Factor & Remainder Theorems | Edexcel IGCSE Further Maths Revision Notes 2019

Factor Theorem

What is the factor theorem?

The factor theorem is used to find the linear factors of a function
- This is closely related to finding the roots (or solutions) of a function or equation
For a function , the factor theorem tells us that
- If $f (a) = 0$ , then $(x - a)$ is a factor of $f (x)$
- If $(x - a)$ is a factor of $f (x)$ , then $f (a) = 0$

How do I use the factor theorem?

Consider the function where is a factor
- Then by the factor theorem we know that $f (a) = 0$
  - I.e., $x = a$ is a solution to the equation $f (x) = 0$
Or consider the function $f (x)$ where $f (a) = 0$
- Then by the factor theorem we know that $(x - a)$ is a factor of $f (x)$
- Therefore $f (x) = (x - a) \times Q (x)$
  - where $Q (x)$ is a function that is also a factor of $f (x)$
- Hence
  - I.e. $Q (x)$ is the quotient when $P (x)$ is divided by $(x - a)$
  - And the remainder is equal to zero
If the linear factor has a coefficient of x (other than 1) you must first factorise out the coefficient
- For the linear factor
  - $f (\frac{c}{b}) = 0$
  - $f (x) = b (x - \frac{c}{b}) \times Q (x)$

Exam Tip

Be careful with the minus sign in a factor
- That means $a$ is a solution to $f (x) = 0$ , not $- a$ !
If you are looking for integer solutions to (where is a polynomial)
- those solutions will always be factors of the constant term in $f (x)$

Worked example

Consider the function

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

. Given that

x = 2

is a solution to the equation

f (x) = 0

, write down a linear factor of

f (x)

By the factor theorem, if

f (a) = 0

then

(x - a)

is a factor of

f (x)

x - 2

Use the factor theorem to determine whether

(x + 1)

is a factor of

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

By the factor theorem,

(x - a)

can only be a factor of

g (x)

g (a) = 0

But be careful – here

a

is equal to

- 1

, not

1

\begin{array}{rcl} g (- 1) & = & 2 {(- 1)}^{3} + 3 {(- 1)}^{2} - (- 1) + 5 \\ = & - 2 + 3 + 1 + 5 \\ = & 7 \end{array}

g (- 1) \neq 0

, so $(x + 1)$ is not a factor of $g (x)$

It is given that

(2 x - 3)

is a factor of

h (x) = 2 x^{3} - b x^{2} + 7 x - 6

. Find the value of

b

(2 x - 3) = 2 (x - \frac{3}{2})

, so

(x - \frac{3}{2})

is a factor of

h (x)

Therefore by the factor theorem,

h (\frac{3}{2}) = 0

\begin{array}{rcl} 2 {(\frac{3}{2})}^{3} - b {(\frac{3}{2})}^{2} + 7 (\frac{3}{2}) - 6 & = & 0 \\ \frac{27}{4} - \frac{9}{4} b + \frac{21}{2} - 6 & = & 0 \\ \frac{45}{4} - \frac{9}{4} b & = & 0 \\ \frac{9}{4} b & = & \frac{45}{4} \\ b & = & \frac{4}{9} \times \frac{45}{4} \end{array}

b = 5

Remainder Theorem

What is the remainder theorem?

The remainder theorem is used to find the remainder when we divide a polynomial function by a linear function
When a polynomial function is divided by a linear function , the value of the remainder $R$ is given by
- Note, if $f (a) = 0$ then $(x - a)$ is a factor of $f (x)$ ; this is the factor theorem

How do I use the remainder theorem?

Consider a polynomial function and a linear function
- - $Q (x)$ is the quotient (also a polynomial function)
  - $R$ is the remainder (a real number)
- This may also be written as $f (x) = Q (x) \times (x - a) + R$
- The remainder theorem tells us that
  - I.e. we don't need to do the algebraic division to find the remainder!
If the linear factor has a coefficient of x (other than 1) then you must first factorise out the coefficient
- For the linear function $(b x - c) = b (x - \frac{c}{b})$
  - $R = f (\frac{c}{b})$

Exam Tip

Be careful with the minus sign in
- You need to put $a$ into $f (x)$ to find the remainder, not $- a$ !

Worked example

Find the remainder when the function

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

is divided by

(x - 2)

We're dividing by

(x - a) = (x - 2)

a = 2

By the remainder theorem the remainder will be

f (a) = f (2)

\begin{array}{rcl} f (2) & = & 2 {(2)}^{4} - 2 {(2)}^{3} - {(2)}^{2} - 3 (2) + 1 \\ = & 32 - 16 - 4 - 6 + 1 \\ = & 7 \end{array}

Remainder = 7

The remainder when

g (x) = 2 x^{3} + x^{2} + b x + 1

is divided by

(2 x + 1)

is 3. Find the value of

b

(2 x + 1) = 2 (x + \frac{1}{2}) = 2 (x - (- \frac{1}{2}))

So here the value of

a

to use is

- \frac{1}{2}

By the remainder theorem the remainder will be equal to

g (- \frac{1}{2})

\begin{array}{rcl} 2 {(- \frac{1}{2})}^{3} + {(- \frac{1}{2})}^{2} + b (- \frac{1}{2}) + 1 & = & 3 \\ - \frac{1}{4} + \frac{1}{4} - \frac{1}{2} b + 1 & = & 3 \\ 1 - \frac{b}{2} & = & 3 \\ - \frac{b}{2} & = & 2 \end{array}

b = - 4

Factor & Remainder Theorems (Edexcel IGCSE Further Maths)

Revision Note

Factor Theorem

What is the factor theorem?

How do I use the factor theorem?

Exam Tip

Worked example

Remainder Theorem

What is the remainder theorem?

How do I use the remainder theorem?

Exam Tip

Worked example

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Author: Roger