Root theory

This chapter delves into the foundational principles of root theory, providing essential tools for the analysis and manipulation of polynomial equations. A thorough understanding of these concepts, including theorems for root identification and characterisation, is critical for success in CMI examinations, where they underpin various algebraic problem-solving tasks.

---

Chapter Contents

| Topic |

|---|-------| | 1 | Remainder theorem | | 2 | Factor theorem | | 3 | Vieta relations | | 4 | Rational roots | | 5 | Integer roots | | 6 | Multiple roots |

---

We begin with Remainder theorem.

Part 1: Remainder theorem

Remainder theorem

Overview

The Remainder Theorem is one of the fastest tools in polynomial algebra. Instead of performing full long division, it lets us find the remainder after division by a linear factor using just substitution. In CMI-style questions, this saves time and also connects directly to the Factor Theorem, root testing, and polynomial structure. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

Find the remainder when a polynomial is divided by $x-a$ .

Use the substitution shortcut for divisors like $x+1$ , $x-2$ , and more generally $ax+b$ .

Apply the Factor Theorem to detect roots and factors.

Avoid sign mistakes in expressions like $x+1$ and $x-1$ .

Use the theorem efficiently in exam-style questions without full division.

---

Core Idea

📖 Remainder theorem

If a polynomial $p(x)$ is divided by $x-a$ , then the remainder is

$\qquad p(a)$

So the remainder can be found by simply substituting $x=a$ into the polynomial.

📐 Main identity

When $p(x)$ is divided by $x-a$ , we can write

$\qquad p(x)=(x-a)q(x)+r$

where $q(x)$ is a polynomial and $r$ is a constant remainder.

Now substitute $x=a$ :

$\qquad p(a)=(a-a)q(a)+r=r$

Hence the remainder is

$\qquad \boxed{p(a)}$

---

Special Cases You Must Know

📐 Quick substitution rules

divisor $x-a$ $\to$ remainder $p(a)$

divisor $x+a$ $\to$ remainder $p(-a)$

divisor $x-1$ $\to$ remainder $p(1)$

divisor $x+1$ $\to$ remainder $p(-1)$

⚠️ Most Common Sign Error

If the divisor is $x+1$ , do not substitute $1$ .

Since
$\qquad x+1=x-(-1)$

the correct substitution is

$\qquad x=-1$

---

Factor Theorem

📐 Factor theorem

A polynomial $p(x)$ has factor $x-a$ if and only if

$\qquad p(a)=0$

So:

if $p(a)=0$ , then $x-a$ is a factor

if $x-a$ is a factor, then $a$ is a root of $p(x)$

❗ Connection

The Remainder Theorem gives the remainder $p(a)$ .

The Factor Theorem is the special case when that remainder is zero.

---

General Linear Divisor

📐 Division by

ax+b

If a polynomial $p(x)$ is divided by $ax+b$ , then the remainder is

$\qquad p\left(-\dfrac{b}{a}\right)$

because $ax+b=0$ gives

$\qquad x=-\dfrac{b}{a}$

Example If the divisor is

2x-3

, then the remainder is

\qquad p\left(\dfrac{3}{2}\right)

---

Minimal Worked Examples

Example 1 Find the remainder when

\qquad p(x)=x^4+2x^2-5

is divided by

x-2

. By the Remainder Theorem,

\qquad \text{remainder}=p(2)

\qquad p(2)=2^4+2(2^2)-5=16+8-5=19

So the remainder is

\boxed{19}

. --- Example 2 Find the remainder when

\qquad p(x)=x^5-3x+4

is divided by

x+1

. Since

x+1=x-(-1)

, the remainder is

\qquad p(-1)=(-1)^5-3(-1)+4=-1+3+4=6

So the remainder is

\boxed{6}

. ---

Fast Evaluation Strategy

💡 CMI Strategy

Convert the divisor into the form $x-a$ mentally.

Substitute carefully with the correct sign.

For large powers at $x=1$ or $x=-1$ , use parity:

(-1)^{\text{even}}=1

(-1)^{\text{odd}}=-1

In longer polynomials, group terms only if it reduces arithmetic mistakes.

If remainder is zero, immediately think: factor, root, and further factorisation.

---

Pattern Recognition

📐 Useful quick values

If dividing by:

$x$ $\to$ remainder $p(0)$

$x-1$ $\to$ remainder $p(1)$

$x+1$ $\to$ remainder $p(-1)$

These are extremely common in exam questions.

💡 Why

x+1

questions are easy

At $x=-1$ , powers alternate:

$(-1)^0=1$

$(-1)^1=-1$

$(-1)^2=1$

$(-1)^3=-1$

So remainders for division by

x+1

are often quick to compute even for high-degree polynomials.

---

Common Errors

⚠️ Avoid These Errors

❌ using $p(1)$ for division by $x+1$

❌ forgetting that the remainder on division by a linear polynomial is a constant

❌ treating the theorem as valid for non-linear divisors like $x^2+1$

❌ concluding factor without checking remainder is exactly zero

❌ arithmetic mistakes while substituting negative values

✅ Correct habits:

rewrite divisor as $x-a$

substitute slowly

use parity for powers of $-1$

connect zero remainder with factor theorem

---

Practice Questions

:::question type="MCQ" question="Find the remainder when

p(x)=2x^3-5x+7

is divided by

x-2

." options=["

9

","

13

","

15

","

17

"] answer="B" hint="Use the substitution

x=2

." solution="By the Remainder Theorem, the remainder is

\qquad p(2)=2(2^3)-5(2)+7=16-10+7=13

Hence the correct option is

\boxed{B}

." ::: :::question type="NAT" question="Find the remainder when

x^5+3x^2-7

is divided by

x+1

." answer="-5" hint="For division by

x+1

, substitute

x=-1

." solution="The remainder is

\qquad p(-1)=(-1)^5+3(-1)^2-7=-1+3-7=-5

Hence the answer is

\boxed{-5}

." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["If a polynomial

p(x)

is divided by

x-a

, the remainder is

p(a)

.","If

p(a)=0

, then

x+a

is a factor of

p(x)

.","If

p(-1)=3

, then the remainder when

p(x)

is divided by

x+1

3

.","If

x-a

is a factor of

p(x)

, then

p(a)=0

."] answer="A,C,D" hint="Use the remainder theorem and factor theorem carefully." solution="1. True. This is exactly the Remainder Theorem.

False. If

p(a)=0

, then

x-a

is a factor, not

x+a

True. Division by

x+1

means substitute

x=-1

True. This is the Factor Theorem.

Hence the correct answer is

\boxed{A,C,D}

." ::: :::question type="SUB" question="Find the value of

k

if the polynomial

p(x)=x^3+kx^2-4x+7

leaves remainder

10

when divided by

x-1

." answer="6" hint="Use the fact that the remainder on division by

x-1

p(1)

." solution="By the Remainder Theorem, the remainder on division by

x-1

\qquad p(1)

So,

\qquad p(1)=1^3+k(1)^2-4(1)+7=1+k-4+7=k+4

We are given that this remainder is

10

, so

\qquad k+4=10

Hence

\qquad k=6

Therefore, the answer is

\boxed{6}

." ::: ---

Summary

❗ Key Takeaways for CMI

The remainder on division by $x-a$ is $p(a)$ .

The divisor $x+1$ corresponds to substitution at $x=-1$ .

Zero remainder means factor.

Factor theorem is the zero-remainder version of remainder theorem.

For divisor $ax+b$ , substitute $x=-\dfrac{b}{a}$ .

Fast substitution is often much better than long division in exam settings.

---

💡 Next Up

Proceeding to Factor theorem.

---

Part 2: Factor theorem

Factor Theorem

Overview

The factor theorem is one of the most powerful bridges between roots and divisibility of polynomials. In exam problems, it is rarely used in isolation. It appears in proving divisibility, constructing polynomials from given roots, identifying repeated roots, and reducing higher-degree expressions by extracting known factors. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

use the factor theorem to test whether $(x-a)$ divides a polynomial,

apply the remainder theorem quickly,

prove that distinct roots give a product factor,

handle repeated roots and multiplicity,

use factor-based reasoning in algebraic and proof-style questions.

---

Core Theorem

📖 Factor Theorem

Let $p(x)$ be a polynomial and let $a$ be a real or complex number.

Then

$\qquad p(a)=0 \iff (x-a)$ is a factor of $p(x)$

In words: a number $a$ is a root of a polynomial if and only if $(x-a)$ divides the polynomial.

📐 Remainder Theorem

When a polynomial $p(x)$ is divided by $(x-a)$ , the remainder is

$\qquad p(a)$

So:

if $p(a)=0$ , then the remainder is $0$ , hence $(x-a)$ divides $p(x)$

if $p(a)\ne 0$ , then $(x-a)$ is not a factor

---

First-Principles Proof Pattern

📐 How the Proof Usually Works

When dividing $p(x)$ by $(x-a)$ , we can write

$\qquad p(x)=(x-a)q(x)+r$

where $r$ is a constant.

Now substitute $x=a$ :

$\qquad p(a)=(a-a)q(a)+r=r$

So if $p(a)=0$ , then $r=0$ , and therefore

$\qquad p(x)=(x-a)q(x)$

Hence $(x-a)$ is a factor of $p(x)$ .

This is exactly the standard logic behind many CMI proof-completion questions. ---

Distinct Roots Give a Product Factor

📐 Two Distinct Roots

If $a$ and $b$ are distinct roots of $p(x)$ , then

$\qquad (x-a)(x-b)$ divides $p(x)$

Reason:

Since $p(a)=0$ , we get

\qquad p(x)=(x-a)q(x)

Now substitute $x=b$ :

\qquad 0=p(b)=(b-a)q(b)

Since $a\ne b$ , we have $b-a\ne 0$ , so

\qquad q(b)=0

Therefore $(x-b)$ divides $q(x)$

Hence $(x-a)(x-b)$ divides $p(x)$

❗ Generalization

If $a_1,a_2,\dots,a_k$ are distinct roots of $p(x)$ , then

$\qquad (x-a_1)(x-a_2)\cdots(x-a_k)$ divides $p(x)$

---

Repeated Roots and Multiplicity

📖 Multiplicity of a Root

If

$\qquad p(x)=(x-a)^m q(x)$

with $q(a)\ne 0$ , then $a$ is called a root of multiplicity $m$ .

Examples:

multiplicity $1$ $\rightarrow$ simple root

multiplicity $2$ $\rightarrow$ repeated root or double root

multiplicity $3$ $\rightarrow$ triple root

📐 Repeated Root Test

If $(x-a)^2$ divides $p(x)$ , then $a$ is a repeated root.

A very useful criterion is:

$\qquad a$ is a repeated root of $p(x)$ if and only if
$\qquad p(a)=0$ and $p'(a)=0$

This criterion becomes very useful in construction problems and discriminant-style questions. ---

How to Use the Theorem in Practice

💡 CMI Strategy

If you are asked whether $(x-a)$ is a factor, compute $p(a)$ immediately.

If you are given a root, write the polynomial as

\qquad p(x)=(x-a)q(x)

If two distinct roots are given, extract one factor first, then use the second root on the quotient.

If a repeated root is suspected, check both

\qquad p(a)=0

and

\qquad p'(a)=0

In proof questions, remember that the remainder after division by $(x-a)$ is a constant.

---

Important Extensions

📐 Polynomial in Two Variables

The same logic works for polynomials in two variables.

If $P(x,y)$ is a polynomial and after substituting $x=f(y)$ we get

$\qquad P(f(y),y)=0$

for all relevant $y$ , then

$\qquad (x-f(y))$ divides $P(x,y)$

This is the two-variable analogue of the factor theorem.

This is exactly the idea behind problems of the form:

if $P(\omega y,y)=0$ , then $(x-\omega y)$ divides $P(x,y)$

symmetry may force another related factor as well

multiplying conjugate factors can give a polynomial with integer coefficients

---

Symmetric Polynomial Insight

📐 A Standard Complex-Factor Pattern

Let $\omega$ be a cube root of unity with

$\qquad \omega^3=1,\qquad \omega\ne 1,\qquad 1+\omega+\omega^2=0$

If a symmetric polynomial $A(x,y)$ satisfies

$\qquad A(\omega y,y)=0$

then $(x-\omega y)$ divides $A(x,y)$ .

Since $A(x,y)=A(y,x)$ , interchanging $x$ and $y$ can force a related factor as well.

A very common resulting integer-coefficient factor is

$\qquad (x-\omega y)(x-\omega^2 y)=x^2+xy+y^2$

This pattern is extremely useful in olympiad-style algebra. ---

Minimal Worked Examples

Example 1 Check whether

(x-2)

divides

\qquad p(x)=x^3-5x+6

Compute

\qquad p(2)=8-10+6=4

Since

p(2)\ne 0

(x-2)

is not a factor. --- Example 2 Check whether

(x+1)

divides

\qquad p(x)=x^3+x^2-x-1

Compute

\qquad p(-1)=-1+1+1-1=0

(x+1)

is a factor. In fact,

\qquad x^3+x^2-x-1=(x+1)(x^2-1)=(x+1)^2(x-1)

--- Example 3 Suppose

2

and

3

are roots of a polynomial

p(x)

. Then

\qquad (x-2)(x-3)

divides

p(x)

p(x)

can be written as

\qquad p(x)=(x-2)(x-3)q(x)

for some polynomial

q(x)

. ---

Common Mistakes

⚠️ Avoid These Errors

❌ Forgetting that the remainder on division by $(x-a)$ is a constant

❌ Confusing “ $a$ is a root” with “ $x-a$ is a root”

❌ Assuming two roots automatically mean degree $2$

❌ Forgetting that repeated roots need higher powers like $(x-a)^2$

❌ Using the factor theorem without checking arithmetic carefully

---

Quick Recognition Sheet

📐 Fast Recall

$p(a)=0 \iff (x-a)$ divides $p(x)$

remainder on division by $(x-a)$ is $p(a)$

if $a,b$ are distinct roots, then $(x-a)(x-b)$ divides $p(x)$

if $a$ is a repeated root, then $(x-a)^2$ divides $p(x)$

repeated root test:

\qquad p(a)=0

and

\qquad p'(a)=0

---

Practice Questions

:::question type="MCQ" question="If

p(3)=0

, then which of the following must be true?" options=["

(x+3)

is a factor of

p(x)

","

(x-3)

is a factor of

p(x)

","

3

is a factor of

p(x)

","

p(x)

is constant"] answer="B" hint="Use the factor theorem directly." solution="By the factor theorem,

\qquad p(a)=0 \iff (x-a)

is a factor. Here

a=3

, so

\qquad (x-3)

is a factor of

p(x)

. Hence the correct option is

\boxed{B}

." ::: :::question type="NAT" question="Find the remainder when

p(x)=x^4-3x^2+2x+5

is divided by

(x-2)

." answer="13" hint="Use the remainder theorem." solution="By the remainder theorem, the remainder on division by

(x-2)

\qquad p(2)

Now,

\qquad p(2)=2^4-3(2^2)+2(2)+5

\qquad =16-12+4+5=13

Hence the remainder is

\boxed{13}

." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["If

p(a)=0

and

p(b)=0

with

a\ne b

, then

(x-a)(x-b)

divides

p(x)

","If

(x-a)

divides

p(x)

, then

p(a)=0

","If

p(a)=5

, then

(x-a)

divides

p(x)

","If

(x-a)^2

divides

p(x)

, then

a

is a repeated root of

p(x)

"] answer="A,B,D" hint="Use factor theorem and multiplicity." solution="1. True. Distinct roots give the product factor.

True. This is the converse direction of the factor theorem.

False. If

p(a)=5

, then the remainder is

5

, not

0

, so

(x-a)

does not divide

p(x)

True. By definition, if

(x-a)^2

divides

p(x)

, then

a

is a repeated root.

Hence the correct answer is

\boxed{A,B,D}

." ::: :::question type="SUB" question="Let

p(x)=x^3+kx^2-4x-4k

. Find all values of

k

for which

(x-2)

is a factor of

p(x)

." answer="

k=0

" hint="Set

p(2)=0

." solution="For

(x-2)

to be a factor, the factor theorem requires

\qquad p(2)=0

Now

\qquad p(2)=2^3+k(2^2)-4(2)-4k

\qquad =8+4k-8-4k=0

So this is true for every real value of

k

. Therefore

(x-2)

is a factor for all values of

k

. Hence the answer is

\boxed{\text{all real }k}

." ::: ---

Summary

❗ Key Takeaways for CMI

The factor theorem converts root information into divisibility information.

The remainder theorem is the fastest way to test linear factors.

Distinct roots give a product of distinct linear factors.

Repeated roots correspond to higher powers like $(x-a)^2$ .

In proof questions, the form

\qquad p(x)=(x-a)q(x)+r

is the standard starting point.

The same idea extends naturally to multivariable polynomial divisibility.

---

💡 Next Up

Proceeding to Vieta relations.

---

Part 3: Vieta relations

Vieta Relations

Overview

Vieta relations connect the coefficients of a polynomial directly to sums and products of its roots. This makes them one of the fastest tools in algebra: instead of finding the roots explicitly, we can still compute expressions involving them. In CMI-style questions, Vieta relations often appear together with symmetry, divisibility, reciprocal roots, and power sums such as

\sum r_i^2

. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

Write Vieta relations for monic and non-monic polynomials.

Compute sums, pairwise products, and full products of roots without solving the polynomial.

Find sums of powers such as $\sum r_i^2$ using symmetric identities.

Compute sums of reciprocals of roots when allowed.

Use divisibility conditions like $(x-1)$ or $(x^2-1)$ to extract coefficient information.

---

Standard Setup

📖 General Polynomial and Its Roots

Let
$\qquad P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$
with $a_n\ne 0$ .

Let its roots be $r_1,r_2,\dots,r_n$ in the complex number system, counted with multiplicity.

Then
$\qquad P(x)=a_n(x-r_1)(x-r_2)\cdots(x-r_n)$

---

Vieta Relations for a Monic Polynomial

📐 Monic Form

If
$\qquad P(x)=x^n+c_1x^{n-1}+c_2x^{n-2}+\cdots+c_n$
has roots $r_1,\dots,r_n$ , then

$r_1+r_2+\cdots+r_n = -c_1$

$\sum_{1\le i<j\le n} r_ir_j = c_2$

$\sum_{1\le i<j<k\le n} r_ir_jr_k = -c_3$

$\cdots$

$r_1r_2\cdots r_n = (-1)^n c_n$

❗ Sign Pattern

The signs alternate:

sum of 1 root $\rightarrow$ negative

sum of 2-root products $\rightarrow$ positive

sum of 3-root products $\rightarrow$ negative

and so on

---

Vieta Relations for a General Polynomial

📐 Non-Monic Form

If
$\qquad P(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_0$
has roots $r_1,\dots,r_n$ , then

$r_1+\cdots+r_n = -\dfrac{a_{n-1}}{a_n}$

$\sum_{i<j} r_ir_j = \dfrac{a_{n-2}}{a_n}$

$\sum_{i<j<k} r_ir_jr_k = -\dfrac{a_{n-3}}{a_n}$

$\cdots$

$r_1r_2\cdots r_n = (-1)^n\dfrac{a_0}{a_n}$

---

Most Useful Derived Identities

📐 Power Sum Identities

Let
$\qquad s_1=r_1+r_2+\cdots+r_n$
and
$\qquad s_2=\sum_{i<j} r_ir_j$

Then

$\qquad \sum_{i=1}^n r_i^2 = s_1^2 - 2s_2$

Also,

$\qquad \sum_{i=1}^n r_i^3 = s_1^3 - 3s_1s_2 + 3s_3$

where
$\qquad s_3=\sum_{i<j<k} r_ir_jr_k$

💡 Most Exam-Relevant One

The formula used most often in short questions is

$\qquad \sum r_i^2 = \left(\sum r_i\right)^2 - 2\sum_{i<j} r_ir_j$

---

Sum of Reciprocals of Roots

📐 Reciprocal Sum

If none of the roots is zero, then

$\qquad \sum_{i=1}^n \dfrac{1}{r_i} <br>= \dfrac{\sum_{i=1}^n \prod_{j\ne i} r_j}{r_1r_2\cdots r_n} <br>= \dfrac{e_{n-1}}{e_n}$

In coefficient form,

$\qquad \sum_{i=1}^n \dfrac{1}{r_i} = -\dfrac{a_1}{a_0}$

⚠️ Very Important Restriction

This formula is valid only when $a_0\ne 0$ , equivalently when $0$ is not a root.

---

Divisibility and Roots

📐 Factor Theorem

For any polynomial $P(x)$ :

$(x-\alpha)$ divides $P(x)$ if and only if $P(\alpha)=0$

$(x^2-1)$ divides $P(x)$ if and only if both $P(1)=0$ and $P(-1)=0$

since

\qquad x^2-1=(x-1)(x+1)

❗ Why This Matters Here

Many Vieta problems do not ask for the roots directly. Instead, they tell you something like:

polynomial is divisible by $x^2-1$

$1$ and $-1$ are roots

roots satisfy a reciprocal condition

and then expect coefficient information from these facts.

---

PYQ Insight 1: Sum of Squares of Roots

❗ 2024-Type Pattern

If
$\qquad P(x)=x^6+10x^5+11x^4+12x^3+13x^2-12x-11$
has roots $z_1,\dots,z_6$ , then

$\qquad z_1+\cdots+z_6 = -10$

and

$\qquad \sum_{i<j} z_iz_j = 11$

Therefore,

$\qquad \sum_{i=1}^6 z_i^2 <br>= \left(\sum z_i\right)^2 - 2\sum_{i<j} z_iz_j$

$\qquad = (-10)^2 - 2(11)=100-22=78$

So the required value is $\boxed{78}$ .

---

PYQ Insight 2: Divisibility by $x^2-1$

❗ 2020-Type Pattern

Suppose
$\qquad P(x)=10x^{400}+ax^{399}+bx^{398}+3x+15$
is divisible by $x^2-1$ .

Then
$\qquad P(1)=0$ and $P(-1)=0$

Now

$\qquad P(1)=10+a+b+3+15=0$
so
$\qquad a+b=-28$

Also

$\qquad P(-1)=10-a+b-3+15=0$
so
$\qquad -a+b=-22$

Solving,
$\qquad b=-25,\quad a=-3$

Now since $a_0=15\ne 0$ , none of the roots is zero, so

$\qquad \sum \dfrac{1}{r_i} = -\dfrac{a_1}{a_0}=-\dfrac{3}{15}=-\dfrac15$

This is exactly the kind of two-step use of Vieta that appears in exams.

---

Minimal Worked Examples

Example 1 If the roots of

\qquad x^2-7x+10=0

are

\alpha,\beta

, then

\qquad \alpha+\beta=7,\qquad \alpha\beta=10

--- Example 2 If the roots of

\qquad 2x^3-5x^2+7x-4=0

are

r_1,r_2,r_3

, then

\qquad r_1+r_2+r_3=\dfrac{5}{2}

\qquad r_1r_2+r_2r_3+r_3r_1=\dfrac{7}{2}

\qquad r_1r_2r_3=\dfrac{4}{2}=2

because for a cubic

\qquad r_1r_2r_3 = -\dfrac{a_0}{a_3}

and here

\qquad -\dfrac{-4}{2}=2

---

Root Transform Tricks

📐 Useful Transformations

If roots are $r_1,\dots,r_n$ , then:

roots of $P(-x)$ are $-r_1,\dots,-r_n$

roots of $P(kx)$ are $\dfrac{r_1}{k},\dots,\dfrac{r_n}{k}$

if none of the roots is zero, then the roots of

\qquad x^nP\left(\dfrac{1}{x}\right)

are

\qquad \dfrac{1}{r_1},\dots,\dfrac{1}{r_n}

These are often useful in advanced reciprocal-root problems. ---

CMI Strategy

💡 How to Attack Vieta Problems

Write the degree and coefficients clearly.

Identify which symmetric expression is being asked.

Use direct Vieta first: $\sum r_i$ , $\sum r_ir_j$ , product of roots.

For $\sum r_i^2$ , do not try to find roots; use

\qquad \sum r_i^2 = (\sum r_i)^2 - 2\sum r_ir_j

For reciprocal sums, check first that zero is not a root.

If divisibility is given, immediately apply the factor theorem.

Keep sign alternation under control; most mistakes here are sign mistakes.

---

Common Mistakes

⚠️ Avoid These Errors

❌ Forgetting the alternating sign pattern in Vieta

❌ Using monic formulas on a non-monic polynomial without dividing by the leading coefficient

❌ Using $\sum \dfrac{1}{r_i} = -\dfrac{a_1}{a_0}$ when $a_0=0$

❌ Confusing $\sum r_i^2$ with $\left(\sum r_i\right)^2$

✅ Remember:

\qquad \left(\sum r_i\right)^2 = \sum r_i^2 + 2\sum_{i<j} r_ir_j

❌ Forgetting that divisibility by $(x^2-1)$ means checking both $1$ and $-1$

---

Practice Questions

:::question type="MCQ" question="If

\alpha,\beta

are the roots of

x^2-5x+6=0

, then

\alpha+\beta

equals" options=["

-5

","

5

","

6

","

-6

"] answer="B" hint="Use the coefficient of

x

." solution="For the monic quadratic

\qquad x^2-5x+6=0

, the sum of roots is the negative of the coefficient of

x

. Thus

\qquad \alpha+\beta = 5

So the correct option is

\boxed{B}

." ::: :::question type="NAT" question="If the roots of

x^3-4x^2+5x-2=0

are

r_1,r_2,r_3

, find

r_1^2+r_2^2+r_3^2

." answer="6" hint="Use

\sum r_i^2 = (\sum r_i)^2 - 2\sum r_ir_j

." solution="From Vieta,

\qquad r_1+r_2+r_3 = 4

and

\qquad r_1r_2+r_2r_3+r_3r_1 = 5

Hence

\qquad r_1^2+r_2^2+r_3^2 = (r_1+r_2+r_3)^2 - 2(r_1r_2+r_2r_3+r_3r_1)

\qquad = 4^2 - 2\cdot 5 = 16-10=6

Therefore the answer is

\boxed{6}

." ::: :::question type="MSQ" question="Which of the following statements are true for a polynomial

a_nx^n+\cdots+a_1x+a_0

with nonzero roots

r_1,\dots,r_n

?" options=["

r_1r_2\cdots r_n = (-1)^n\dfrac{a_0}{a_n}

","

\sum \dfrac{1}{r_i} = -\dfrac{a_1}{a_0}

","If

(x-1)

divides

P(x)

, then

P(1)=0

","If

(x^2-1)

divides

P(x)

, then only

P(1)=0

is needed"] answer="A,B,C" hint="Use Vieta and the factor theorem." solution="1. True. This is the standard product-of-roots formula.

True, provided

a_0\ne 0

, which is equivalent to saying the roots are nonzero.

True. This is the factor theorem.

False. Since

x^2-1=(x-1)(x+1)

, we need both

P(1)=0

and

P(-1)=0

Hence the correct answer is

\boxed{A,B,C}

." ::: :::question type="SUB" question="Let the roots of

x^4+2x^3-7x^2+3x+5=0

z_1,z_2,z_3,z_4

. Find

\sum z_i^2

." answer="

18

" hint="First find

\sum z_i

and

\sum_{i<j} z_iz_j

." solution="For

\qquad x^4+2x^3-7x^2+3x+5=0

, by Vieta,

\qquad z_1+z_2+z_3+z_4 = -2

and

\qquad \sum_{i<j} z_iz_j = -7

Now

\qquad \sum z_i^2 = \left(\sum z_i\right)^2 - 2\sum_{i<j} z_iz_j

\qquad \sum z_i^2 = (-2)^2 - 2(-7)=4+14=18

Therefore the required value is

\boxed{18}

." ::: ---

Summary

❗ Key Takeaways for CMI

Vieta relations connect coefficients directly with symmetric sums of roots.

For non-monic polynomials, divide by the leading coefficient mentally.

The fastest derived identity is

\qquad \sum r_i^2 = (\sum r_i)^2 - 2\sum r_ir_j

The reciprocal sum formula is

\qquad \sum \dfrac{1}{r_i} = -\dfrac{a_1}{a_0}

,
but only when

a_0\ne 0

Divisibility conditions and Vieta often work together.

Most errors in this topic come from sign mistakes, not concept mistakes.

---

💡 Next Up

Proceeding to Rational roots.

---

Part 4: Rational roots

Rational Roots

Overview

The idea of a rational root is simple: it is a root that can be written in the form

\qquad \dfrac{p}{q}

where

p

and

q

are integers and

q \ne 0

. But in polynomial equations, especially CMI-style algebra, rational roots are useful not only for finding solutions, but also for testing factorisation, reducing degree, and proving non-existence of easy roots. This topic mainly revolves around the Rational Root Theorem, careful substitution, and the link between roots and factors. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

identify what a rational root is

apply the Rational Root Theorem correctly

list all possible rational roots of a polynomial

test candidates efficiently using substitution

use a rational root to factor a polynomial and reduce its degree

---

Definition

📖 Rational Root

A rational root of a polynomial equation is a root of the form

$\qquad \dfrac{p}{q}$

where $p,q \in \mathbb{Z}$ , $q \ne 0$ , and the fraction is in lowest terms.

---

Core Link Between Roots and Factors

📐 Factor Theorem

If $P(x)$ is a polynomial and

$\qquad P(r)=0$

then $(x-r)$ is a factor of $P(x)$ .

So every rational root immediately gives a linear factor.

❗ Why Rational Roots Matter

If you find one rational root of a polynomial of degree $3$ or higher, then:

you get a factor

you can divide the polynomial by that factor

the remaining problem becomes simpler

---

Rational Root Theorem

📐 Rational Root Theorem

Let

$\qquad P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$

be a polynomial with integer coefficients.

If $\dfrac{p}{q}$ in lowest terms is a rational root of $P(x)=0$ , then

$p$ divides the constant term $a_0$

$q$ divides the leading coefficient $a_n$

So every possible rational root must be of the form

\qquad \pm \dfrac{\text{factor of }a_0}{\text{factor of }a_n}

---

How to Generate Candidates

💡 Candidate List Method

For a polynomial with integer coefficients:

write all factors of the constant term

write all factors of the leading coefficient

form all possible fractions

include both positive and negative values

test each candidate in the polynomial

This gives all possible rational roots, not guaranteed roots.

---

Important Special Cases

📐 Monic Polynomial Case

If the polynomial is monic, that is, the leading coefficient is $1$ , then every rational root must actually be an integer dividing the constant term.

For example, if

$\qquad P(x)=x^3-5x^2+2x+8$

then any rational root must be among the integer divisors of $8$ :

$\qquad \pm 1,\pm 2,\pm 4,\pm 8$

📐 Zero as a Root

If the constant term is $0$ , then

$\qquad x=0$

is automatically a root, and $x$ is a factor of the polynomial.

---

Minimal Worked Examples

Example 1 Find all possible rational roots of

\qquad 2x^3-3x^2-8x+3=0

Here:

constant term $=3$

leading coefficient $=2$

Factors of

3

are

\qquad \pm 1,\pm 3

Factors of

2

are

\qquad \pm 1,\pm 2

So the possible rational roots are

\qquad \pm 1,\ \pm 3,\ \pm \dfrac12,\ \pm \dfrac32

These are only candidates. Each must be tested. --- Example 2 Test whether

x=2

is a root of

\qquad x^3-4x^2+x+6

Substitute

x=2

\qquad P(2)=2^3-4(2^2)+2+6

\qquad =8-16+2+6=0

x=2

is a root, hence

(x-2)

is a factor. ---

Efficient Testing Ideas

💡 How to Test Fast

When testing candidate roots:

start with small integers first: $\pm 1,\pm 2$

look for sign cancellation or easy mental arithmetic

if the polynomial is monic, test divisors of the constant term first

once a root is found, factor immediately instead of testing all remaining candidates blindly

---

Rational Root Theorem Does Not Say

⚠️ Avoid These Misunderstandings

The Rational Root Theorem does not say:

every polynomial has a rational root

every candidate obtained is an actual root

irrational or complex roots do not exist

It only says:

if a rational root exists, then it must come from the candidate list.

---

Relation with Factorisation

📐 After Finding a Rational Root

If $r$ is a rational root of $P(x)$ , then

$\qquad P(x)=(x-r)Q(x)$

for some polynomial $Q(x)$ .

This is the usual next step in solving cubic and quartic equations by algebraic methods.

---

Common Error Patterns

⚠️ Avoid These Errors

❌ forgetting to reduce $\dfrac{p}{q}$ to lowest terms before applying the theorem

❌ listing only positive candidates and forgetting negatives

❌ confusing possible rational roots with actual roots

❌ making arithmetic mistakes while testing candidates

❌ forgetting that the theorem requires integer coefficients

---

When the Theorem Applies

❗ Applicability

Use the Rational Root Theorem when:

the polynomial has integer coefficients

you want to find rational roots or show likely absence of simple rational roots

you want to begin factorisation of a cubic or quartic

Use caution if coefficients are not integers. In such cases, first clear denominators if convenient.

---

Strategy for CMI-Type Questions

💡 CMI Strategy

First inspect whether the polynomial is monic.

Write the constant term and leading coefficient separately.

Generate the full rational candidate set carefully.

Test easy candidates first.

Once a root is found, factor and reduce the degree.

If no candidate works, conclude there is no rational root from the theorem.

---

Practice Questions

:::question type="MCQ" question="Which of the following can be a rational root of the polynomial

2x^3+5x^2-x-3=0

?" options=["

\dfrac{5}{2}

","

\dfrac{3}{2}

","

\dfrac{4}{3}

","

\dfrac{5}{3}

"] answer="B" hint="Use the Rational Root Theorem: numerator divides the constant term, denominator divides the leading coefficient." solution="For the polynomial

\qquad 2x^3+5x^2-x-3

the constant term is

-3

and the leading coefficient is

2

. So a rational root, if it exists, must be of the form

\qquad \pm \dfrac{\text{factor of }3}{\text{factor of }2}

Hence the possible rational roots are

\qquad \pm 1,\ \pm 3,\ \pm \dfrac12,\ \pm \dfrac32

Among the given options, only

\dfrac32

appears in the candidate list. Therefore the correct option is

\boxed{B}

." ::: :::question type="NAT" question="How many possible rational roots does the polynomial equation

x^3-7x+6=0

have according to the Rational Root Theorem?" answer="8" hint="The polynomial is monic, so possible rational roots are just the integer divisors of the constant term, with both signs." solution="The polynomial is

\qquad x^3-7x+6

Since it is monic, any rational root must be an integer dividing the constant term

6

. The divisors of

6

are

\qquad 1,\ 2,\ 3,\ 6

Including both positive and negative values, the possible rational roots are

\qquad \pm 1,\ \pm 2,\ \pm 3,\ \pm 6

This gives a total of

\boxed{8}

possible rational roots." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["If

\dfrac{p}{q}

is a rational root in lowest terms of a polynomial with integer coefficients, then

p

divides the constant term","If

\dfrac{p}{q}

is a rational root in lowest terms of a polynomial with integer coefficients, then

q

divides the leading coefficient","Every candidate from the Rational Root Theorem is an actual root","If a polynomial is monic with integer coefficients, then every rational root must be an integer"] answer="A,B,D" hint="Recall exactly what the theorem guarantees." solution="1. True. This is one part of the Rational Root Theorem.

True. This is the other part of the theorem.

False. The theorem gives only possible rational roots, not guaranteed roots.

True. If the leading coefficient is

1

, then

q

must divide

1

, so

q=\pm 1

. Hence every rational root must be an integer.

Therefore the correct answer is

\boxed{A,B,D}

." ::: :::question type="SUB" question="List all possible rational roots of the polynomial

3x^3-2x^2-5x+10=0

." answer="

\pm 1,\\ \pm 2,\\ \pm 5,\\ \pm 10,\\ \pm \dfrac13,\\ \pm \dfrac23,\\ \pm \dfrac53,\\ \pm \dfrac{10}{3}

" hint="Numerators come from factors of the constant term and denominators from factors of the leading coefficient." solution="For

\qquad 3x^3-2x^2-5x+10=0

the constant term is

10

, and the leading coefficient is

3

. Factors of

10

are

\qquad \pm 1,\ \pm 2,\ \pm 5,\ \pm 10

Factors of

3

are

\qquad \pm 1,\ \pm 3

Hence all possible rational roots are

\qquad \pm 1,\ \pm 2,\ \pm 5,\ \pm 10,\ \pm \dfrac13,\ \pm \dfrac23,\ \pm \dfrac53,\ \pm \dfrac{10}{3}

These are the possible rational roots according to the Rational Root Theorem. They are not necessarily actual roots. Therefore the required list is

\boxed{\pm 1,\ \pm 2,\ \pm 5,\ \pm 10,\ \pm \dfrac13,\ \pm \dfrac23,\ \pm \dfrac53,\ \pm \dfrac{10}{3}}

." ::: ---

Summary

❗ Key Takeaways for CMI

A rational root is a root of the form $\dfrac{p}{q}$ with integers $p,q$ and $q\ne 0$ .

For a polynomial with integer coefficients, any rational root must satisfy:

- numerator divides the constant term
- denominator divides the leading coefficient

The theorem gives a candidate list, not actual roots.

For monic polynomials, every rational root must be an integer divisor of the constant term.

Once a rational root is found, factor the polynomial and reduce the degree.

The theorem is a filtering tool and a factorisation tool, not a guarantee of solvability.

---

💡 Next Up

Proceeding to Integer roots.

---

Part 5: Integer roots

Integer Roots

Overview

When a polynomial has integer coefficients, its integer roots are highly restricted. In CMI-style problems, the main idea is that an integer root cannot be arbitrary: it must interact in a rigid way with the constant term and with Vieta's relations. This topic is not just about testing divisors, but about combining Factor Theorem, Remainder Theorem, and Vieta's formulas intelligently. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

use the Factor Theorem to test whether an integer is a root.

apply the integer-root condition for monic polynomials with integer coefficients.

use divisibility of the constant term to shortlist possible integer roots.

connect integer roots with Vieta's formulas in quadratic and cubic equations.

rule out impossible rational roots in monic integer polynomials.

---

Core Idea

📖 Integer Root

An integer root of a polynomial $f(x)$ is an integer $m$ such that

$\qquad f(m)=0$

By the Factor Theorem, this is equivalent to saying that

$\qquad x-m$

is a factor of $f(x)$ .

📐 Factor Theorem

If $f(a)=0$ , then

$\qquad x-a$

is a factor of $f(x)$ .

Conversely, if $(x-a)$ divides $f(x)$ , then

$\qquad f(a)=0$

---

Remainder Theorem

📐 Remainder Theorem

When a polynomial $f(x)$ is divided by $x-a$ , the remainder is

$\qquad f(a)$

So:

if $f(a)=0$ , then $a$ is a root

if $f(a)\ne 0$ , then $a$ is not a root

💡 Why This Matters

To test whether an integer $m$ is a root, you do not need full factorisation. Just compute

$\qquad f(m)$

If it is $0$ , then $m$ is a root.

---

Main Integer Root Rule

❗ Most Important Theorem

Let

$\qquad f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$

have integer coefficients.

If an integer $m$ is a root of $f(x)$ , then

$\qquad m \mid a_0$

That is, the integer root must divide the constant term.

📐 Monic Polynomial Version

If

$\qquad f(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$

has integer coefficients and $m$ is an integer root, then

$\qquad m \mid a_0$

So the only possible integer roots are the positive and negative divisors of the constant term.

---

Why Integer Roots Divide the Constant Term

💡 Quick Proof Idea

Suppose $m$ is a root of

$\qquad f(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$

Then

$\qquad m^n+a_{n-1}m^{n-1}+\cdots+a_1m+a_0=0$

So

$\qquad a_0 = -m(m^{n-1}+a_{n-1}m^{n-2}+\cdots+a_1)$

Hence $a_0$ is a multiple of $m$ . Therefore

$\qquad m \mid a_0$

---

Consequences for Monic Integer Polynomials

❗ Very Useful Consequences

For a monic polynomial with integer coefficients:

every integer root must divide the constant term

if the constant term is prime, the only possible integer roots are

\qquad \pm 1,\ \pm p

if the constant term is $\pm 1$ , then the only possible integer roots are

\qquad \pm 1

if the constant term is $0$ , then

\qquad x=0

is definitely a root

---

Rational Root Observation

⚠️ Important Extension

For a monic polynomial with integer coefficients, any rational root must actually be an integer.

So if

$\qquad f(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_0$

has integer coefficients, then roots like

$\qquad \dfrac12,\ \dfrac23,\ \dfrac{7}{5}$

cannot occur as rational roots.

📐 Reason

This is a special case of the Rational Root Theorem.

For a monic polynomial, if $\dfrac{p}{q}$ in lowest terms is a rational root, then

$p$ divides the constant term

$q$ divides the leading coefficient

But the leading coefficient is

1

, so

\qquad q=1

Hence the rational root is an integer.

---

Vieta's Formulas and Integer Roots

📐 Quadratic Case

If

$\qquad f(x)=x^2+bx+c$

has roots $r,s$ , then

$\qquad r+s=-b$
$\qquad rs=c$

So if both roots are integers, they are integer factor pairs of $c$ whose sum is $-b$ .

📐 Cubic Case

If

$\qquad f(x)=x^3+ax^2+bx+c$

has roots $p,q,r$ , then

$\qquad p+q+r=-a$
$\qquad pq+qr+rp=b$
$\qquad pqr=-c$

So if one root is an integer, it must divide $c$ .
If all three roots are integers, then their product is exactly $-c$ .

---

Polynomial Decomposition Around a Root Candidate

📐 Useful Identity

For any polynomial $f(x)$ and any number $a$ ,

$\qquad f(x)=(x-a)Q(x)+f(a)$

This is just the division algorithm.

So if the coefficients of $f(x)$ are integers and $a$ is an integer, then $Q(x)$ also has integer coefficients when dividing by $x-a$ .

This is often useful when $f(a)$ is known but $a$ is not necessarily a root.

---

Minimal Worked Examples

Example 1 Find all possible integer roots of

\qquad x^3-7x^2+10x-8=0

Since the polynomial is monic and the constant term is

-8

, any integer root must divide

-8

. So the only possibilities are

\qquad \pm 1,\ \pm 2,\ \pm 4,\ \pm 8

These are the only candidates worth testing. --- Example 2 Can

\dfrac12

be a root of a monic polynomial with integer coefficients? No. Because any rational root of a monic polynomial with integer coefficients must be an integer. Since

\qquad \dfrac12

is not an integer, it cannot be such a root. ---

Standard Solving Strategy

💡 How to Attack Integer-Root Problems

identify the constant term.

list all integer divisors of the constant term.

test them using the Remainder Theorem.

once one root is found, factor out $(x-r)$ .

solve the lower-degree polynomial that remains.

use Vieta's relations if the problem asks about sums, products, or possible values.

---

Common Mistakes

⚠️ Avoid These Errors

❌ Thinking any integer can be a root

✅ Integer roots must divide the constant term.

❌ Forgetting signs

✅ If

d

divides

c

, then both

d

and

-d

are candidates.

❌ Confusing integer root condition with full factorisation

✅ Divisibility gives only candidates, not automatic roots.

❌ Forgetting that monic integer polynomials cannot have non-integer rational roots

✅ For monic polynomials, rational root

\Rightarrow

integer root.

❌ Using Vieta without checking coefficient signs

✅ For

\qquad x^3+ax^2+bx+c

the product of roots is

\qquad -c

---

CMI Strategy

💡 Exam Thinking

Use divisibility first, not expansion.

If the constant term is small or highly structured, exploit its divisors.

If the polynomial is monic, eliminate fractional rational roots immediately.

In cubic problems, combine:

- root divides constant term
- sum/product from Vieta

If a value like $f(1)$ is given, rewrite using

\qquad f(x)=(x-1)Q(x)+f(1)

---

Practice Questions

:::question type="MCQ" question="If

f(x)=x^3+2x^2-5x-6

, then which of the following can be an integer root of

f(x)

?" options=["

4

","

5

","

6

","

7

"] answer="C" hint="An integer root of a monic polynomial must divide the constant term." solution="The polynomial is monic and its constant term is

-6

. So any integer root must divide

-6

. The possible integer roots are

\qquad \pm 1,\ \pm 2,\ \pm 3,\ \pm 6

Among the options, only

6

is possible as an integer root candidate. Therefore the correct answer is

\boxed{C}

." ::: :::question type="NAT" question="How many positive integer roots can a monic polynomial with integer coefficients and constant term

12

possibly have as candidates before testing?" answer="6" hint="Count the positive divisors of

12

." solution="For a monic polynomial with integer coefficients, every integer root must divide the constant term. So if the constant term is

12

, the positive integer root candidates are the positive divisors of

12

\qquad 1,2,3,4,6,12

There are

6

such candidates. Hence the answer is

\boxed{6}

." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["If an integer

m

is a root of a polynomial with integer coefficients, then

m

divides the constant term.","Every divisor of the constant term is automatically a root.","If a monic polynomial with integer coefficients has a rational root, then that root must be an integer.","For

x^3+ax^2+bx+c

, the product of the roots is

-c

."] answer="A,C,D" hint="Separate root conditions from root candidates." solution="1. True. This is the standard integer-root divisibility condition.

False. Dividing the constant term only makes a number a candidate, not necessarily an actual root.

True. By the Rational Root Theorem, a rational root of a monic integer polynomial must be an integer.

True. By Vieta's formula for

\qquad x^3+ax^2+bx+c

, the product of the roots is

\qquad -c

. Hence the correct answer is

\boxed{A,C,D}

." ::: :::question type="SUB" question="Let

f(x)=x^3+ax^2+bx+c

where

a,b,c

are integers. Show that if

r

is a rational root of

f(x)

, then

r

must be an integer." answer="Any rational root of a monic polynomial with integer coefficients must be an integer." hint="Use the Rational Root Theorem with leading coefficient

1

." solution="Let

\qquad r=\dfrac{p}{q}

be a rational root of

f(x)

in lowest terms, where

p,q

are integers and

\gcd(p,q)=1

. Since

\qquad f(x)=x^3+ax^2+bx+c

is a monic polynomial with integer coefficients, the Rational Root Theorem gives:

$p$ divides the constant term $c$

$q$ divides the leading coefficient

But the leading coefficient is

1

, so

q

must divide

1

. Hence

\qquad q=1

Therefore

\qquad r=\dfrac{p}{1}=p

which is an integer. So any rational root of a monic polynomial with integer coefficients must be an integer. Hence the required statement is proved." ::: ---

Summary

❗ Key Takeaways for CMI

An integer root of an integer-coefficient polynomial must divide the constant term.

For monic integer polynomials, rational roots must actually be integers.

Factor Theorem and Remainder Theorem are the fastest tools for testing root candidates.

Vieta's formulas connect roots with coefficients and are especially useful in cubic problems.

Divisibility gives possible roots, not guaranteed roots.

In root-theory questions, combine divisibility, factorisation, and Vieta instead of using only one idea.

---

💡 Next Up

Proceeding to Multiple roots.

---

Part 6: Multiple roots

Multiple Roots

Overview

A multiple root is much more than “a root repeated twice”. It changes the algebra, the graph, and the derivative structure of a polynomial. In CMI-style questions, this topic is often tested through factorisation, derivative sharing, uniqueness of real roots, and monic integer polynomials. The central idea is:

\qquad \text{a repeated root is a common root of } p(x) \text{ and } p'(x)

This makes derivatives a natural tool in root theory. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

define multiplicity of a root precisely

detect repeated roots using $p(x)$ and $p'(x)$

understand how multiplicity changes after differentiation

connect multiple roots with graph behaviour

use multiplicity ideas in monic, rational, and integer-coefficient polynomials

---

Core Definition

📖 Multiplicity of a Root

A number $r$ is called a root of multiplicity $m$ of a polynomial $p(x)$ if

$\qquad p(x) = (x-r)^m q(x)$

for some polynomial $q(x)$ such that

$\qquad q(r) \ne 0$

Here:

$m=1$ means a simple root

$m\ge 2$ means a multiple root

📐 Immediate Consequences

If $r$ is a root of multiplicity $m$ , then

$p(r)=0$

$p'(r)=0$ if $m\ge 2$

$p''(r)=0$ if $m\ge 3$

in general, the first $m-1$ derivatives vanish at $r$

That is,

\qquad p(r)=p'(r)=\cdots=p^{(m-1)}(r)=0

but

\qquad p^{(m)}(r)\ne 0

---

The Main Test for Repeated Roots

📐 Repeated Root Criterion

A polynomial $p(x)$ has a repeated root $r$ if and only if

$\qquad p(r)=0 \quad \text{and} \quad p'(r)=0$

Equivalently:

$\qquad p(x) \text{ has a repeated root } \iff \gcd(p,p') \text{ is non-constant}$

Why this works If

\qquad p(x)=(x-r)^m q(x), \quad m\ge 2

then on differentiating,

\qquad p'(x)=m(x-r)^{m-1}q(x) + (x-r)^m q'(x)

(x-r)

is still a factor of

p'(x)

, hence

p'(r)=0

. Conversely, if

p(r)=0

and

p'(r)=0

, then

(x-r)

divides both

p

and

p'

, so the root is not simple. ---

How Differentiation Changes Multiplicity

📐 Multiplicity Drop Rule

If $r$ is a root of multiplicity $m$ of $p(x)$ , then $r$ is a root of multiplicity $m-1$ of $p'(x)$ .

Examples:

if $r$ is a double root of $p$ , then $r$ is a simple root of $p'$

if $r$ is a triple root of $p$ , then $r$ is a double root of $p'$

if $r$ is a quadruple root of $p$ , then $r$ is a triple root of $p'$

---

Graph Meaning

💡 How the Graph Behaves at a Multiple Root

If $r$ is a root of $p(x)$ :

odd multiplicity $\Rightarrow$ graph crosses the $x$ -axis

even multiplicity $\Rightarrow$ graph touches the $x$ -axis and turns back

Special cases:

simple root: crosses normally

double root: touches and turns

triple root: crosses but flattens

This is not just graphical intuition. It is often the quickest way to guess whether a unique real root must be repeated. ---

Even Degree and “Exactly One Real Root”

❗ A Very Important Structural Fact

For a polynomial of even degree with real coefficients, non-real roots occur in conjugate pairs.

So if such a polynomial has exactly one distinct real root, then that real root must have even multiplicity.

For a quartic polynomial, if there is exactly one real number $r$ such that $p(r)=0$ , then the multiplicity of $r$ must be:

$2$ , or

In particular,

r

is a multiple root.

💡 Why This Matters

Once you know the unique real root is multiple, you immediately know

$\qquad p(r)=0 \quad \text{and} \quad p'(r)=0$

So the derivative becomes a root-detection tool.

---

Quadratic Case

📐 Repeated Root in a Quadratic

For

$\qquad ax^2+bx+c=0, \quad a\ne 0$

a repeated root occurs exactly when the discriminant is zero:

$\qquad b^2-4ac=0$

In that case, the double root is

$\qquad x=-\dfrac{b}{2a}$

This is the simplest model of the repeated-root phenomenon. ---

Rational and Integer Coefficients

📐 Monic Integer Polynomial Fact

If a monic polynomial with integer coefficients has a rational root, then that root must be an integer.

Reason:
By the Rational Root Theorem, if

$\qquad x=\dfrac{u}{v}$

in lowest terms is a root of

$\qquad x^n+a_{n-1}x^{n-1}+\cdots+a_0$

with integer coefficients, then

$u \mid a_0$

$v \mid 1$

v=1

, hence the root is an integer.

❗ Practical Exam Use

In monic integer polynomials, once you prove a root is rational, you have automatically proved it is an integer.

---

Standard Factor Forms

📐 Useful Models

double root at $r$ :

\qquad p(x)=(x-r)^2 q(x)

triple root at $r$ :

\qquad p(x)=(x-r)^3 q(x)

quartic with exactly one distinct real root:

\qquad p(x)=(x-r)^2(x^2+ux+v)

\qquad p(x)=(x-r)^4

In the first case, the quadratic factor must have no real root if

r

is the only real root.

---

Minimal Worked Examples

Example 1 Show that

x=2

is a repeated root of

\qquad p(x)=x^3-6x^2+12x-8

We factor:

\qquad p(x)=(x-2)^3

x=2

is a root of multiplicity

3

. Also,

\qquad p'(x)=3x^2-12x+12=3(x-2)^2

Hence

\qquad p(2)=0 \quad \text{and} \quad p'(2)=0

--- Example 2 For what value of

k

does

\qquad x^2-2kx+9=0

have a repeated root? A repeated root in a quadratic means discriminant zero:

\qquad (-2k)^2 - 4\cdot 1 \cdot 9 = 0

\qquad 4k^2-36=0

\qquad k^2=9

\qquad k=\pm 3

---

CMI Strategy

💡 How to Attack Multiple-Root Problems

first ask whether the root is simple or repeated

if repeated, write $p(r)=0$ and $p'(r)=0$

for quadratics, use discriminant zero

for monic integer polynomials, prove rational first, then conclude integer

if a polynomial has exactly one real root and even degree, expect even multiplicity

if needed, rewrite the polynomial as $(x-r)^m q(x)$ and compare coefficients

---

Common Mistakes

⚠️ Avoid These Errors

❌ thinking “one real root” means multiplicity $1$

❌ forgetting that non-real roots come in conjugate pairs for real coefficients

❌ using only $p(r)=0$ and forgetting $p'(r)=0$ for repeated roots

❌ assuming every touching point is a double root without checking algebraically

❌ forgetting that for monic integer polynomials, rational roots are automatically integers

---

Practice Questions

:::question type="MCQ" question="If

r

is a double root of a polynomial

p(x)

, which of the following must be true?" options=["

p(r)\ne 0

and

p'(r)=0

","

p(r)=0

and

p'(r)=0

","

p(r)=0

and

p''(r)=0

always","

p'(r)\ne 0

"] answer="B" hint="Use the repeated root criterion." solution="If

r

is a double root, then

(x-r)^2

divides

p(x)

. Therefore

\qquad p(r)=0

and also

\qquad p'(r)=0

. So the correct option is

\boxed{B}

." ::: :::question type="NAT" question="The polynomial

p(x)=x^3-3x^2+3x-1

has a root of multiplicity what?" answer="3" hint="Recognize a binomial expansion." solution="We observe that

\qquad x^3-3x^2+3x-1=(x-1)^3

Hence

x=1

is the only root, and it has multiplicity

\boxed{3}

." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["If

r

is a repeated root of

p(x)

, then

r

is a root of

p'(x)

.","A quartic polynomial with real coefficients and exactly one distinct real root must have that root with even multiplicity.","If a monic polynomial with integer coefficients has a rational root, then that root is an integer.","If

p(r)=0

, then

r

must be a repeated root."] answer="A,B,C" hint="Separate the repeated-root test from the ordinary root condition." solution="1. True. A repeated root is a common root of

p

and

p'

True. In even degree with real coefficients, non-real roots occur in conjugate pairs, so the lone real root must consume an even multiplicity.

True. By the Rational Root Theorem, a rational root of a monic integer polynomial must be an integer.

False. A simple root also satisfies

p(r)=0

, so this condition alone does not imply repetition.

Hence the correct answer is

\boxed{A,B,C}

." ::: :::question type="SUB" question="Prove that if

r

is a root of multiplicity

m\ge 2

of a polynomial

p(x)

, then

r

is a root of multiplicity

m-1

p'(x)

." answer="If

p(x)=(x-r)^m q(x)

with

q(r)\ne 0

, then differentiating shows

(x-r)^{m-1}

divides

p'(x)

but

(x-r)^m

does not." hint="Write

p(x)

in factorized form and differentiate using the product rule." solution="Since

r

is a root of multiplicity

m

, we can write

\qquad p(x)=(x-r)^m q(x)

where

q(r)\ne 0

. Differentiate using the product rule:

\qquad p'(x)=m(x-r)^{m-1}q(x) + (x-r)^m q'(x)

Factor out

(x-r)^{m-1}

\qquad p'(x)=(x-r)^{m-1}\big(mq(x)+(x-r)q'(x)\big)

(x-r)^{m-1}

divides

p'(x)

. Now evaluate the bracket at

x=r

\qquad mq(r)+(r-r)q'(r)=mq(r)

Since

m\ge 2

and

q(r)\ne 0

, this quantity is nonzero. Therefore

(x-r)

does not divide the bracket. Hence

(x-r)^{m-1}

is the highest power of

(x-r)

dividing

p'(x)

. So

r

is a root of multiplicity

\boxed{m-1}

p'(x)

. " ::: ---

Summary

❗ Key Takeaways for CMI

A root $r$ has multiplicity $m$ if $p(x)=(x-r)^m q(x)$ with $q(r)\ne 0$ .

A repeated root is exactly a common root of $p(x)$ and $p'(x)$ .

Differentiation reduces multiplicity by $1$ .

In even-degree real polynomials, a unique real root must have even multiplicity.

For monic integer polynomials, rational roots are automatically integers.

In root theory problems, the pair $p(r)=0$ and $p'(r)=0$ is often the decisive step.

Chapter Summary

❗ Root theory — Key Points

Remainder & Factor Theorems: $P(a)$ is the remainder when $P(x)$ is divided by $(x-a)$ . Consequently, $(x-a)$ is a factor of $P(x)$ if and only if $P(a)=0$ . These are fundamental for evaluating polynomials and identifying linear factors.

Vieta's Relations: These critical relationships connect the coefficients of a polynomial to symmetric expressions of its roots (sums, products, and sums of products). They are indispensable for solving problems involving root properties without explicit calculation of the roots.

Rational Root Theorem: For a polynomial with integer coefficients, any rational root $p/q$ (in simplest form) must satisfy that $p$ divides the constant term and $q$ divides the leading coefficient. This theorem significantly narrows the search space for rational roots.

Integer Root Theorem: A special case of the Rational Root Theorem, stating that any integer root of a polynomial with integer coefficients must be a divisor of the constant term.

Multiple Roots: If $x=a$ is a multiple root of a polynomial $P(x)$ , then $P(a)=0$ and $P'(a)=0$ . This derivative test is a powerful tool for identifying repeated roots and simplifying polynomial factorization.

---

Chapter Review Questions

:::question type="MCQ" question="Let $\alpha, \beta, \gamma$ be the roots of the polynomial equation $x^3 - 6x^2 + 11x - 6 = 0$ . Determine the value of $\alpha^2 + \beta^2 + \gamma^2$ ." options=["6","10","14","22"] answer="14" hint="Use Vieta's relations to express the sum of squares in terms of elementary symmetric polynomials." solution="From Vieta's relations:
$\alpha + \beta + \gamma = 6$
$\alpha\beta + \beta\gamma + \gamma\alpha = 11$
We know that $\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)$ .
Substituting the values:
$\alpha^2 + \beta^2 + \gamma^2 = (6)^2 - 2(11) = 36 - 22 = 14$ ."
:::

:::question type="NAT" question="If the polynomial $P(x) = x^3 - kx^2 + 5x - 3$ leaves a remainder of 3 when divided by $(x-2)$ , what is the value of $k$ ?" answer="3" hint="Apply the Remainder Theorem directly." solution="By the Remainder Theorem, $P(2)$ must be equal to the remainder when $P(x)$ is divided by $(x-2)$ .
Given remainder is 3, so $P(2) = 3$ .
Substitute $x=2$ into $P(x)$ :
$2^3 - k(2^2) + 5(2) - 3 = 3$
$8 - 4k + 10 - 3 = 3$
$15 - 4k = 3$
$4k = 12$
$k = 3$ ."
:::

:::question type="MCQ" question="For what value of $a$ does the polynomial $P(x) = x^3 - 6x^2 + 12x + a$ have a multiple root?" options=["-8","-4","0","4"] answer="-8" hint="If $x=r$ is a multiple root, then $P(r)=0$ and $P'(r)=0$ ." solution="If $P(x)$ has a multiple root at $x=r$ , then $P(r)=0$ and $P'(r)=0$ .
First, find the derivative of $P(x)$ :
$P'(x) = \frac{d}{dx}(x^3 - 6x^2 + 12x + a) = 3x^2 - 12x + 12$ .
Set $P'(x) = 0$ to find potential multiple roots:
$3x^2 - 12x + 12 = 0$
$x^2 - 4x + 4 = 0$
$(x-2)^2 = 0$
This implies that $x=2$ is the only potential multiple root.
For $x=2$ to be a multiple root of $P(x)$ , we must have $P(2)=0$ .
Substitute $x=2$ into $P(x)$ :
$P(2) = (2)^3 - 6(2)^2 + 12(2) + a = 8 - 6(4) + 24 + a = 8 - 24 + 24 + a = 8 + a$ .
Setting $P(2)=0$ :
$8 + a = 0 \implies a = -8$ ."
:::

:::question type="NAT" question="Find the sum of all integer roots of the polynomial $P(x) = x^4 - 2x^3 - 7x^2 + 8x + 12$ ." answer="2" hint="Use the Integer Root Theorem to identify potential roots, then test them." solution="According to the Integer Root Theorem, any integer root of $P(x)$ must be a divisor of the constant term, 12.
The divisors of 12 are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$ .
Let's test these values:
$P(1) = 1 - 2 - 7 + 8 + 12 = 12 \neq 0$
$P(-1) = (-1)^4 - 2(-1)^3 - 7(-1)^2 + 8(-1) + 12 = 1 + 2 - 7 - 8 + 12 = 0$ . So $x=-1$ is a root.
$P(2) = (2)^4 - 2(2)^3 - 7(2)^2 + 8(2) + 12 = 16 - 16 - 28 + 16 + 12 = 0$ . So $x=2$ is a root.
$P(-2) = (-2)^4 - 2(-2)^3 - 7(-2)^2 + 8(-2) + 12 = 16 + 16 - 28 - 16 + 12 = 0$ . So $x=-2$ is a root.
$P(3) = (3)^4 - 2(3)^3 - 7(3)^2 + 8(3) + 12 = 81 - 54 - 63 + 24 + 12 = 0$ . So $x=3$ is a root.
The integer roots are $-1, 2, -2, 3$ .
The sum of these integer roots is $(-1) + 2 + (-2) + 3 = 2$ ."
:::

---

What's Next?

💡 Continue Your CMI Journey

This chapter has equipped you with essential tools for analyzing polynomial equations. Building upon this foundation, your CMI journey will extend to related areas. The study of Polynomial Functions will delve deeper into their graphical properties, transformations, and applications. The concept of roots naturally leads to Complex Numbers, where every polynomial equation has a full set of solutions. Furthermore, Calculus provides powerful analytical methods, such as differentiation, to efficiently identify multiple roots, determine local extrema, and understand the precise behavior of polynomial functions.

Root theory

Root theory

Chapter Contents

| Topic |

Part 1: Remainder theorem

Remainder theorem

Overview

Learning Objectives

Core Idea

Special Cases You Must Know

Factor Theorem

General Linear Divisor

Minimal Worked Examples

Fast Evaluation Strategy

Pattern Recognition

Common Errors

Practice Questions

Summary

Part 2: Factor theorem

Factor Theorem

Overview

Learning Objectives

Core Theorem

First-Principles Proof Pattern

Distinct Roots Give a Product Factor

Repeated Roots and Multiplicity

How to Use the Theorem in Practice

Important Extensions

Symmetric Polynomial Insight

Minimal Worked Examples

Common Mistakes

Quick Recognition Sheet

Practice Questions

Summary

Part 3: Vieta relations

Vieta Relations

Overview

Learning Objectives

Standard Setup

Vieta Relations for a Monic Polynomial

Vieta Relations for a General Polynomial

Most Useful Derived Identities

Sum of Reciprocals of Roots

Divisibility and Roots

PYQ Insight 1: Sum of Squares of Roots

PYQ Insight 2: Divisibility by x2−1x^2-1x2−1

Minimal Worked Examples

Root Transform Tricks

CMI Strategy

Common Mistakes

Practice Questions

Summary

Part 4: Rational roots

Rational Roots

Overview

Learning Objectives

Definition

Core Link Between Roots and Factors

Rational Root Theorem

How to Generate Candidates

Important Special Cases

Minimal Worked Examples

Efficient Testing Ideas

Rational Root Theorem Does Not Say

Relation with Factorisation

Common Error Patterns

When the Theorem Applies

Strategy for CMI-Type Questions

Practice Questions

Summary

Part 5: Integer roots

Integer Roots

Overview

Learning Objectives

Core Idea

Remainder Theorem

Main Integer Root Rule

Why Integer Roots Divide the Constant Term

Consequences for Monic Integer Polynomials

Rational Root Observation

PYQ Insight 2: Divisibility by $x^2-1$