Algebraic Division and the Factor Theorem

Dividing polynomials and finding factors systematically

SPEC 2.6 • ALGEBRA AND FUNCTIONS

Why Divide Polynomials?

We can factorise quadratics easily, but what about cubic polynomials like

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

?

The GoalTo factorise cubics completely, we need two key tools:

The Factor Theorem — to find factors

Polynomial Division — to break down the cubic

Key Concept

Just as

15 \div 3 = 5

tells us that

15 = 3 \times 5

, polynomial division lets us write:

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

The Division Identity

When we divide polynomials, we get a quotient and possibly a remainder:

Division Identity

f (x) = divisor \times quotient + remainder

When dividing

f (x)

by

(x - a)

:

f (x) = (x - a) \times Q (x) + R

Example

When we divide

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

by

(x - 2)

:

x^{3} + 2 x^{2} - 5 x - 6 = (x - 2) (x^{2} + 4 x + 3) + 0

Polynomial Long Division

The process follows the same steps as numerical long division:

Method

Divide the leading term of the dividend by the leading term of the divisor

Multiply the entire divisor by this result

Subtract from the dividend

Bring down the next term

Repeat until the remainder has lower degree than the divisor

Key Point

When dividing by

(a x + b)

, divide by the leading term

a x

, not just

x

!

Worked Example01

Divide $x^{3} + 2 x^{2} - 5 x - 6$ by $(x - 2)$ .

Solution

x²+ 4x+ 3

x³+ 2x²− 5x− 6

− 2x²

4x²− 5x

4x²− 8x

3x− 6

So the remainder is 0.

Answer

x^{3} + 2 x^{2} - 5 x - 6 = (x - 2) (x^{2} + 4 x + 3)

Worked Example02

Divide $2 x^{3} - 5 x^{2} - 16 x + 10$ by $(x - 4)$ .

Solution

2x²+ 3x− 4

2x³− 5x²− 16x+ 10

− 8x²

3x²− 16x

3x²− 12x

−4x+ 10

−4x+ 16

So the remainder is −6.

Answer

2 x^{3} - 5 x^{2} - 16 x + 10 = (x - 4) (2 x^{2} + 3 x - 4) - 6

Polynomials with Missing Terms

When a polynomial is missing a power of

x

, we must include it with coefficient 0.

Critical Rule

Always write missing terms as 0 times that power!

x^{3} + 8

becomes

x^{3} + 0 x^{2} + 0 x + 8

x^{3} - 5 x + 2

becomes

x^{3} + 0 x^{2} - 5 x + 2

x^{4} + 3 x^{2} - 1

becomes

x^{4} + 0 x^{3} + 3 x^{2} + 0 x - 1

Why This Matters

Without placeholder zeros, the columns won't align properly during subtraction!

Worked Example03

Divide $x^{3} - 27$ by $(x - 3)$ .

Solution

First: Rewrite as

x^{3} + 0 x^{2} + 0 x - 27

x²+ 3x+ 9

x³+ 0x²+ 0x− 27

− 3x²

3x²+ 0x

3x²− 9x

9x− 27

This is the difference of cubes: $a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2})$

Answer

x^{3} - 27 = (x - 3) (x^{2} + 3 x + 9)

Divisors Like (2x − 1)

When the divisor has a leading coefficient other than 1, we divide by that coefficient:

Key DifferenceFor divisor

(2 x - 1)

, divide leading terms by

2 x

, not

x

:

6 x^{3} \div 2 x = 3 x^{2}

4 x^{2} \div 2 x = 2 x

10 x \div 2 x = 5

Remember

The process is identical — just be careful with the division step!

Worked Example04

Divide $6 x^{3} + 7 x^{2} - 15 x + 4$ by $(2 x - 1)$ .

Solution

3x²+ 5x− 5

6x³+ 7x²− 15x+ 4

− 3x²

10x²− 15x

10x²− 5x

−10x+ 4

−10x+ 5

So the remainder is −1.

Answer

6 x^{3} + 7 x^{2} - 15 x + 4 = (2 x - 1) (3 x^{2} + 5 x - 5) - 1

Dividing by Quadratics

When the divisor is a quadratic, the quotient has lower degree:

Degree Ruledegree(quotient) = degree(dividend) − degree(divisor)

Cubic ÷ Quadratic = Linear quotient (+ possible remainder)

Quartic ÷ Quadratic = Quadratic quotient

Important

The remainder can now be up to degree 1 (linear), since it must be lower degree than the quadratic divisor.

Worked Example05

Divide $x^{3} + 2 x^{2} - x + 3$ by $(x^{2} + 1)$ .

Solution

Write divisor as

x^{2} + 0 x + 1

for alignment.

+ 2

x³+ 2x²− x+ 3

x³+ 0x²+ x

2x²− 2x+ 3

2x²+ 0x+ 2

−2x+ 1

So the remainder is

- 2 x + 1

(linear).

Answer

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

The Factor Theorem

Factor TheoremFor a polynomial

f (x)

:

If

f (a) = 0

, then

(x - a)

is a factor of

f (x)

Conversely: If

(x - a)

is a factor, then

f (a) = 0

Understanding

The factor theorem connects roots and factors:

x = a

is a root of

f (x) = 0

⟺

(x - a)

is a factor of

f (x)

Watch the Signs!

If

f (2) = 0

, then

(x - 2)

is a factor

If

f (- 3) = 0

, then

(x + 3)

is a factor

The Remainder Theorem

Remainder TheoremWhen

f (x)

is divided by

(x - a)

:

Remainder =

f (a)

Key Insight

You can find the remainder without doing the division!

Just substitute

x = a

into

f (x)

.

Example

Find the remainder when

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

is divided by

(x - 2)

:

f (2) = 8 - 8 + 5 = 5

→ Remainder = 5

Connection

The Factor Theorem is a special case: when remainder = 0,

(x - a)

is a factor.

Finding Factors

StrategyTest factors of the constant term

For

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

, test:

x = \pm 1, \pm 2, \pm 3, \pm 6

Quick Checks

For $x = 1$ : Add all coefficients. If sum = 0, then

(x - 1)

is a factor.

For $x = - 1$ : Alternate signs and add. If result = 0, then

(x + 1)

is a factor.

Worked Example06

Factorise $f (x) = x^{3} - 6 x^{2} + 11 x - 6$ completely.

Solution

•

1. Test

x = 1

:

f (1) = 1 - 6 + 11 - 6 = 0

✓ So

(x - 1)

is a factor

•

2. Divide to get:

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

•

3. Factorise quadratic:

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

Answer

f (x) = (x - 1) (x - 2) (x - 3)

Finding Unknown Coefficients

Method

If

(x - a)

is a factor, substitute

x = a

into

f (x)

Set the result equal to 0 (or given remainder)

Solve for the unknown coefficient

Two Unknowns

Need two equations — usually two factors, or a factor and a remainder.

Worked Example07

The polynomial $f (x) = x^{3} + a x^{2} + b x - 12$ has factors $(x - 2)$ and $(x + 3)$ . Find $a$ and $b$ .

Solution

•

1.

f (2) = 0

:

8 + 4 a + 2 b - 12 = 0

→

2 a + b = 2

... (1)

•

2.

f (- 3) = 0

:

- 27 + 9 a - 3 b - 12 = 0

→

3 a - b = 13

... (2)

•

3. Add (1) + (2):

5 a = 15

→

a = 3

•

4. From (1):

b = 2 - 6 = - 4

Answer

a = 3

,

b = - 4

Practice

Question 1: Divide

4 x^{3} - 8 x^{2} + x - 2

by

(2 x - 1)

.

Solution

2x²− 3x− 1

4x³− 8x²+ x− 2

− 2x²

−6x²+ x

−6x²+ 3x

−2x− 2

−2x+ 1

So the remainder is −3.

Answer

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

Practice

Question 2: Divide

x^{4} + x^{3} - 7 x^{2} - x + 6

by

(x^{2} - 1)

.

Solution

Write

x^{2} - 1 = x^{2} + 0 x - 1

x²+ x− 6

x⁴+ x³− 7x²− x+ 6

x⁴+ 0x³− x²

x³− 6x²− x

x³+ 0x²− x

−6x²+ 0x+ 6

Can factor further: $= (x - 1) (x + 1) (x + 3) (x - 2)$

Answer

x^{4} + x^{3} - 7 x^{2} - x + 6 = (x^{2} - 1) (x^{2} + x - 6)

Practice

Question 3: Given that when

f (x) = x^{3} + p x + 6

is divided by

(x + 2)

, the remainder is 4. Find

p

.

Solution

•

1. By Remainder Theorem: remainder =

f (- 2)

•

2.

f (- 2) = (- 2)^{3} + p (- 2) + 6 = - 8 - 2 p + 6 = - 2 - 2 p

•

3.

- 2 - 2 p = 4

→

- 2 p = 6

→

p = - 3

Answer

p = - 3

Summary

Polynomial Long Division

Divide leading terms, multiply, subtract, bring down — repeat.

Missing terms: Use 0 as placeholder coefficient.

Different Divisors

(2x − 1): Divide by 2x, not x

Quadratic: Remainder can be linear

Key Theorems

Factor Theorem:

f (a) = 0

⟺

(x - a)

is a factor

Remainder Theorem: Remainder =

f (a)

when dividing by

(x - a)