1. Remainder and Factor Theorem
Name : __________________( )
Class : _________
Date : _________

2. 3x3 4x2 x 2 1 1 1 1 Dividing by a Monomial
If the divisor only has one term, split the polynomial up into a fraction for each term.
divisor
Now reduce each fraction.
3x3
4x2 x 2 1 1 1 1

3. Subtract (which changes the sign of each term in the polynomial)
Long Division
If the divisor has more than one term, perform long division. You do the same steps with polynomial division as with integers. Let's do two problems, one with integers you know how to do and one with polynomials and copy the steps.
Subtract (which changes the sign of each term in the polynomial)
Now multiply by the divisor and put the answer below. x
+ 11 2 1
Bring down the next number or term
Multiply and put below
Now divide 3 into or x into 11x
First divide 3 into or x into x2
x x2 + 8x - 5
Remainder added here over divisor 64
x2 – 3x
subtract 5 8
11x
- 5 32
11x - 33
This is the remainder 26 28
So we found the answer to the problem x2 + 8x – 5  x – 3 or the problem written another way:

4. Remainder added here over divisor
Let's Try Another One
If any powers of terms are missing you should write them in with zeros in front to keep all of your columns straight.
Subtract (which changes the sign of each term in the polynomial) y
- 2
Write out with long division including 0y for missing term
Multiply and put below
Bring down the next term
Multiply and put below
y y2 + 0y + 8
Divide y into y2
Divide y into -2y
Remainder added here over divisor
y2 + 2y
subtract
-2y
+ 8
- 2y - 4
This is the remainder 12

5. Reminder:
For a polynomial function , the factor theorem says that:
if then is a factor
e.g. 1 Find one linear factor of
is a factor

6. x and 2x are linear factors of
Now suppose
Possible linear factors are now
x and 2x are linear factors of

7. x and 2x are linear factors of
Now suppose
Possible linear factors are now
x and 2x are linear factors of
1 and 3 are the factors of 3

8. x and 2x are linear factors of
Now suppose
Possible linear factors are now
x and 2x are linear factors of
1 and 3 are the factors of 3
To check if is a factor we use x = 1
The value of x comes from letting
So, to check we let
So,

9. So, to factorise
Consider:
So,
Starting with the easiest:
( The only difference is that the signs of the 1st and 3rd terms change. )
Can you see that either?

10. So,
We can now complete the factorisation by inspection:

11. So,
We can now complete the factorisation by inspection:

12. So,
We can now complete the factorisation by inspection:

13. So,
We can now complete the factorisation by inspection:
Checking the quadratic term:

14. So,
We can now complete the factorisation by inspection:
Checking the quadratic term:

15. So,
We can now complete the factorisation by inspection:
Checking the linear term:

16. So,
We can now complete the factorisation by inspection:
Checking the linear term:

17. So,
We can now complete the factorisation by inspection:
The quadratic factor has no linear factors, so the factorisation is complete.

18. e.g. 2 Factorise
Solution: Let
Possible factors are
So,
BUT we can save time by noticing that, since all the terms of are positive,
and
cannot be factors.
Also, you may have spotted that the coefficients show that isn’t a factor.

19. We can start with
So,

20. We can start with
So,

21. We can start with
So,

22. We can start with
So,

23. We can start with
So,
Check the linear term

24. We can start with
So,
Check the linear term

25. We can start with
So,
There are no more factors.

26. SUMMARY
Factorising Cubic Functions
Use the factor theorem to find one linear factor
For checking for a factor of the form
let
and solve to find the value of x to substitute.
Use inspection to find the quadratic factor.
Check for further factors of the quadratic.

27. Exercises
1. Factorise the following cubic:
2. Show that is a factor of the following cubic and then factorise it completely:

28. 1. Let
Possible factors are
Solution:
So,

29. So,
There are no more factors.
2. Let
Solution: Show is a factor
So,

30. The slides that follow remind you about the remainder theorem.
The theorem is extended to consider linear terms of the form in the same way as the factor theorem.

31. 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. 1 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

32. 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,

33. SUMMARY
Extending the Remainder theorem
The remainder when a polynomial is divided by is given by R where

34. We can start with
So,
There are no more factors.

35. To find the remainder when a polynomial is divided by
solve to find the value of x
let
substitute into to find the remainder

36. e.g. 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,