sum of the cubes
is 10. What is the largest real value that x + y can have? (Source: American Invitational Mathematics
Examination (AIME))
5. Find x, y, and z if
x+y+z =8
2
x + y 2 + z 2 = 62
1
x
+
1
y
+
1
z
= 90
and x ≥ y ≥ z
6.The polynomial x3 − ax2 + bx − 2010 has three positive integer zeros. What is the smallest possible value
of a?(Source: AMC 10 )
7. Let a, b, c, and d be complex numbers satisfying
a + b + c + d = a3 + b3 + c3 + d3 = 0.
Prove that a pair of the a, b, c, and d must add up to 0. (Source: Tournament of Towns)
8 The cubic polynomial x3 + ax2 + bx + c has real coefficients
and three real roots r, s, and t such that
√
r ≥ s ≥ t . Show that k = a2 − 3b ≥ 0 and that k ≤ r − t. (Source: United States of a America
Mathematical Olympiad (USAMO))
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