“Teach A Level Maths”
Vol. 1: AS Core Modules
22: Division and
The Remainder Theorem
© Christine Crisp
Division and the Remainder Theorem
Module C1
Module C2
AQA
Edexcel
OCR
MEI/OCR
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Division and the Remainder Theorem
Division
We’ll first look at what happens when we divide
numbers.
e.g. 7  2  3 1
2
Remainder
7
1
This can be written as
 3 
2
2
Quotient
3 is called the quotient
1 is the remainder
Algebraic expressions can be divided in a similar way
Division and the Remainder Theorem
e.g. 1 Find the quotient and remainder when x 3  1
is divided by x 2
Solution:
x3 1
x
Quotient
2

x3
x
2
x 

1
x2
1
x2
Remainder
3
Bx  C
2
x

3
x

5
e.g. 2 Write
in the form A 
3
3
x
x
3
3
2
x

3
x

5
2
x
3x  5
Solution:


3
3
x
x
x3
3x  5
2 
x3
Division and the Remainder Theorem
Exercises
1. Find the quotient and remainder when x 3  7 x 2  4
is divided by x
3
2
3
2
x
7
x
4
x

7
x

4
Solution:



x
x
x
x
4
2
 x  7x 
x
2
The quotient is x  7 x and the remainder is -4
2. Write
Solution:
x 3  2x 2  4x  3
x2
x 3  2x 2  4x  3
x
2
in the form Ax  B 

x3
2

2x 2
2

Cx  D
x2
 4x  3
x
x
x2
 4x  3
x  2 
x2
Division and the Remainder Theorem
Dividing by an expression of the form x - a can be
done in 2 ways
e.g.1 Divide 2 x  6 by x  1
Solution: Method 1 Long division.
Write the division as follows:
x
2x  2
•
x  1 2x  6
Divide the 1st term of the numerator by the 1st
term of the denominator.
Division and the Remainder Theorem
Dividing by an expression of the form x - a can be
done in 2 ways
e.g.1 Divide 2 x  6 by x  1
Solution: Method 1 Long division.
Write the division as follows:
x
2x  2
•
2
x  1 2x  6
Write this answer above the polynomial being
divided.
Division and the Remainder Theorem
Dividing by an expression of the form x - a can be
done in 2 ways
e.g.1 Divide 2 x  6 by x  1
Solution: Method 1 Long division.
Write the division as follows:
2
x  1 2x  6
2x  2
2( x  1)
 2x  2
8
•
•
Multiply x – 1 by this number . . .
and write the answer below
Subtract: 6 – (– 2) = 8
The quotient is 2 and
So, 2 x  6  2  8
the remainder is 8.
x 1
x 1
Division and the Remainder Theorem
Dividing by an expression of the form x - a can be
done in 2 ways
e.g.1 Divide 2 x  6 by x  1
Solution: Method 2 ( Inspection )
•
Write
•
( x  1)
2x  6

x 1
x 1
Copy the denominator onto the top line
Division and the Remainder Theorem
Dividing by an expression of the form x - a can be
done in 2 ways
e.g.1 Divide 2 x  6 by x  1
Solution: Method 2 ( Inspection )
•
Write
2x  2
x
2x  6
2 ( x  1)

x 1
x 1
•
Divide the 1st term of the numerator . . .
•
. . . by the 1st term of the denominator
Multiply . . .
so the 1st term at the top is now correct
2x = 2x
Division and the Remainder Theorem
Dividing by an expression of the form x - a can be
done in 2 ways
e.g.1 Divide 2 x  6 by x  1
Solution: Method 2 ( Inspection )
•
Write
•
Adjust the constant term . . .
•
2 x 66 2 ( x  1)  8

x 1
x 1
+6 = 2 + 8
2 x  6 2( x  1)
8
Separate the 2 terms:


x 1
x 1
x 1
The quotient is 2 and the remainder is 8
Division and the Remainder Theorem
Method 1 is very complicated for harder
divisions, so from now on we will use
Method 2 only.
Division and the Remainder Theorem
e.g.2 Divide
2x 2  5x  3
by x  1
Solution:
2x 2  5x  3

x 1
( x  1)
x 1
Write the denominator on the top line
Division and the Remainder Theorem
e.g.2 Divide
2x 2  5x  3
by x  1
Solution:
2x 2  5x  3

x 1
2x ( x  1)
Correct the 1st term.
2x 2  2 x 2
x 1
Division and the Remainder Theorem
e.g.2 Divide
2x 2  5x  3
by x  1
Solution:
2 x 2  5 xx  3

x 1
•
2x ( x  1)  7 ( x  1)
x 1
Copy the denominator and correct the next term.
 5 x  2 x  7 x
Division and the Remainder Theorem
e.g.2 Divide
2x 2  5x  3
by x  1
Solution:
2 x 2  5 x  33

x 1
2x ( x  1)  7 ( x  1)  4
x 1
•
Correct the last term . . .
3 7 4
•
Check the numerator.
2x 2  2x  7x  7  4  2x 2  5x  3
So,
4
2x 2  5x  3
 2x  7 
x 1
x 1
Division and the Remainder Theorem
e.g.2 Divide
2x 2  5x  3
by x  1
Solution:
2x 2  5x  3

x 1
2x ( x  1)  7 ( x  1)  4
x 1
•
Write the denominator on the top line
•
Correct the 1st term.
•
Copy the denominator and correct the next term.
•
Correct the last term . . .
•
Check the numerator.
So,
4
2x 2  5x  3
 2x  7 
x 1
x 1
Division and the Remainder Theorem
e.g.3 Divide
x 3  2x 2  3
by x  1
Solution:
x 3  2x 2
x 1
3

x 2 ( x  1)
x 1
x3  x3
Tip: As there is no linear x-term leave a space
Division and the Remainder Theorem
e.g.3 Divide
x 3  2x 2  3
by x  1
Solution:
x 3  2x 2
x 1
3

x 2 ( x  1)  x ( x  1)
x 1
 2x 2   x 2  x 2
Division and the Remainder Theorem
e.g.3 Divide
x 3  2x 2  3
by x  1
Solution:
x 3  2x 2
x 1
3

x 2 ( x  1)  x ( x  1) 1 ( x  1)
x 1
Be careful!
 x( 1)   x
0  x  x
Division and the Remainder Theorem
e.g.3 Divide
x 3  2x 2  3
by x  1
Solution:
x 3  2x 2
x 1
3

x 2 ( x  1)  x ( x  1) 1 ( x  1) 2
x 1
3  1  2
Division and the Remainder Theorem
e.g.3 Divide
x 3  2x 2  3
by x  1
Solution:
x 3  2x 2
x 1

3

x 2 ( x  1)  x ( x  1) 1 ( x  1) 2
x 1
x 3  2x 2  3
2
2
 x  x 1
x 1
x 1
The quotient is x 2  x  1
and the remainder is 2
Division and the Remainder Theorem
Exercises
1. Divide
x 2  4 x  4 by x + 2
Solution:
x ( x  2) 2 ( x  2)  8
x2
8
x 2  4x  4
x

2



x2
x2
The quotient is x + 2 and the remainder is  8
x 2  4x  4

x2
2. Divide x 3  3 x 2  4 x  1 by x  1
The solution is on the next slide
Division and the Remainder Theorem
Solution:
x 3  3x 2  4x  1

x 1

x 2 ( x  1)  2 x ( x  1)  6 ( x  1)  7
x 1
x 3  3x 2  4x  1
7
2
 x  2x  6 
x 1
x 1
The quotient is x 2  2 x  6 and the remainder is 7
Division and the Remainder Theorem
The remainder theorem gives the remainder when a
polynomial is divided by a linear factor
It doesn’t enable us to find the quotient
e.g. Find the remainder when x 3  7 x 2  4 is
divided by x - 1
The method is the same as that for the factor
theorem
Let
f ( x)  x 3  7 x 2  4
f (1)  (1) 3  7(1) 2  4
4
The remainder is 4
The remainder theorem says that if we divide a
polynomial f ( x ) by x – a, the remainder is given by f (a )
Division and the Remainder Theorem
Proof of the Remainder theorem
Let f ( x ) be a polynomial that is divided by x - a
The quotient is another polynomial and the
remainder is a constant.
We can write
f ( x)
R
 g( x ) 
xa
xa
Multiplying by x – a gives
f ( x )  ( x  a ) g( x )  R
f (a )  (a  a ) g(a )  R
So,

f (a )  R
Division and the Remainder Theorem
e.g.1 Find the remainder when x 3  3 x 2  4 x  1 is
divided by x  2
Solution: Let
f ( x)  x 3  3 x 2  4 x  1
So, a  2

R  f ( 2)
f (2)  (2) 3  3(2) 2  4(2)  1
 8  12  8  1

R  13
Tip: Use the remainder theorem to check the
remainder when using long division. If the remainder
is correct the quotient will be too!
Division and the Remainder Theorem
e.g.2 Find the remainder when x 3  x 2  2 x  4 is
divided by 2 x  1
Solution: Let
f ( x)  x 3  x 2  2 x  4
To find the value of a, we let 2x + 1 = 0. The value
of x gives the value of a.
2 x  1  0  x   12 so, R  f  12


 
3
2
1
1
1
f   2     2     2   2  12  4
f   12    18  14  1  4
1  2  8  32
R
8
21
 R
8
Division and the Remainder Theorem
Exercises
1.
Find the remainder when x 3  4 x 2  3 x  5 is
divided by x + 1
3
2
Solution: Let f ( x )  x  4 x  3 x  5



2.
f (1)  (1) 3  4(1) 2  3(1)  5
R  1  4  3  5
R  5
If x + 2 is a factor of x 3  ax2  bx  6 and if the
remainder on division by x – 1 is – 3, find the
values of a and b.
Division and the Remainder Theorem
Exercises
2.
If x + 2 is a factor of x 3  ax2  bx  6 and if the
remainder on division by x – 1 is – 3, find the
values of a and b.
Solution: Let
f ( x)  x 3  ax2  bx  6
f (2)  0  (2) 3  a(2) 2  b(2)  6  0

4a  2b  14  0

f (1)  3 

1  a  b  6  3
(1) + (2) 

Substitute in (2)
2a  b  7 - - - (1)
a  b  2 - - - (2)
3a  9  a  3
b  1
Division and the Remainder Theorem
Division and the Remainder Theorem
The following slides contain repeats of
information on earlier slides, shown without
colour, so that they can be printed and
photocopied.
For most purposes the slides can be printed
as “Handouts” with up to 6 slides per sheet.
Division and the Remainder Theorem
Dividing by an expression of the form x - a can be
done in 2 ways:
Method 1: Long Division
Method 2: Inspection
Method 2 is usually easier but an example of
method 1 is given next.
Division and the Remainder Theorem
e.g.1 Divide 2 x  6 by x  1
Solution: Method 1 Long division.
Write the division as follows:
x
2x  2
•
•
•
•
2
x  1 2x  6
2x  2
2( x  1)
 2x  2
8
Divide the 1st term of the numerator by the 1st
term of the denominator.
Multiply x – 1 by this number . . .
Write this answer above the polynomial being
divided.
Subtract: 6 – (– 2) = 8 and write the answer below
The quotient is 2 and
So, 2 x  6  2  8
the remainder is 8.
x 1
x 1
Division and the Remainder Theorem
e.g.1 Divide 2 x  6 by x  1
Solution: Method 2 ( Inspection )
•
Write
•
•
Copy the denominator onto the top line
Divide the 1st term of the numerator
by the 1st term of the denominator
Multiply so the 1st term at the top is now correct
Adjust the constant term
2( x  1)
8
Separate the 2 terms: 2 x  6 

•
•
•
2x  6
2 ( x  1)  8

x 1
x 1
x 1
x 1
The quotient is 2 and the remainder is 8
x 1
Division and the Remainder Theorem
The remainder theorem says that if we divide a
polynomial f ( x ) by x – a, the remainder is given by f (a )
Proof of the Remainder theorem
Let f ( x ) be a polynomial that is divided by x - a
The quotient is another polynomial and the
remainder is a constant.
We can write
f ( x)
R
 g( x ) 
xa
xa
Multiplying by x – a gives
So,
f ( x )  ( x  a ) g( x )  R
f (a )  (a  a ) g(a )  R
R
Division and the Remainder Theorem
e.g.1 Find the remainder when x 3  3 x 2  4 x  1 is
divided by x  2
Solution: Let
f ( x)  x 3  3 x 2  4 x  1
So, a  2

R  f ( 2)
f (2)  (2) 3  3(2) 2  4(2)  1
 8  12  8  1

R  13
Tip: Use the remainder theorem to check the
remainder when using long division. If the remainder
is correct the quotient will be too!
Division and the Remainder Theorem
e.g.2 Find the remainder when x 3  x 2  2 x  4 is
divided by 2 x  1
Solution: Let
f ( x)  x 3  x 2  2 x  4
To find the value of a, we let 2x + 1 = 0. The
value of x gives the value of a.
2 x  1  0  x   12 so, R  f  12


 
3
2
1
1
1
f   2     2     2   2  12  4
f   12    18  14  1  4
1  2  8  32
R
8

21
R
8