MathematicsClass 10ICSE

Find a Polynomial's Remainder Without Doing Any Long Division

Use the Remainder Theorem to find remainders in one step, test factors with the Factor Theorem, find unknown coefficients from factor and remainder conditions, and factorise cubic polynomials completely.

How can you find a remainder without dividing?

When a polynomial

f (x)

is divided by

x - a

, the result can always be written as

f (x) = (x - a) \times q (x) + r

where

q (x)

is the quotient and

r

is a number. **Put

x = a

**: the first term becomes zero, leaving

f (a) = r

. **So the remainder is simply

f (a)

** — no long division needed.

This chapter covers the Remainder Theorem, the Factor Theorem, unknown coefficients and complete factorisation of cubics.

How do you use the Remainder Theorem to find the remainder when a polynomial is divided by a linear expression?

**Find the value of

x

that makes the divisor zero and substitute it into the polynomial; the result is the remainder.

- Divisor

x - a

** — remainder

f (a)

- **Divisor

x + a

** — remainder

f (- a)

- **Divisor

a x - b

** — remainder

f (\frac{b}{a})

Worked examples with

f (x) = 2 x^{3} - 3 x^{2} + 4 x - 5

\div (x - 2) : f (2) = 16 - 12 + 8 - 5 = 7

\div (x + 1) : f (- 1) = - 2 - 3 - 4 - 5 = - 14

\div (2 x - 1) : f (\frac{1}{2}) = \frac{2}{8} - \frac{3}{4} + 2 - 5 = - \frac{7}{2}

An everyday example. **Packing

50

laddoos into boxes of

8

** leaves

2

over, because

50 = 8 \times 6 + 2

— a remainder found without packing a single box.

The substance. **For

x + 1

, substitute

- 1

, not

+ 1

**; the value used is the one that makes the divisor zero.

How do you use the Factor Theorem to test whether a linear expression is a factor?

**A linear expression

x - a

is a factor of

f (x)

exactly when

f (a) = 0

, that is, when the remainder is zero.

Worked example 1.** Is

x - 3

a factor of

x^{3} - 4 x^{2} + x + 6

f (3) = 27 - 36 + 3 + 6 = 0 yes, it is a factor

Worked example 2. Is

x + 2

a factor of

x^{3} + 3 x^{2} - 4 x - 10

f (- 2) = - 8 + 12 + 8 - 10 = 2 \neq = 0 no

Worked example 3. Is

2 x + 1

a factor of

2 x^{3} + x^{2} - 2 x - 1

f (- \frac{1}{2}) = - \frac{1}{4} + \frac{1}{4} + 1 - 1 = 0 yes

An everyday example. **

48

laddoos shared equally among

8

cousins leave none over**, so

8

is a factor of

48

— a remainder of zero means a factor.

The link. The Factor Theorem is the Remainder Theorem with the remainder equal to zero.

How do you find an unknown coefficient from a factor or a given remainder?

**Substitute the value that makes the divisor zero, set the result equal to

0

for a factor or to the given remainder, and solve; two conditions give two equations for two unknowns.

Worked example 1 — factor.**

x - 2

is a factor of

x^{3} + k x^{2} - x + 6

f (2) = 8 + 4 k - 2 + 6 = 12 + 4 k = 0 k = - 3

Worked example 2 — remainder.

2 x^{3} + a x^{2} + 3 x - 5

leaves remainder

4

when divided by

x - 1

f (1) = 2 + a + 3 - 5 = a = 4

Worked example 3 — two unknowns.

x - 2

is a factor of

x^{3} + a x^{2} + b x + 6

, and dividing by

x - 3

leaves remainder

3

f (2) = 8 + 4 a + 2 b + 6 = 0 \Rightarrow 2 a + b = - 7

f (3) = 27 + 9 a + 3 b + 6 = 3 \Rightarrow 3 a + b = - 10

Subtracting,

a = - 3

, and then

b = - 1

.

Check:

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

gives

f (2) = 0

and

f (3) = 27 - 27 - 3 + 6 = 3

.

An everyday example. If a packing clerk knows a stock must divide exactly into boxes, the missing quantity can be worked out from that condition alone.

The substance. Each condition gives one equation, so two unknowns always need two facts.

How do you factorise a cubic polynomial completely using the Factor Theorem?

Find one factor by trying small values that divide the constant term, divide to get a quadratic, and factorise that quadratic.

Worked example 1. Factorise

2 x^{3} + x^{2} - 13 x + 6

.

Step 1 — find a factor. Try

x = 2

16 + 4 - 26 + 6 = 0

, so

x - 2

is a factor.

Step 2 — find the quotient. Write

2 x^{3} + x^{2} - 13 x + 6 = (x - 2) (2 x^{2} + p x - 3)

. Comparing the

x^{2}

terms:

p - 4 = 1

, so

p = 5

.

Step 3 — factorise the quadratic.

2 x^{2} + 5 x - 3 = (2 x - 1) (x + 3)

2 x^{3} + x^{2} - 13 x + 6 = (x - 2) (2 x - 1) (x + 3)

Worked example 2. Factorise

x^{3} - 6 x^{2} + 11 x - 6

f (1) = 1 - 6 + 11 - 6 = 0 x^{3} - 6 x^{2} + 11 x - 6 = (x - 1) (x^{2} - 5 x + 6) = (x - 1) (x - 2) (x - 3)

An everyday example. **A cuboid carton with volume

x^{3} - 6 x^{2} + 11 x - 6

** could have sides

x - 1

x - 2

and

x - 3

— factorising reveals its dimensions.

The substance. Try the factors of the constant term first: for

6

, test

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

Exam tip

What earns full marks on the Remainder and Factor Theorems?

State which theorem you are using, show the substitution in full, and write the final factorised form as a product.

- **Write the value of

x

that makes the divisor zero
- Show every term** of

f (a)

before adding
- Conclude in words: is a factor, or is not a factor
- For unknowns, write each condition as a separate equation
- For cubics, show how the quotient was found
- Check by expanding the factors when time allows

The trap. Substituting

x = 2

for the divisor

x + 2

. **For

x + 2

, use

x = - 2

.**

Did you know

Why is 10 raised to any power, minus 1, always divisible by 9?

Numbers like

9

99

999

and

9999

are all divisible by

9

. The Factor Theorem explains why this never fails.

Take

f (x) = x^{n} - 1

. Then

f (1) = 1 - 1 = 0

, so **

x - 1

is always a factor.** Put

x = 10

1 0^{n} - 1 = (10 - 1) \times q (10) = 9 \times q (10)

The same reasoning with

x = 7

shows that

7^{n} - 1

is always divisible by

6

— for example,

7^{2} - 1 = 48

Exam relevance

How do the Remainder and Factor Theorems lead into JEE Main?

This is foundation work for Class 11 Complex Numbers and Quadratic Equations and Binomial Theorem, both JEE Main chapters.

What gets built on. Knowing that

x - a

is a factor exactly when

a

is a root links polynomials to their roots, which JEE Main uses to build equations and relate roots to coefficients. Remainder problems reappear in harder forms, such as finding the remainder of a large power when divided by a number using the Binomial Theorem, and finding a remainder when dividing by a quadratic.

Question types. Multiple-choice and numerical-value questions on remainders and unknown coefficients.

The trap that costs marks. Using the wrong sign when substituting for a divisor of the form

x + a

Key takeaways

What must you be able to do from this part?

- Remainder Theorem: dividing

f (x)

x - a

leaves

f (a)

- **For

x + a

** use

f (- a)

; **for

a x - b

** use

f (\frac{b}{a})

- Factor Theorem:

x - a

is a factor when

f (a) = 0

- Unknown coefficients: one equation per condition;

a = - 3

b = - 1

in the example
- Factorising cubics: find one factor, get the quadratic quotient, factorise it
- **

2 x^{3} + x^{2} - 13 x + 6 = (x - 2) (2 x - 1) (x + 3)

**

Factorise

x^{3} + 2 x^{2} - 5 x - 6

completely, then expand your answer to check it.

How can you find a remainder without dividing?

How do you use the Remainder Theorem to find the remainder when a polynomial is divided by a linear expression?

How do you use the Factor Theorem to test whether a linear expression is a factor?

How do you find an unknown coefficient from a factor or a given remainder?

How do you factorise a cubic polynomial completely using the Factor Theorem?

What earns full marks on the Remainder and Factor Theorems?

Why is 10 raised to any power, minus 1, always divisible by 9?

How do the Remainder and Factor Theorems lead into JEE Main?

What must you be able to do from this part?

Related study notes

Ready to put this into practice?