1. The document discusses factorizing polynomials by using the remainder and factor theorems. 2. The remainder theorem states that the remainder when a polynomial f(x) is divided by x - a is f(a). 3. The factor theorem states that if f(a) = 0, then x - a is a factor of f(x).

1. IGCSEFM Factor Theorem
Dr J Frost (jfrost@tiffin.kingston.sch.uk)
Last modified: 22nd February 2016
Objectives: (from the specification)

2. Dividing Polynomials
You’re used to dividing whole numbers to find a remainder…
Bro Definition: Recall that a
polynomial is just an
expression of the form
𝑎 + 𝑏𝑥 + 𝑐𝑥2
+ 𝑑𝑥3
+ ⋯.
Quadratics and cubics are
examples of polynomials.
7
3
= 2 +
1
3
dividend
divisor
quotient
remainder
?
?
?
?
It’s actually possible to do it when dividing polynomials too…
𝑥2
𝑥 − 1
= 𝑥 +
1
𝑥 − 1
Quotient
Remainder

3. We’re trying to work out the remainder when we divide a
polynomial 𝑓 𝑥 by 𝑥 − 𝑎
Therefore:
𝑓 𝑎 = 𝑟
What if 𝑓 𝑎 = 0?
The remainder is 0, thus 𝒙 − 𝒂 must be a factor of 𝒇(𝒙)
Remainder and Factor Theorem
𝑓 𝑥
𝑥 − 𝑎
= 𝑞 +
𝑟
𝑥 − 𝑎
Multiplying both sides by 𝑥 − 𝑎:
𝑓 𝑥 = 𝑥 − 𝑎 𝑞 + 𝑟
Quotient
Remainder
i.e. We get the remainder
when dividing by 𝑥 − 𝑎 by
just subbing 𝑎 into the original
function.
!
?
?
?

4. Remainder and Factor Theorem
Remainder Theorem
For a polynomial 𝑓(𝑥), the remainder when
𝑓(𝑥) is divided by 𝑥 − 𝑎 is 𝑓 𝑎 .
Factor Theorem
If 𝑓 𝑎 = 0, then by above, the remainder is
0. Thus (𝑥 − 𝑎) is a factor of 𝑓 𝑥 .
!
!
What is the remainder when 𝑥2
− 3𝑥 + 2 is divided by 𝑥 − 2?
𝒇 𝟐 = 𝟐𝟐 − 𝟑 𝟐 + 𝟐 = 𝟎
(Since no remainder, 𝒙 − 𝟐 must be a factor of 𝒙𝟐 − 𝟑𝒙 + 𝟐)
What is the remainder when 𝑥3
+ 2𝑥 is divided by 𝑥 + 3?
𝒇 −𝟑 = −𝟑 𝟑 + 𝟐 −𝟑 = −𝟑𝟑
?
?

5. Examples
Remainder when 𝑥2 + 1 is divided by 𝑥 − 2?
𝑓 2 = 5
Remainder when 𝑥3 − 𝑥 is divided by 𝑥 + 1?
𝑓 −1 = 0
Remainder when 𝑥2 + 1 is divided by 2𝑥 − 1?
𝑓
1
2
=
5
4
Remainder when 𝑥2 − 𝑥 is divided by 3𝑥 + 4?
𝑓 −
4
3
=
28
9
?
?
?
?
Bro Hint:
What value of
𝑥 makes the
thing we’re
dividing by 0?

6. Test Your Understanding So Far…
Find the remainder when 𝑥2 + 𝑥 is divided by 𝑥 − 2.
𝒇 𝟐 = 𝟐𝟐 + 𝟐 = 𝟔
Find the remainder when 𝑥2 − 5𝑥 + 3 is divided by 𝑥 + 1.
𝒇 −𝟏 = −𝟏 𝟐
− 𝟓 −𝟏 + 𝟑 = 𝟗
Find the remainder when 2𝑥3 − 𝑥2 is divided by 2𝑥 + 1.
𝒇 −
𝟏
𝟐
= 𝟐 −
𝟏
𝟐
𝟑
− −
𝟏
𝟐
𝟐
= −
𝟏
𝟐
1
2
3
?
?
?

7. Show that 𝑥 − 2 is a factor of 𝑥3
+ 𝑥2
− 4𝑥 − 4
Further Examples
𝒇 𝟐 = 𝟖 + 𝟒 − 𝟖 − 𝟒 = 𝟎
?
Set 2 Paper 1 Q14
𝒇 𝟓 = 𝟓𝟑 − 𝟔 𝟓𝟐 + 𝟓𝒂 − 𝟐𝟎 = 𝟎
𝟓𝒂 − 𝟒𝟓 = 𝟎
𝒂 = 𝟗
?

8. Test Your Understanding
Jan 2013 Paper 2 Q22
𝒇 𝟐 = 𝟖 + 𝟒𝒂 + 𝟐𝒃 + 𝟐𝟒 = 𝟎 → 𝟐𝒂 + 𝒃 = −𝟏𝟔
𝒇 −𝟑 = −𝟐𝟕 + 𝟗𝒂 − 𝟑𝒃 + 𝟐𝟒 = 𝟎 → 𝟑𝒂 − 𝒃 = 𝟏
∴ 𝒂 = −𝟑, 𝒃 = −𝟏𝟎
Jan 2012 Paper 2 Q16a
𝒇 𝟏 = 𝟏𝟑 − 𝟐𝟏 + 𝟐𝟎 = 𝟎
𝒇 𝟒 = 𝟒𝟑 − 𝟐𝟏 𝟒 + 𝟐𝟎 = 𝟎
?
?

9. Exercise 1
Find the remainder when the polynomials are
divided by the linear function, and write
“factor” if a factor.
𝑥3 − 8𝑥 + 7 𝑥 − 1 𝟎 (𝒇𝒂𝒄𝒕𝒐𝒓)
𝑥3 − 7𝑥2 − 5𝑥 + 1 𝑥 + 1 − 𝟐
𝑥3
+ 𝑥2
− 4𝑥 − 5 𝑥 + 2 − 𝟏
𝑥3
− 6𝑥2
+ 10𝑥 − 4 𝑥 − 2 − 𝟏𝟎
𝑥3 + 27 𝑥 + 3 𝟎 (𝒇𝒂𝒄𝒕𝒐𝒓)
𝑥3
− 𝑎𝑥2
+ 𝑎2
𝑥 − 𝑎3
𝑥 − 𝑎 𝟎 (𝒇𝒂𝒄𝒕𝒐𝒓)
Show that 𝑥 − 2 is a factor of 𝑥3 − 4𝑥
𝒇 𝟐 = 𝟐𝟑
− 𝟒 𝟐 = 𝟎
𝑓 𝑥 = 𝑥3
+ 2𝑥2
+ 𝑎𝑥 − 76
𝑥 − 4 is a factor of 𝑓(𝑥).
Work out the value of 𝑎.
𝟐𝟎 + 𝟒𝒂 = 𝟎 → 𝒂 = −𝟓
𝑓 𝑥 = 𝑥3 + 𝑎𝑥2 − 2𝑥 + 3
(𝑥 − 1) is a factor of 𝑓 𝑥 . Determine the
value of 𝑎.
𝒇 𝟏 = 𝟎 → 𝒂 = −𝟐
𝑥 − 1 and 𝑥 + 2 are factors of
𝑎𝑥3 − 2𝑥2 − 5𝑥 + 𝑏. Determine the
values of 𝑎 and 𝑏.
𝒂 = 𝟏, 𝒃 = 𝟔
(𝑥 + 3) and 𝑥 + 2 are factors of 𝑥3
+
𝑎𝑥2 + 𝑥 + 𝑏. Determine 𝑎 and 𝑏.
𝒂 = 𝟒, 𝒃 = −𝟔
[June 2013 Paper 2 Q21] (𝑥 − 𝑎) is a
factor of 2𝑥3 − 7𝑎𝑥 + 3𝑎. Work out
the largest possible value of 𝑎.
𝒇 𝒂 = 𝟐𝒂𝟑
− 𝟕𝒂𝟐
+ 𝟑𝒂 = 𝟎
𝒂 𝟐𝒂𝟐
− 𝟕𝒂 + 𝟑 = 𝟎
𝒂 𝟐𝒂 − 𝟏 𝒂 − 𝟑 = 𝟎
Largest value is 3.
𝑥2 − 4 is a factor of
𝑥3
+ 𝑎𝑥2
+ 𝑏𝑥 − 20.
Find the values of 𝑎 and 𝑏.
Note that 𝒙𝟐
− 𝟒 = 𝒙 + 𝟐 𝒙 − 𝟐 so
𝒇 −𝟐 = 𝟎 and 𝒇 𝟐 = 𝟎.
𝒂 = 𝟓, 𝒃 = −𝟒
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1
2
3
4
5
6
a
b
c
d
e
f
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N
7

10. Fully Factorising
Fully factorise 𝑥3
+ 6𝑥2
+ 5𝑥 − 12
What’s a dumb but moderately effective way of finding the factors?
If 𝒇 𝒙 = 𝒙𝟑
+ 𝟔𝒙𝟐
+ 𝟓𝒙 − 𝟏𝟐, then say try
𝒇 𝟏 , 𝒇 𝟐 , 𝒇 𝟑 , 𝒇 −𝟏 , 𝒇 −𝟐 , 𝒇(−𝟑) and see where the
remainder is 0.
𝒇 𝟏 = 𝟎 so 𝒙 − 𝟏 is a factor.
𝒇 −𝟑 = 𝟎 so 𝒙 + 𝟑 is a factor.
𝒇 −𝟒 = 𝟎 so 𝒙 + 𝟒 is a factor.
𝒙𝟑 + 𝟔𝒙𝟐 + 𝟓𝒙 − 𝟏𝟐 = 𝒙 − 𝟏 𝒙 + 𝟑 𝒙 + 𝟒
But we had to try a lot of values. Is there a better way?
?
?

11. 4 2 3 . 0 0 0 0
11
3
3 3
9 3
8 .
8 8
5 0
1. We found how many whole
number of times (i.e. the
quotient) the divisor went into
the dividend.
2. We multiplied the quotient
by the dividend.
3. …in order to find the
remainder.
4. Find we ‘brought down’ the
next number.
Normal Long Division

12. 𝑥 − 1 𝑥3
+ 6𝑥2
+ 5𝑥 − 12
𝑥2
𝑥3 − 𝑥2
7𝑥2 + 5𝑥
+7𝑥
7𝑥2
− 7𝑥
12𝑥 − 12
+12
12𝑥 − 12
0
The Anti-Idiot Test:
You should get a remainder of
0 at the end if you know it’s
supposed to divide exactly.
Divide just the
first terms, i.e. 𝑥3
by 𝑥2
.
Multiply whole
times it went in by
the 𝑥 we divided
by so we can find
remainder.
Bring down
extra term.
And repeat!

13. Finishing off the question
Fully factorise 𝑥3
+ 6𝑥2
+ 5𝑥 − 12
𝑥 − 1 𝑥3
+ 6𝑥2
+ 5𝑥 − 12
𝑥2 +7𝑥 +12
𝑥3
+ 6𝑥2
+ 5𝑥 − 12 = 𝑥 − 1 𝑥2
+ 7𝑥 + 12
= 𝑥 − 1 𝑥 + 3 𝑥 + 4
?
?

14. Quicker Way
Fully factorise 𝑥3
+ 6𝑥2
+ 5𝑥 − 12
We established 𝑥 − 1 was a factor.
We can immediately tell two of the terms of the other bracket (think
about the expansion)
(𝑥 − 1)(𝑥2 + ? 𝑥 + 12)
? ?
We then know that the two brackets this larger bracket factorises to
must end with two numbers that multiply to give 12.
(𝑥 + 3) and (𝑥 + 4) sounds like a sensible guess, so we then could try
𝑓(−3) and 𝑓(−4) to see if we were right.

15. Another Example
Fully factorise 𝑥3
− 2𝑥2
− 5𝑥 + 6
Try to find an initial factor by using the factor theorem.
𝒇 𝟏 = 𝟏 − 𝟐 − 𝟓 + 𝟔 = 𝟎
Therefore 𝒙 − 𝟏 is a factor.
Then divide by this factor you found.
𝑥 − 1 𝑥3
− 2𝑥2
− 5𝑥 + 6
𝑥2 −𝑥 −6
𝑥3
− 𝑥2
−𝑥2
− 5𝑥
−𝑥2
+ 𝑥
−6𝑥 + 6
−6𝑥 + 6
Then:
𝑥3 − 2𝑥2 − 5𝑥 + 6
= 𝑥 − 1 𝑥2 − 𝑥 − 6
= 𝑥 − 1 𝑥 + 2 𝑥 − 3
? ? ?
?
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16. Test Your Understanding
Fully factorise 𝑥3
− 3𝑥2
− 4𝑥 + 12
Recap:
1. Try 𝑓 … for a few values to establish an
initial factor.
2. Do long division by your factor to find the
remaining expression.
3. Factorise (by normal quadratic factorisation)
this expression you get.
= 𝑥 + 2 𝑥 − 2 𝑥 − 3
?
If you finish quickly:
Solve 𝑥3 + 3𝑥2 − 6𝑥 − 8 = 0
𝑥 + 1 𝑥 + 4 𝑥 − 2 = 0 → 𝑥 = −1 𝑜𝑟 − 4 𝑜𝑟 2
?

17. Exercises
Factorise the following:
𝑥3 − 6𝑥2 + 11𝑥 − 6 = 𝒙 − 𝟏 𝒙 − 𝟐 𝒙 − 𝟑
𝑥3 + 2𝑥2 − 𝑥 − 2 = 𝒙 + 𝟏 𝒙 + 𝟐 𝒙 − 𝟏
𝑥3 + 7𝑥2 + 14𝑥 + 8 = 𝒙 + 𝟏 𝒙 + 𝟐 𝒙 + 𝟒
𝑥3
− 3𝑥2
− 10𝑥 + 24 = 𝒙 + 𝟑 𝒙 − 𝟐 𝒙 − 𝟒
𝑥3 + 2𝑥2 − 13𝑥 + 10 = 𝒙 + 𝟓 𝒙 − 𝟏 𝒙 − 𝟐
Solve the following:
𝑥3 − 𝑥2 − 4𝑥 + 4 = 0 → 𝒙 = −𝟐, 𝟐, 𝟏
𝑥3 − 𝑥2 − 14𝑥 + 24 = 0 → 𝒙 = −𝟒, 𝟑, 𝟐
𝑥3 + 5𝑥2 + 6𝑥 = 0 → 𝒙 = 𝟎, −𝟐, −𝟑
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1
a
b
c
d
e
a
b
c
2