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

2. Division and the Remainder Theorem Module C1 AQA Edexcel OCRMEI/OCR Module C2 "Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages"

3. Division and the Remainder Theorem We’ll first look at what happens when we divide numbers. e.g. Quotient This can be written as Remainder 3 is called the quotient 1 is the remainder Algebraic expressions can be divided in a similar way Division

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

5. Division and the Remainder Theorem 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

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

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

8. Division and the Remainder Theorem Write the division as follows: e.g.1 Divide by Dividing by an expression of the form x - a can be done in 2 ways 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.

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

10. Division and the Remainder Theorem Solution: Method 2 ( Inspection ) Divide the 1 st term of the numerator... Dividing by an expression of the form x - a can be done in 2 ways e.g.1 Divide by Write Multiply...... by the 1 st term of the denominator so the 1st term at the top is now correct 2x = 2x

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

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

13. Division and the Remainder Theorem Solution: e.g.2 Divide by Write the denominator on the top line

14. Division and the Remainder Theorem Solution: e.g.2 Divide by Correct the 1 st term.

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

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

17. Division and the Remainder Theorem Solution: e.g.2 Divide by Correct the last term... Check the numerator. So, Write the denominator on the top line Correct the 1 st term. Copy the denominator and correct the next term.

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

19. Division and the Remainder Theorem Solution: e.g.3 Divide by

20. Division and the Remainder Theorem Solution: e.g.3 Divide by Be careful!

21. Division and the Remainder Theorem Solution: e.g.3 Divide by 11

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

23. Division and the Remainder Theorem Divide by x + 2 1. Solution: The quotient is x + 2 and the remainder is The solution is on the next slide Exercises2. Divide by

24. Division and the Remainder Theorem Solution: The quotient is and the remainder is

25. Division and the Remainder Theorem e.g. Find the remainder when is divided by x - 1 The remainder theorem gives the remainder when a polynomial is divided by a linear factor It doesn’t enable us to find the quotient 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. Division and the Remainder Theorem 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. Division and the Remainder Theorem 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. Division and the Remainder Theorem 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. Division and the Remainder TheoremExercises Find the remainder when is divided by x + 1 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. Division and the Remainder TheoremExercises 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 - - - (2) - - - (1) (1) + (2) Substitute in (2)

31. Division and the Remainder Theorem

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. 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.

34. Division and the Remainder Theorem 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 1 st term of the numerator by the 1 st term of the denominator. Write this answer above the polynomial being divided.

35. Division and the Remainder Theorem 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 1 st term of the numerator Multiply by the 1 st term of the denominator so the 1st term at the top is now correct

36. Division and the Remainder Theorem 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. Division and the Remainder Theorem 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. Division and the Remainder Theorem 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,