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Zeros of Polynomials: Star problems

Ten optional challenges to stretch your reasoning. Work on paper, use hints when you need them, and check the answer or full solution when you are ready. You can skip these problems and continue the course.

1 of 3 stars: Stretch
2 of 3 stars: Challenge
3 of 3 stars: Deep challenge

Stars indicate difficulty within this set.

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Problem 1 of 10

Problem 1 The smallest polynomial

Difficulty: 1 of 3 stars, Stretch

A monic polynomial $P$ has rational coefficients. The number $2+\sqrt3$ is a zero of multiplicity at least $2$ , and $1+2i$ is a zero. Find the smallest possible degree of $P$ , determine the unique polynomial of that degree in factored form, and find $P(1)$ .

Justify why the required conjugate zeros must also have the required multiplicities.
1. Hint 1
  Apply the irrational-conjugate and complex-conjugate root principles before counting the degree.
2. Hint 2
  The quadratic with roots $2\pm\sqrt3$ is $x^2-4x+1$ . Divide repeatedly by this quadratic, keeping track of the remaining multiplicity.
Answer

Degree $6$ ; $P(x)=(x^2-4x+1)^2(x^2-2x+5)$ ; $P(1)=16$ .

Full solution

For a rational-coefficient polynomial, division by $q(x)=x^2-4x+1$ leaves a remainder $ux+v$ with rational $u,v$ .
Substitution of $2+\sqrt3$ gives
$u(2+\sqrt3)+v=0.$
Irrationality forces $u=v=0$ , so $q$ divides $P$ .
The root $2+\sqrt3$ is simple in $q$ ; therefore, after one division, it is still a zero of the rational-coefficient quotient.
The same argument shows that $q^2$ divides $P$ .
Thus $2-\sqrt3$ also occurs at least twice.

Because the coefficients are real, conjugating $P(1+2i)=0$ gives $P(1-2i)=0$ .
These two zeros are distinct from the four zeros already counted with multiplicity.
Their quadratic factor is
$(x-1)^2+4=x^2-2x+5.$
Hence the degree is at least $6$ .

The displayed product has rational coefficients, is monic, and meets every requirement with degree $6$ .
No additional factor is possible at that degree, proving uniqueness.
At $x=1$ , the factors give
$(-2)^2\cdot4=16.$

Answer

Degree $6$ ; $P(x)=(x^2-4x+1)^2(x^2-2x+5)$ ; $P(1)=16$ .

Key idea

Conjugate-root requirements constrain multiplicities as well as the list of distinct zeros.
Problem 1: Mark complete
Problem 2 A sum with an unexpected sign

Difficulty: 1 of 3 stars, Stretch

Let $r,s,t$ be the three complex zeros, counted with multiplicity, of $P(x)=x^3-5x^2+7x-1$ . Without finding the zeros individually, evaluate
$\frac1{(r-2)^2}+\frac1{(s-2)^2}+\frac1{(t-2)^2}.$
Use your result to prove that $P$ does not have three real zeros.
1. Hint 1
  Shift each zero by $2$ before taking reciprocals.
2. Hint 2
  Compute $P(u+2)$ , then substitute $u=1/v$ and clear $v^3$ . Apply Vieta to the polynomial whose zeros are the three reciprocals.
Answer

The sum is $-1$ . Consequently, two of the zeros are nonreal.

Full solution

Since $P(2)=1$ , none of the denominators vanishes.
Writing $x=u+2$ gives
$P(u+2)=u^3+u^2-u+1.$
Thus the shifted zeros $r-2,s-2,t-2$ are nonzero zeros of this cubic.

For $v=1/u$ , division by $u^3$ transforms the equation into
$1+v-v^2+v^3=0.$
Its three zeros, with multiplicity, are therefore $A=1/(r-2)$ , $B=1/(s-2)$ , and $C=1/(t-2)$ .
Vieta gives $A+B+C=1$ and
$AB+BC+CA=1.$
Consequently,
$A^2+B^2+C^2=(A+B+C)^2-2(AB+BC+CA)=1-2=-1.$

If all three original zeros were real, then $A,B,C$ would be real, and their squared sum could not be negative.
So at least one zero is nonreal.
Nonreal zeros of a real-coefficient polynomial occur in conjugate pairs; in this cubic, exactly two zeros are therefore nonreal.

Answer

The sum is $-1$ . Consequently, two of the zeros are nonreal.

Key idea

A symmetric expression can reveal the nature of unknown roots without solving for them.
Problem 2: Mark complete
Problem 3 Recovering multiplicities from a graph

Difficulty: 1 of 3 stars, Stretch

A monic real polynomial has degree $8$ and no complex zeros other than the real numbers $-2,1,3$ . Its graph crosses the horizontal axis at $-2$ and $3$ , and touches the axis without crossing at $1$ . The coefficient of $x^7$ is $-11$ . Determine the polynomial and prove uniqueness.
1. Hint 1
  Let the multiplicities of $-2,1,3$ be $a,b,c$ . Crossing and touching determine their parity.
2. Hint 2
  Combine $a+b+c=8$ with the sum of all eight zeros, counted with multiplicity. Eliminate $b$ first.
Answer

$P(x)=(x+2)(x-1)^4(x-3)^3$ .

Full solution

At a real zero, an odd multiplicity reverses the sign of a polynomial, while an even multiplicity preserves it.
Thus $a,c$ are positive odd integers and $b$ is a positive even integer.
The fundamental theorem of algebra gives $a+b+c=8$ , because the statement excludes every other zero.

For a monic degree- $8$ polynomial, the coefficient of $x^7$ is the negative of the sum of the eight zeros, counted with multiplicity.
Hence
$-2a+b+3c=11.$
Substituting $b=8-a-c$ yields $2c=3a+3$ .

Since $b\ge2$ and $c\ge1$ , the positive odd integer $a$ can only be $1,3,$ or $5$ .
These give $c=3,6,$ or $9$ , respectively.
The middle value is not odd, and the last exceeds the available degree.
Therefore
$a=1,c=3,b=4.$
The proposed product is monic of degree $8$ , has exactly the required crossing behavior, and its root sum is $-2+4+9=11$ .
This verifies existence as well as uniqueness.

Answer

$P(x)=(x+2)(x-1)^4(x-3)^3$ .

Key idea

Graph behavior and coefficient information impose different constraints that can determine multiplicities together.
Problem 3: Mark complete
Problem 4 Squaring every zero

Difficulty: 2 of 3 stars, Challenge

The zeros of $P(x)=x^4-2x^3-7x^2+8x+12$ , counted with multiplicity, are $r_1,r_2,r_3,r_4$ . Construct the monic polynomial $Q$ of degree $4$ whose zeros are $r_1^2,r_2^2,r_3^2,r_4^2$ .

Use the even and odd parts of $P$ to construct $Q$ before factoring $P$ . Then identify the distinct zeros of $Q$ and their multiplicities, and explain why a repeated zero of $Q$ need not come from a repeated zero of $P$ .
1. Hint 1
  Examine $P(t)P(-t)$ ; all odd powers of $t$ cancel.
2. Hint 2
  Write $P(t)=(t^4-7t^2+12)-2t(t^2-4)$ and use a difference of squares.
Answer

$Q(z)=(z-4)^2(z-1)(z-9)=z^4-18z^3+105z^2-232z+144$ . Its zeros are $1,4,9$ , with multiplicities $1,2,1$ .

Full solution

Because the degree is even, $P(t)P(-t)=\prod_{j=1}^4(t-r_j)(t+r_j)=\prod_{j=1}^4(t^2-r_j^2)$ .
Thus replacing $t^2$ by $z$ in this product produces exactly the desired monic polynomial, including multiplicities.

Separating the even and odd parts gives
$P(t)P(-t)=(t^4-7t^2+12)^2-4t^2(t^2-4)^2.$
Therefore
$Q(z)=(z^2-7z+12)^2-4z(z-4)^2.$
Since $z^2-7z+12=(z-3)(z-4)$ , this becomes
$(z-4)^2((z-3)^2-4z)=(z-4)^2(z-1)(z-9).$
Expansion gives the stated coefficients.

The zero $4$ occurs twice in $Q$ .
In fact, $P(x)=(x^2-4)(x^2-2x-3)=(x-2)(x+2)(x-3)(x+1)$ , whose four zeros are all simple.
The distinct zeros $2$ and $-2$ merge under squaring.
A transformation of roots can create multiplicity even when the starting polynomial has none.

Answer

$Q(z)=(z-4)^2(z-1)(z-9)=z^4-18z^3+105z^2-232z+144$ . Its zeros are $1,4,9$ , with multiplicities $1,2,1$ .

Key idea

When transforming roots, count how many original roots map to the same output.
Problem 4: Mark complete
Problem 5 Lifting zeros through a parabola

Difficulty: 2 of 3 stars, Challenge

Let $P(t)=(t+2)(t-1)(t-5)$ . For a real parameter $k$ , define the degree- $6$ polynomial $Q_k(x)=P(x^2-4x+k)$ .

For every real $k$ , determine the number of distinct real zeros of $Q_k$ and all their multiplicities. Also give the number of nonreal zeros, counted with multiplicity. Explain the changes using the least value of the inner quadratic, without expanding the degree- $6$ polynomial.
1. Hint 1
  A zero of $Q_k$ occurs exactly when $x^2-4x+k$ equals one of the three zeros of $P$ .
2. Hint 2
  Write the inner quadratic as $(x-2)^2+k-4$ . For each target value, decide whether its two preimages are distinct real numbers, one repeated real number, or a nonreal pair.
Answer

For $k<2, k=2, 2<k<5, k=5, 5<k<9, k=9, k>9$ , the distinct real-zero counts are respectively $6,5,4,3,2,1,0$ . At each boundary $k=2,5,9$ , $x=2$ has multiplicity $2$ ; all other real zeros are simple. The nonreal-zero counts with multiplicity are respectively $0,0,2,2,4,4,6$ .

Full solution

The factorization of $P$ gives
$Q_k(x)=((x-2)^2+k-2)((x-2)^2+k-5)((x-2)^2+k-9).$
Thus the three root equations are
$(x-2)^2=2-k,5-k,9-k.$
They correspond geometrically to intersections of the inner parabola, whose minimum is $k-4$ , with the heights $-2,1,5$ .

For each threshold $h\in\{2,5,9\}$ , if $k<h$ the corresponding factor supplies the two distinct real zeros $2\pm\sqrt{h-k}$ .
If $k=h$ , it is $(x-2)^2$ and supplies a double zero at $2$ .
If $k>h$ , it supplies the nonreal pair $2\pm i\sqrt{k-h}$ .
Zeros from different factors cannot coincide, since that would require $(x-2)^2$ to equal two different numbers.

Counting these contributions gives, in increasing parameter order, $6,5,4,3,2,1,0$ distinct real zeros across the seven cases in the answer.
The associated nonreal counts are $0,0,2,2,4,4,6$ .
At a boundary the one double zero contributes two to the total multiplicity; all other zeros are simple.
Hence the real and nonreal multiplicities always total $6$ , as required.
The root count changes when the vertex of the inner parabola reaches one of the three target heights.

Answer

For $k<2, k=2, 2<k<5, k=5, 5<k<9, k=9, k>9$ , the distinct real-zero counts are respectively $6,5,4,3,2,1,0$ . At each boundary $k=2,5,9$ , $x=2$ has multiplicity $2$ ; all other real zeros are simple. The nonreal-zero counts with multiplicity are respectively $0,0,2,2,4,4,6$ .

Key idea

In a composed polynomial, each outer zero supplies preimages whose number and multiplicity depend on the inner function.
Problem 5: Mark complete
Problem 6 Three evenly spaced zeros

Difficulty: 2 of 3 stars, Challenge

Prove that the zeros of the real-coefficient cubic $x^3+Ax^2+Bx+C$ , counted with multiplicity, can be arranged in an arithmetic progression of complex numbers if and only if
$2A^3-9AB+27C=0.$
Here an arithmetic progression means $m-d,m,m+d$ , where $m,d$ may be complex and $d=0$ is allowed. Apply your criterion to find every real $k$ for which $x^3-6x^2+kx+12$ has such zeros. Find the zeros in that case and say whether they are real.
1. Hint 1
  The middle zero of an arithmetic progression must be one third of the sum of all three zeros.
2. Hint 2
  Conversely, if $m=-A/3$ is a zero, the two remaining zeros sum to $2m$ , so they can be written as $m-d$ and $m+d$ .
Answer

The criterion is necessary and sufficient. The application gives $k=2$ , with zeros $2-\sqrt{10},2,2+\sqrt{10}$ , all real.

Full solution

If the zeros are $m-d,m,m+d$ , their sum is $3m=-A$ , so $m=-A/3$ is itself a zero.
Substitution into the cubic gives
$P(-A/3)=(2A^3-9AB+27C)/27.$
This proves necessity.

For sufficiency, suppose the displayed coefficient expression is zero.
Then $m=-A/3$ is a zero, so factor $P(x)=(x-m)Q(x)$ with $Q$ monic quadratic.
Its two complex zeros, say $u,v$ , exist and are counted with multiplicity.
Their sum is $-A-m=2m$ .
Setting $d=(v-u)/2$ gives $u=m-d$ and $v=m+d$ .
This includes a repeated or triple zero without any special division by $d$ .

In the application, the middle zero must be $2$ .
Evaluating the cubic at $2$ yields $2k-4$ , so $k=2$ is necessary and, by the criterion, sufficient.
At that parameter, setting $x=y+2$ gives
$y^3-10y=y(y^2-10).$
The three zeros are therefore $2$ and $2\pm\sqrt{10}$ , all real.

Answer

The criterion is necessary and sufficient. The application gives $k=2$ , with zeros $2-\sqrt{10},2,2+\sqrt{10}$ , all real.

Key idea

A geometric arrangement of roots can often be expressed as one strategically chosen polynomial evaluation.
Problem 6: Mark complete
Problem 7 Four zeros on a circle

Difficulty: 2 of 3 stars, Challenge

For each real $k$ , let $P_k(z)=z^4-2z^3+kz^2-2z+1$ . Find all $k$ for which all four complex zeros, counted with multiplicity, have modulus $1$ . At the boundary values of your answer, list the zeros and their multiplicities.
1. Hint 1
  The constant term shows that $z=0$ is never a zero. Divide by $z^2$ and set $t=z+1/z$ .
2. Hint 2
  For a complex number $z$ of modulus $1$ , $1/z=\overline z$ , so $t$ is real and lies in $[-2,2]$ . Conversely, analyze $z^2-tz+1=0$ when $t\in[-2,2]$ .
Answer

$2\le k\le3$ . At $k=2$ : $1$ has multiplicity $2$ , and $i,-i$ are simple. At $k=3$ : $(1+i\sqrt3)/2$ and $(1-i\sqrt3)/2$ each have multiplicity $2$ .

Full solution

Dividing by $z^2$ gives
$(z+1/z)^2-2(z+1/z)+k-2=0.$
Thus the two values of $t=z+1/z$ are the zeros of $t^2-2t+k-2$ .
If $|z|=1$ , then
$t=z+\overline z=2\operatorname{Re}z\in[-2,2].$
Conversely, for real $t$ in this interval, the roots of $z^2-tz+1$ are $(t\pm i\sqrt{4-t^2})/2$ , both of modulus $1$ ; the formula includes the repeated roots at the endpoints.

The two $t$ -values are $1\pm\sqrt{3-k}$ .
They are real exactly when $k\le3$ .
For both to lie in $[-2,2]$ , the upper value must be at most $2$ , which requires $\sqrt{3-k}\le1$ , or $k\ge2$ .
This also guarantees the lower value is at least $0$ , hence at least $-2$ .
Both conditions are therefore equivalent to
$2\le k\le3.$

At $k=2$ , the $t$ -values are $2,0$ , giving
$P_2(z)=(z-1)^2(z^2+1).$
At $k=3$ , both $t$ -values equal $1$ , giving
$P_3(z)=(z^2-z+1)^2.$
These factorizations establish the stated multiplicities.

Answer

$2\le k\le3$ . At $k=2$ : $1$ has multiplicity $2$ , and $i,-i$ are simple. At $k=3$ : $(1+i\sqrt3)/2$ and $(1-i\sqrt3)/2$ each have multiplicity $2$ .

Key idea

A reciprocal substitution links unit-modulus complex roots to a bounded real interval.
Problem 7: Mark complete
Problem 8 Three moments of four roots

Difficulty: 3 of 3 stars, Deep challenge

Real numbers $a,b,c,d$ satisfy
$a+b+c+d=0,\qquad a^2+b^2+c^2+d^2=12,\qquad a^3+b^3+c^3+d^3=0.$
Prove that the four numbers, counted with repetition, can be grouped into opposite pairs. Then find sharp lower and upper bounds for $abcd$ , and identify every equality case.
1. Hint 1
  Use the first two sums to find the pairwise product sum. Expand the cube of the first sum to recover the sum of the triple products.
2. Hint 2
  Build the monic polynomial having these four numbers as its roots. Which coefficients vanish? Then write the roots as $u,-u,v,-v$ .
Answer

The roots are $u,-u,v,-v$ with $u^2+v^2=6$ . Hence $0\le abcd\le9$ . The lower case is any permutation of $(0,0,\sqrt6,-\sqrt6)$ ; the upper case is any permutation of $(\sqrt3,\sqrt3,-\sqrt3,-\sqrt3)$ .

Full solution

Let $e_2$ be the sum of all six pairwise products and $e_3$ the sum of all four triple products.
Squaring the zero sum gives $0=12+2e_2$ , so $e_2=-6$ .
Expanding a cube and collecting terms gives the identity $a^3+b^3+c^3+d^3=e_1^3-3e_1e_2+3e_3$ , where
$e_1=a+b+c+d.$
The given sums therefore imply $e_3=0$ .

The monic polynomial with these roots is consequently $P(x)=x^4-6x^2+abcd$ , which is even.
A zero $r$ of an even polynomial has $-r$ as a zero with the same multiplicity: replacing $x$ by $-x$ preserves the polynomial and its linear factors.
Any zero at $0$ has even multiplicity because only even powers occur.
Since all four given roots are real, they can thus be written $u,-u,v,-v$ .

The square-sum condition gives $u^2+v^2=6$ , while
$abcd=u^2v^2\ge0.$
Also $(u^2-v^2)^2\ge0$ implies
$u^2v^2\le(6/2)^2=9.$
Equality at the lower end means one squared value is $0$ and the other $6$ ; equality at the upper end means both are $3$ .
The resulting listed roots satisfy all three original sums.

Answer

The roots are $u,-u,v,-v$ with $u^2+v^2=6$ . Hence $0\le abcd\le9$ . The lower case is any permutation of $(0,0,\sqrt6,-\sqrt6)$ ; the upper case is any permutation of $(\sqrt3,\sqrt3,-\sqrt3,-\sqrt3)$ .

Key idea

Power sums can force symmetry of the entire root set before individual roots are known.
Problem 8: Mark complete
Problem 9 The distances between three unknown roots

Difficulty: 3 of 3 stars, Deep challenge

The polynomial $P(x)=x^3-3x-1$ has real zeros $a,b,c$ . Prove that they are distinct, and construct the monic cubic whose zeros are $(a-b)^2,(b-c)^2,(c-a)^2$ . Do this without finding $a,b,c$ individually.

You may use the fact that a continuous polynomial graph with opposite signs at two inputs has a zero between them.
1. Hint 1
  First locate three zeros in disjoint intervals. Next set $u=a-b,v=b-c,w=c-a$ and exploit $u+v+w=0$ .
2. Hint 2
  For the product of squared differences, use $(a-b)(a-c)=3a^2-3$ , then multiply the three corresponding identities. Evaluate $P(1)$ and $P(-1)$ to simplify the product.
Answer

The required polynomial is $Q(t)=t^3-18t^2+81t-81$ .

Full solution

The values $P(-2)=-3$ , $P(-1)=1$ , $P(0)=-1$ , $P(1)=-3$ , and $P(2)=1$ locate a zero in each of $(-2,-1),(-1,0),(1,2)$ .
A cubic has at most three distinct zeros, so these account for all of them and prove distinctness.

Vieta gives $a+b+c=0$ , $ab+bc+ca=-3$ , and $abc=1$ .
Thus
$a^2+b^2+c^2=6.$
Set
$u=a-b,v=b-c,w=c-a.$
Then $u+v+w=0$ and
$u^2+v^2+w^2=3(a^2+b^2+c^2)=18.$
Hence $uv+vw+wu=-9$ , and
$u^2v^2+v^2w^2+w^2u^2=(uv+vw+wu)^2-2uvw(u+v+w)=81.$

To find the remaining coefficient, $b+c=-a$ and $bc=a^2-3$ give
$(a-b)(a-c)=3a^2-3.$
Multiplying the three analogous identities yields
$-(a-b)^2(b-c)^2(c-a)^2=27\prod(a^2-1).$
Here $\prod(a-1)=-P(1)=3$ and $\prod(a+1)=-P(-1)=-1$ , so the product of squared differences is $81$ .
Vieta now gives
$Q(t)=t^3-18t^2+81t-81.$
The minus sign in the product identity comes from reversing one factor in each of three unordered pairs.

Answer

The required polynomial is $Q(t)=t^3-18t^2+81t-81$ .

Key idea

Build a polynomial for derived quantities by computing their symmetric sums directly.
Problem 9: Mark complete
Problem 10 A root set closed under two moves

Difficulty: 3 of 3 stars, Deep challenge

Seek a monic polynomial $P$ with real coefficients having $3$ as a zero. Every complex zero $r$ of $P$ must satisfy $r\ne0,1$ , and both $1-r$ and $1/r$ must also be zeros. These closure conditions concern which numbers are zeros, without prescribing their multiplicities.

Find the smallest possible degree and the unique polynomial of that degree, in factored form. Determine its constant term and its next-to-leading coefficient without fully expanding. Prove that your list of zeros is closed under both moves.
1. Hint 1
  Start from $3$ and repeatedly apply the two moves until no new number appears.
2. Hint 2
  Organize the resulting numbers into pairs with sum $1$ . Check reciprocals as a separate pairing.
Answer

Degree $6$ ; $P(x)=(x^2-x-6)(x^2-x+2/9)(x^2-x-3/4)$ . Its constant term is $1$ and its $x^5$ coefficient is $-3$ .

Full solution

Starting from $3$ , the two closure rules force $-2$ and $1/3$ .
They then force $-1/2$ , $2/3$ , and $3/2$ .
These are six distinct numbers, none equal to $0$ or $1$ , so every eligible polynomial has degree at least $6$ .

The six numbers can be paired under $r\mapsto1-r$ as $(3,-2)$ , $(1/3,2/3)$ , and $(-1/2,3/2)$ .
They can also be paired under $r\mapsto1/r$ as $(3,1/3)$ , $(-2,-1/2)$ , and $(2/3,3/2)$ .
Thus applying either move to any number in the set stays inside the set; no further zero is forced.

The monic product over these six distinct linear factors therefore works and has the minimum degree.
Pairing the factors with roots summing to $1$ gives the displayed three quadratics.
At degree $6$ each of the six forced zeros must occur once, so the polynomial is unique.
Each pair has sum $1$ , making the total root sum $3$ and the coefficient of $x^5$ equal to $-3$ .
Multiplying the three quadratic constants gives $(-6)(2/9)(-3/4)=1$ , the constant term.

Answer

Degree $6$ ; $P(x)=(x^2-x-6)(x^2-x+2/9)(x^2-x-3/4)$ . Its constant term is $1$ and its $x^5$ coefficient is $-3$ .

Key idea

When roots obey transformations, determine the entire forced orbit before trying to compute coefficients.
Problem 10: Mark complete