1. “Teach A Level Maths” Vol. 1: AS Core Modules
22: Division and
The Remainder Theorem
© Christine Crisp

2. Module C1 Module C2 AQA Edexcel MEI/OCR OCR
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3. Division
We’ll first look at what happens when we divide numbers.
e.g.
Remainder
This can be written as
Quotient
3 is called the quotient
1 is the remainder
Algebraic expressions can be divided in a similar way

4. e.g. 1 Find the quotient and remainder when
is divided by
Solution:
Remainder
Quotient
e.g. 2 Write in the form
Solution:

5. Exercises
Find the quotient and remainder when is divided by x 1.
Solution:
The quotient is and the remainder is -4 2.
Write in the form
Solution:

6. Dividing by an expression of the form x - a can be done in 2 ways
e.g.1 Divide by
Solution: Method 1 Long division.
Write the division as follows:
Divide the 1st term of the numerator by the 1st term of the denominator.

7. Dividing by an expression of the form x - a can be done in 2 ways
e.g.1 Divide by
Solution: Method 1 Long division.
Write the division as follows:
Write this answer above the polynomial being divided.

8. Dividing by an expression of the form x - a can be done in 2 ways
e.g.1 Divide by
Solution: Method 1 Long division.
Write the division as follows: 8
Multiply x – 1 by this number . . .
and write the answer below
Subtract:
6 – (– 2) = 8
So,
The quotient is 2 and the remainder is 8.

9. Dividing by an expression of the form x - a can be done in 2 ways
e.g.1 Divide by
Solution: Method 2 ( Inspection )
Write
Copy the denominator onto the top line

10. Dividing by an expression of the form x - a can be done in 2 ways
e.g.1 Divide by
Solution: Method 2 ( Inspection )
Write
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

11. Dividing by an expression of the form x - a can be done in 2 ways
e.g.1 Divide by
Solution: Method 2 ( Inspection )
Write
Adjust the constant term . . .
+ 6 =
- 2
+ 8
Separate the 2 terms:
The quotient is 2 and the remainder is 8

12. Method 1 is very complicated for harder divisions, so from now on we will use Method 2 only.

13. e.g.2 Divide by
Solution:
Write the denominator on the top line

14. e.g.2 Divide by
Solution:
Correct the 1st term.

15. e.g.2 Divide by
Solution:
Copy the denominator and correct the next term.

16. e.g.2 Divide by
Solution:
Correct the last term . . .
Check the numerator.
So,

17. e.g.2 Divide by
Solution:
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,

18. e.g.3 Divide by
Solution:
Tip: As there is no linear x-term leave a space

19. e.g.3 Divide by
Solution:

20. e.g.3 Divide by
Solution:
Be careful!

21. e.g.3 Divide by
Solution: -1

22. e.g.3 Divide by
Solution:
The quotient is and the remainder is

23. Exercises
Divide by x + 2 1.
Solution:
The quotient is x + 2 and the remainder is
2. Divide by
The solution is on the next slide

24. Solution:
The quotient is and the remainder is

25. 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 is divided by x - 1
The method is the same as that for the factor theorem
The remainder is 4
The remainder theorem says that if we divide a polynomial by x – a, the remainder is given by

26. Proof of the Remainder theorem
Let be a polynomial that is divided by x - a
The quotient is another polynomial and the remainder is a constant.
We can write
Multiplying by x – a gives
So,

27. e.g.1 Find the remainder when is divided by
Solution: Let
So,
Tip: Use the remainder theorem to check the remainder when using long division. If the remainder is correct the quotient will be too!

28. e.g.2 Find the remainder when is divided by
Solution: Let
To find the value of a, we let 2x + 1 = 0. The value of x gives the value of a.
so,

29. Exercises 1.
Find the remainder when is divided by x + 1
Solution: Let 2.
If x + 2 is a factor of and if the remainder on division by x – 1 is – 3, find the values of a and b.

30. Exercises 2.
If x + 2 is a factor of and if the remainder on division by x – 1 is – 3, find the values of a and b.
Solution: Let
- - - (1)
- - - (2)
(1) + (2)
Substitute in (2)

32. 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.

33. 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.

34. Write the division as follows:
e.g.1 Divide by
Multiply x – 1 by this number . . .
and write the answer below
Subtract:
6 – (– 2) = 8 8
The quotient is 2 and the remainder is 8.
So,
Solution: Method 1 Long division.
Divide the 1st term of the numerator by the 1st term of the denominator.
Write this answer above the polynomial being divided.

35. e.g.1 Divide by
Write
Adjust the constant term
Separate the 2 terms:
The quotient is 2 and the remainder is 8
Solution: Method 2 ( Inspection )
Copy the denominator onto the top line
Divide the 1st term of the numerator
Multiply
by the 1st term of the denominator
so the 1st term at the top is now correct

36. The remainder theorem says that if we divide a polynomial by x – a, the remainder is given by
Proof of the Remainder theorem
Let be a polynomial that is divided by x - a
The quotient is another polynomial and the remainder is a constant.
We can write
Multiplying by x – a gives
So,

37. Solution: Let
So,
Tip: Use the remainder theorem to check the remainder when using long division. If the remainder is correct the quotient will be too!
e.g.1 Find the remainder when is divided by

38. e.g.2 Find the remainder when is divided by
Solution: Let
To find the value of a, we let 2x + 1 = 0. The value of x gives the value of a.
so,