Algebra · Chapter 14

Polynomial Roots

Write a polynomial through its roots, find remainders without long division, flip and shift roots, and climb power sums with Newton's identities.

What's this chapter about?

14.1 Polynomial Root Representation
14.2 Symmetric Polynomials
14.3 Polynomial Manipulations
14.4 Practice Problems
14.5 Additional Problems

You'll need

open on YouTube

14.10%

Polynomial Root Representation

In this section you will be able to write a polynomial through its roots, and find a remainder by plugging in instead of dividing.

Dividing $x^{51} + 51$ by $x + 1$ by long division takes 51 rounds. Plugging in $x = -1$ takes one.

A polynomial is pinned down by its roots and its leading coefficient: $P(x) = a_n(x - r_1)(x - r_2)\cdots(x - r_n)$ . So any product like $(k - r_1)(k - r_2)\cdots(k - r_n)$ is just $\frac{P(k)}{a_n}$ .

Dividing by $x - r$ leaves a number as the remainder, and that number is $P(r)$ . Synthetic division shows it: the last cell of the bottom row always matches $P(r)$ . When it is $0$ , you have found a factor.

Exploreplay first, formulas later

P(x) = x^{3} - 6x^{2} + 11x - 6

divided by

x - 4

x³1

x²-6

x11

constant-6

Synthetic division of P(x) by x - 4: coefficients, products and the bottom row
4	1	-6	11	-6
4

Shortcut: plug in x = 4

P(4) = 4^{3} - 6 \cdot 4^{2} + 11 \cdot 4 - 6

Remainder = ?

Try this: Press Play to divide x³ − 6x² + 11x − 6 by x − 4. What is the remainder? Now press Evaluate: is P(4) the same?
Try this: Set r to 1, 2 and 3. What does a remainder of 0 tell you about x − r?
Try this: Try r = 0. Why is the remainder always the constant term?

Name it

Theorem 14.1.1 · Representation of a Polynomial in Terms of Its Roots

A polynomial

P(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0

can be written as

a_n(x - r_1)(x - r_2)(x - r_3)\cdots(x - r_n),

where $r_1, r_2, r_3, \ldots, r_n$ are the $n$ roots of the polynomial.

Corollary 14.1.2 · Representation of a Monic Polynomial in Terms of Its Roots

A polynomial

P(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0

(the leading coefficient is $1$ ) can be written as

(x - r_1)(x - r_2)(x - r_3)\cdots(x - r_n),

where $r_1, r_2, r_3, \ldots, r_n$ are the $n$ roots of the polynomial.

Concept 14.1.3 · Evaluating a Product of Differences

When a problem asks for an expression like $(k - r)(k - s)(k - t)$ , where $r$ , $s$ and $t$ are the roots of a monic polynomial $P$ , it is just $P(k)$ , by the representation above. The same works for a polynomial that is not monic, except the product is

\frac{P(k)}{a_n},

where $a_n$ is the coefficient of the $x^n$ term.

Degree3

Lead a1

P(x) = (x - 1)(x - 2)(x - 3) = x^{3} - 6x^{2} + 11x - 6

r = 1,\; s = 2,\; t = 3

sum

r + s + t = -\frac{b}{a} = -\frac{-6}{1} = 6

sum of products of 2

rs + rt + st = \frac{c}{a} = \frac{11}{1} = 11

product

rst = -\frac{d}{a} = -\frac{-6}{1} = 6

evaluate

P(\frac{9}{2}) = (\frac{9}{2} - 1)(\frac{9}{2} - 2)(\frac{9}{2} - 3) = \frac{105}{8}

Theorem 14.1.4 · Fundamental Theorem of Algebra

A polynomial of degree $n$ (its largest term has power $n$ ) has $n$ complex roots, counted with multiplicity. For example, a double root counts as $2$ roots.

Theorem 14.1.5 · Remainder Theorem

The remainder when a polynomial $P(x)$ is divided by $x - r$ is $P(r)$ .

Corollary 14.1.6 · Factor Theorem

$x - r$ divides a polynomial $P(x)$ if $P(r) = 0$ . This is a direct consequence of the Remainder Theorem.

Watch itpredict, then reveal

Example 14.1

AHSME

Find the remainder when $x^{51} + 51$ is divided by $x + 1$ .

P(x) = x^{51} + 51

divided by

x + 1

Synthetic division of P(x) by x + 1: coefficients, products and the bottom row
-1	1	0 ⋯ 050 zeros	51
-1

Shortcut: plug in x = -1

P(-1) = (-1)^{51} + 51

Remainder = ?

1
Long division by $x + 1$ would take $51$ rounds. Which theorem lets you skip it?

Predict first:
Click a choice or type its letter (A–E).

open on YouTube

Example 14.2

USAMO

In the polynomial $x^4 - 18x^3 + kx^2 + 200x - 1984 = 0$ , the product of $2$ of its roots is $-32$ . Find $k$ .

1
By Vieta, what is the product of all four roots?

Predict first:
Your prediction

open on YouTube

Example 14.3

AIME

For distinct complex numbers $z_1, z_2, \ldots, z_{673}$ , the polynomial

(x - z_1)^3(x - z_2)^3 \cdots (x - z_{673})^3

can be expressed as $x^{2019} + 20x^{2018} + 19x^{2017} + g(x)$ , where $g(x)$ is a polynomial with complex coefficients and with degree at most $2016$ . The sum

\left| \sum_{1 \le j < k \le 673} z_j z_k \right|

can be expressed in the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .

1
The polynomial has $2019$ roots: each $z_j$ three times. Their sum is $-20$ by Vieta. What is $S = z_1 + z_2 + \cdots + z_{673}$ ?

Predict first:
Your prediction

open on YouTube

Try it

Problem 14.5.3(additional)

AHSME

Let $P(x)$ be a polynomial such that when $P(x)$ is divided by $x - 19$ , the remainder is $99$ , and when $P(x)$ is divided by $x - 99$ , the remainder is $19$ . What is the remainder when $P(x)$ is divided by $(x - 19)(x - 99)$ ?

Problem 14.4.1

AMC 10

This problem also appears in the book as 13.2.8.

Let $f$ be a function for which $f\left(\frac{x}{3}\right) = x^2 + x + 1$ . Find the sum of all values of $z$ for which $f(3z) = 7$ .

Contest cornerreal contest problems

Problem 14.4.3

AMC 10

This problem also appears in the book as 13.2.13.

For certain real numbers $a$ , $b$ , and $c$ , the polynomial

g(x) = x^3 + ax^2 + x + 10

has three distinct roots, and each root of $g(x)$ is also a root of the polynomial

f(x) = x^4 + x^3 + bx^2 + 100x + c.

What is $f(1)$ ?

Problem 14.4.4

AMC 10/12

Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$ , the remainder is $x + 2$ , and when $P(x)$ is divided by the polynomial $x^2 + 1$ , the remainder is $2x + 1$ . There is a unique polynomial of least degree with these two properties. What is the sum of the squares of the coefficients of that polynomial?

Problem 14.4.5

AMC 10

Let $P(x)$ be the unique polynomial of minimal degree with the following properties:

$P(x)$ has leading coefficient $1$ ,
$1$ is a root of $P(x) - 1$ ,
$2$ is a root of $P(x - 2)$ ,
$3$ is a root of $P(3x)$ , and
$4$ is a root of $4P(x)$ .

The roots of $P(x)$ are integers, with one exception. The root that is not an integer can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m + n$ ?

Checkpoint for section 14.1

Get 2 of 3 right to master this section. One try each, no hints.

14.20%

Symmetric Polynomials

In this section you will be able to solve symmetric polynomials of even degree by dividing by the middle power and substituting y = x + 1/x.

$x + \frac1x = \sqrt5$ . Find $x^{11} - 7x^7 + x^3$ . You will never need $x$ itself.

Look at $x^4 - 5x^3 + 8x^2 - 5x + 1$ : the coefficients read the same forwards and backwards. That is a symmetric polynomial.

Divide by $x^2$ and the terms pair up: $x^2 + \frac{1}{x^2}$ and $x + \frac1x$ . Both are polynomials in $y = x + \frac1x$ , so a degree-4 equation becomes a quadratic.

Exploreplay first, formulas later

Try this: Press Divide by x². Which terms pair up as x^k and 1/x^k?
Try this: Rewrite in y. What quadratic in y is left? Solve it.
Try this: Each y gives a quadratic in x. Why does y = 2 give only x = 1?

Name it

Definition 14.2.1 · Symmetric Polynomials

A polynomial

P(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0

is symmetric if

a_n = a_0,\quad a_{n-1} = a_1,\quad a_{n-2} = a_2,\quad a_{n-3} = a_3,\ \ldots

Basically, opposite coefficients are equal.

Concept 14.2.2 · Solving Symmetric Polynomials of Even Degree

To solve a symmetric polynomial $P(x) = a_0 x^n + a_1 x^{n-1} + \cdots + a_1 x + a_0$ of even degree:

Divide by $x^{n/2}$ .
Group the $x^k$ and $\frac{1}{x^k}$ terms together.
Make the substitution

y = x + \frac{1}{x}

and write every term in your expression with $y$ .

Solve the reduced polynomial.

Watch itpredict, then reveal

Example 14.4

AMC 10

The real number $x$ satisfies the equation $x + \frac{1}{x} = \sqrt{5}$ . What is the value of $x^{11} - 7x^7 + x^3$ ?

1
Factor out $x^7$ . What is left inside the bracket?

open on YouTube

Example 14.5

Omega Learn

Solve for both roots of $10000x^4 - 5000x^3 + 825x^2 - 50x + 1 = 0$ .

1
The coefficients are not symmetric, but compare them: $10000$ with $1$ , and $5000$ with $50$ . Which $k$ gives $10000 = k^2 \cdot 1$ and $5000 = k \cdot 50$ ?

Predict first:
Your prediction

open on YouTube

Contest cornerreal contest problems

Problem 14.5.7(additional)

AIMEChallenge

This problem also appears in the book as 12.6.25.

The equation $2000x^6 + 100x^5 + 10x^3 + x - 2 = 0$ has exactly two real roots, one of which is $\frac{m + \sqrt{n}}{r}$ , where $m$ , $n$ and $r$ are integers, $m$ and $r$ are relatively prime, and $r > 0$ . Find $m + n + r$ .

Checkpoint for section 14.2

Get 2 of 3 right to master this section. One try each, no hints.

14.30%

Polynomial Manipulations

In this section you will be able to build new polynomials whose roots are flipped or shifted, and compute power sums of roots with Newton sums.

Need $\frac1r + \frac1s + \frac1t$ ? Flip the coefficients and read the answer off a new polynomial.

You often don’t need the roots, only a polynomial that has the right roots. Reverse the coefficients and every root $r$ becomes $\frac1r$ . Replace $x$ by $x - k$ and every root moves up by $k$ .

For sums of powers $r^k + s^k + t^k$ , Newton sums give each one from the coefficients and the sums before it.

Exploreplay first, formulas later

Try this: Press Next row. S₁ uses only a₃ and a₂. Which coefficient joins in for S₂?
Try this: The roots are 1, 2 and 3. Does each S_k equal 1 + 2^k + 3^k? Compare with the check column.
Try this: Add a fifth row. Why does the k·a term disappear once k is larger than the degree?

Name it

Concept 14.3.1 · Reciprocal Roots

P(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0

has roots $r_1, r_2, r_3, \ldots, r_n$ , then

Q(x) = a_0 x^n + a_1 x^{n-1} + \cdots + a_{n-1}x + a_n

has roots

\frac{1}{r_1}, \frac{1}{r_2}, \ldots, \frac{1}{r_n}.

Flip the coefficients of a polynomial and its roots become the reciprocals of the original roots.

Degree3

Lead a1

P(x) = (x + 2)(x - 1)(x - 4) = x^{3} - 3x^{2} - 6x + 8

r = -2,\; s = 1,\; t = 4

sum

r + s + t = -\frac{b}{a} = -\frac{-3}{1} = 3

sum of products of 2

rs + rt + st = \frac{c}{a} = \frac{-6}{1} = -6

product

rst = -\frac{d}{a} = -\frac{8}{1} = -8

reciprocal

Q(x) = 8x^{3} - 6x^{2} - 3x + 1

\text{roots } -\frac{1}{2},\ 1,\ \frac{1}{4}

Concept 14.3.2 · Roots That Are More or Less

P(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0

has roots $r_1, r_2, r_3, \ldots, r_n$ , then

Q(x) = a_n (x - k)^n + a_{n-1}(x - k)^{n-1} + \cdots + a_1(x - k) + a_0

has roots

r_1 + k, r_2 + k, r_3 + k, \ldots, r_n + k.

Degree3

Lead a1

Shift k2

P(x) = (x + 2)x(x - 1) = x^{3} + x^{2} - 2x

r = -2,\; s = 0,\; t = 1

sum

r + s + t = -\frac{b}{a} = -\frac{1}{1} = -1

sum of products of 2

rs + rt + st = \frac{c}{a} = \frac{-2}{1} = -2

product

rst = -\frac{d}{a} = -\frac{0}{1} = 0

shifted

Q(x) = P(x - 2) = x^{3} - 5x^{2} + 6x

\text{roots } 0,\ 2,\ 3

Remark 14.3.4

Polynomial manipulations are useful when evaluating complicated expressions in the roots. For example, to evaluate

\frac{1}{(r - 3)^3} + \frac{1}{(s - 3)^3} + \frac{1}{(t - 3)^3}

for a polynomial with roots $r, s, t$ , don't expand and bash with Vieta's formulas. Instead, construct a new polynomial with roots

\frac{1}{r - 3}, \frac{1}{s - 3}, \frac{1}{t - 3}.

Then use standard Vieta tricks to find the sum of the cubes of its roots. One way is the factorization $a^3 + b^3 + c^3 - 3abc = (a + b + c)\left(a^2 + b^2 + c^2 - ab - bc - ca\right)$ in the new polynomial.

Theorem 14.3.5 · Newton Sums

Let $P(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0$ have roots $r_1, r_2, \ldots, r_n$ , and let

S_k = r_1^k + r_2^k + \cdots + r_n^k.

Then

\begin{aligned} a_n S_1 + a_{n-1} &= 0 \\ a_n S_2 + a_{n-1}S_1 + 2a_{n-2} &= 0 \\ a_n S_3 + a_{n-1}S_2 + a_{n-2}S_1 + 3a_{n-3} &= 0 \end{aligned}

and so on. Here $a_j = 0$ whenever $j$ is negative (that can show up once $k > n$ ).

In words:

Multiply $S_k$ by the leading coefficient $a_n$ .
Multiply $S_{k-1}$ by the next coefficient $a_{n-1}$ , then $S_{k-2}$ by $a_{n-2}$ , and so on, adding the products.
Stop when you reach $S_0$ : instead of $S_0$ , multiply that last coefficient by $k$ . (If the coefficients run out first, just stop.)
Set the sum equal to $0$ .

Watch itpredict, then reveal

Example 14.6

AMC 12

Let $g(x)$ be a polynomial with leading coefficient $1$ , whose three roots are the reciprocals of the three roots of $f(x) = x^3 + ax^2 + bx + c$ , where $1 < a < b < c$ . What is $g(1)$ in terms of $a$ , $b$ , and $c$ ?

P(x) = x^{3} + 2x^{2} + 3x + 4

r \approx -1.651,\; s \approx -0.175 + 1.547i,\; t \approx -0.175 - 1.547i

sum

r + s + t = -\frac{b}{a} = -\frac{2}{1} = -2

sum of products of 2

rs + rt + st = \frac{c}{a} = \frac{3}{1} = 3

product

rst = -\frac{d}{a} = -\frac{4}{1} = -4

evaluate

Q(1) = ?

reciprocal

Q(x) = x^{3} + \frac{3}{4}x^{2} + \frac{1}{2}x + \frac{1}{4}

\text{roots } \approx -0.606,\ \approx -0.072 - 0.638i,\ \approx -0.072 + 0.638i

1
By Concept 14.3.1, flip the coefficients of $f$ . Which polynomial has roots $\frac1r, \frac1s, \frac1t$ ?

Predict first:
Click a choice or type its letter (A–E).

open on YouTube

Try it

Problem 14.5.1(additional)

AHSME

Given the equations $x^2 + kx + 6 = 0$ and $x^2 - kx + 6 = 0$ . If, when the roots of the equation are suitably listed, each root of the second equation is $5$ more than the corresponding root of the first equation, then $k$ equals:

Problem 14.4.2

AIME

This problem also appears in the book as 13.2.12.

Suppose that the roots of $x^3 + 3x^2 + 4x - 11 = 0$ are $a$ , $b$ , and $c$ , and that the roots of $x^3 + rx^2 + sx + t = 0$ are $a + b$ , $b + c$ , and $c + a$ . Find $t$ .

Contest cornerreal contest problems

Problem 14.5.4(additional)

AIME

The real root of the equation $8x^3 - 3x^2 - 3x - 1 = 0$ can be written in the form $\frac{\sqrt[3]{a} + \sqrt[3]{b} + 1}{c}$ , where $a$ , $b$ , and $c$ are positive integers. Find $a + b + c$ .

Problem 14.5.5(additional)

AIME

Let $x_1 < x_2 < x_3$ be the three real roots of the equation $\sqrt{2014}\,x^3 - 4029x^2 + 2 = 0$ . Find $x_2(x_1 + x_3)$ .

Problem 14.5.2(additional)

AMC 10

What is the product of all the roots of the equation

\sqrt{5|x| + 8} = \sqrt{x^2 - 16}?

Checkpoint for section 14.3

Get 2 of 3 right to master this section. One try each, no hints.

14.4

Practice Problems

In this section you will be able to use roots, remainders and signs together on a harder contest problem.

The other practice problems sit in the lessons where their tools live. This one needs the sign of a fraction of two polynomials: where it is zero, where it is undefined, and where it flips.

Want every problem from this chapter on one page? Open the problem set.

Contest cornerreal contest problems

Problem 14.4.6

AMC 12Challenge

Polynomials $P(x)$ and $Q(x)$ each have degree $3$ and leading coefficient $1$ , and their roots are all elements of $\{1, 2, 3, 4, 5\}$ . The function $f(x) = \frac{P(x)}{Q(x)}$ has the property that there exist real numbers $a < b < c < d$ such that the set of all real numbers $x$ such that $f(x) \le 0$ consists of the closed interval $[a, b]$ together with the open interval $(c, d)$ . How many functions $f(x)$ are possible?

14.5

Additional Problems

In this section you will be able to mix every idea of the chapter on challenge problems.

Double roots, compositions $P(P(x))$ and functional equations. Each problem has three hints if you get stuck.

Look for a hidden factor first: a root you can plug in, a square, or a shift that makes the polynomial simpler.

Contest cornerreal contest problems

Problem 14.5.6(additional)

AMC 12

The graph of $y = x^6 - 10x^5 + 29x^4 - 4x^3 + ax^2$ lies above the line $y = bx + c$ except at three values of $x$ , where the graph and the line intersect. What is the largest of these values?

Problem 14.5.8(additional)

AIMEChallenge

Let $P(x)$ be a nonzero polynomial such that $(x - 1)P(x + 1) = (x + 2)P(x)$ for every real $x$ , and $\left(P(2)\right)^2 = P(3)$ . Then $P\left(\frac{7}{2}\right) = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .

Problem 14.5.9(additional)

AIMEChallenge

Let $P(x)$ be a quadratic polynomial with complex coefficients whose $x^2$ coefficient is $1$ . Suppose the equation $P(P(x)) = 0$ has four distinct solutions, $x = 3, 4, a, b$ . Find the sum of all possible values of $(a + b)^2$ .

Problem 14.5.10(additional)

AMC 10Challenge

A quadratic polynomial with real coefficients and leading coefficient $1$ is called disrespectful if the equation $p(p(x)) = 0$ is satisfied by exactly three real numbers. Among all the disrespectful quadratic polynomials, there is a unique such polynomial $\tilde{p}(x)$ for which the sum of the roots is maximized. What is $\tilde{p}(1)$ ?