Topic 1: Multiplying (expanding) and dividing
Multiplying
Expanding is when the distributive law is used to remove brackets, taking a product and writing this
as a sum. Factorising is the opposite of expanding, where a sum is written as a product.
Product
Sum
Sum
Product
3a + 3b = 3(a + b)
Factorising
3(a + b) = 3a + 3b
Expanding
To expand brackets, everything in the brackets must be multiplied by the term outside the
brackets.
Multiplying by a constant
When an algebraic expression is multiplied by a constant, every term in the polynomial must be
multiplied by the constant. If the constant is a negative, take care getting the sign of the answer. If
more than one multiplication is being performed, finish the question by adding or subtracting like
terms.
For example:
2( f + 2 g − i )
=2 × f + 2 × 2 g − 2 × i
= 2 f + 4 g − 2i
−3( x 2 − 4 x − 9)
= −3 × x 2 − 3 × −4 x − 3 × −9
=
−3 x 2 + 12 x + 27
3(25 − 4 x 2 ) − 5(2 x 2 + 3 x − 10)
=
75 − 12 x 2 − 10 x 2 − 15 x + 50
=
−12 x 2 − 10 x 2 − 15 x + 50 + 75
(Take Care! 75 is added)
=
−22 x 2 − 15 x + 125
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2
( 6a + 9b − 4 )
3
2
2
2
= × 6a + × 9b − × 4
3
3
3
8
= 4a + 6b −
3
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Multiplying by a single term containing a variable(s)
When a polynomial is multiplied by a term containing a variable, multiplication of variables is
required. When multiplying variables, skills in exponents may be required.
Examples:
a (b + 2c + 8)
=a × b + a × 2c + a × 8
=ab + 2ac + 8a
3 x( x 2 + x − 3)
=3 x × x 2 + 3 x × x + 3 x × − 3
= 3x3 + 3x 2 − 9 x
3ab(4a - b)
=×
3ab 4a - 3ab × b
−3 x 2 ( x 2 − x + 6)
=
−3 x 2 × x 2 − 3 x 2 × − x + − 3 x 2 × 6
= 12a 2b - 3ab 2
=
−3 x 4 + 3 x3 − 18 x 2
5m3 (am + bn - 4)
= 5m3 × am + 5m3 × bn - 5m3 × 4
=5am 4 + 5bm3 n - 20m3

3 4
x2
3
3
x
x
+
- 4x 

3
x

1
3
2
3
3
3 x2
3
- ×4 x
=
× 3x 4 - × x 3 + ×
x
x
x 3
x
= 9 x 3 - 3 x 2 + x - 12
•
•
Take care with signs
Terms involving numbers (coefficients) and pronumerals are usually written with the
number (coefficient) in front, followed by the variables in alphabetical order.
Multiplying one polynomial by another polynomial
When a polynomial is multiplied by another polynomial, it is important to use a method that
ensures that every term from one polynomial is multiplied by every term in the other polynomial.
After practice, you may be able to do these mentally.
To help in this process, one of two methods is used. Remember that when no operation is shown,
multiplication is implied.
The first method involves multiplying the second polynomial by each individual term in the first
polynomial. This is achieved by multiplying the second polynomial by each term of the first
polynomial. These are multiplied out and like terms (if any) are collected. In the question
a + 3 2b + 1 the second term is multiplied by a and then by 3.This method works well in any
(
)(
situation.
)
( a + 3)( 2b + 1)
= a ( 2b + 1) + 3 ( 2b + 1)
= 2ab + a + 6b + 3
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Note: There are no like terms so this cannot be simplified any further.
The second method involves multiplying every term in the first polynomial with every term in the
second polynomial. If two binomials are used, the acronym FOIL is useful;
F – multiply the First terms
O – multiply the Outer terms
I – multiply the Inner terms
L – multiply the Last terms
Multiply the First terms: a x 2b = 2ab
Multiply the Last terms: 3 x 1 = 3
(a + 3)(2b + 1)
Multiply the Inside terms: 3 x 2b = 6b
Multiply the Outside terms: a x 1 =a
2 b
(a + 3)(2b + 1)= 2ab + a + 6b + 3
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Example: Expand
Method 1:
Method 2:
F
L
(a + 4)(b + 1)
(a + 4)(b + 1)
Split the 1st factor into a and 4
= a (b + 1) + 4(b + 1)
= ab + a + 4b + 4
I
O
F = a × b = ab
O = a ×1 = a
I = 4 × b = 4b
L = 4 ×1 = 4
= ab + a + 4b + 4
F
( x + 3)( x − 5)
= x( x − 5) + 3( x − 5)
L
( x + 3)( x − 5)
= x 2 − 5 x + 3 x − 15
I
= x 2 − 2 x − 15
O
F = x × x = x2
−5 x
O=
x × −5 =
I = 3 × x = 3x
−15
L=
3× − 5 =
= x 2 − 5 x + 3 x − 15
= x 2 − 2 x − 15
(2 x − 1)(3 − 4 x)
= 2 x(3 − 4 x) − 1(3 − 4 x)
F
L
(2 x − 1)(3 − 4 x)
= 6 x − 8x2 − 3 + 4 x
I
=
−8 x 2 + 10 x − 3
O
F = 2x × 3 = 6x
−8 x 2
O=
2x × − 4x =
I=
−
1× 3 =
−
3
−
+4 x
L=
1× − 4 x =
= 6 x − 8x2 − 3 + 4 x
=
−8 x 2 + 10 x − 3
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(2 x − 3)( x 2 − 2 x + 4)
= 2 x( x 2 − 2 x + 4) − 3( x 2 − 2 x + 4)
= 2 x3 − 4 x 2 + 8 x − 3 x 2 + 6 x − 12
= 2 x3 − 7 x 2 + 14 x − 12
The FOIL method only works (best) for two
binomial factors.
( x − 2 y + 4 z )(4 x + y − 2 z )
= x(4 x + y − 2 z ) − 2 y (4 x + y − 2 z ) + 4 z (4 x + y − 2 z )
= 4 x 2 + xy − 2 xz − 8 xy − 2 y 2 + 4 yz + 16 xz + 4 yz − 8 z 2
= 4 x 2 − 7 xy + 14 xz − 2 y 2 + 8 yz − 8 z 2
3(3 x − 5)( x 2 − x − 1)
expand brackets first
= 3[3 x( x 2 − x − 1) − 5( x 2 − x − 1)]
= 3[3 x3 − 3 x 2 − 3 x − 5 x 2 + 5 x + 5]
= 3(3 x3 − 8 x 2 + 2 x + 5)
= 9 x3 − 24 x 2 + 6 x + 15
The video below covers some examples similar to those above. Click on the image below to start
downloading the video. Depending on your internet connection, this may take a minute or two.
Video ‘Multiplying (expanding)’
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Identities
Special binomial multiplications – Perfect Square
In this section, the special case of a binomial squared, written as (a + b) 2 , is considered. Because
this can be used for any binomial factor, it is given the special name of an identity.
Developing the identity:
( a + b) 2
(a + b)(a + b)
=
= a ( a + b) + b( a + b)
= a 2 + ab + ba + b 2
=a 2 + 2ab + b 2
(a + b) 2 =a 2 + 2ab + b 2
When expanding a perfect square, there are two options:
(1) use the identity or (2) just expand using method 1.
For example:
Using the identity
Expanding using method 1
=
a x=
using the identity
,b 3
( x + 3) 2
=
( x + 3)( x + 3)
= x( x + 3) + 3( x + 3)
( x + 3) 2
= x22 + 2 × x × 3+ 322
+2 ab
a
+b
= x 2 + 3x + 3x + 9
= x + 6x + 9
2
= x2 + 6x + 9
(3 x + 5) 2
using the identity
=
a 3=
x, b 5
= (3 x) 2 + 2 × 3 x × 5+ 522
a2
+2 ab
= 9 x + 30 x + 25
2
+b
(3 x + 5) 2
= 3 x(3 x + 5) + 5(3 x + 5)
= 9 x 2 + 15 x + 15 x + 25
= 9 x 2 + 30 x + 25
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(2 x + 5 y ) 2
using the identity
a 2=
x, b 5 y
=
= (2 x) 2 + 2 × 2 x × 5 y + (5 y ) 2
+2 ab
a2
(2 x + 5 y ) 2
= 2 x(2 x + 5 y ) + 5 y (2 x + 5 y )
=4 x 2 + 10 xy + 10 xy + 25 y 2
+ b2
= 4 x 2 + 20 x + 25 y 2
using brackets will ensure both
coefficient and pronumeral are squared
= 4 x 2 + 20 x + 25 y 2
There is similar identity based on subtraction;
( a − b) 2
(a − b)(a − b)
=
= a ( a − b) − b( a − b)
= a 2 − ab − ba + b 2
=a 2 − 2ab + b 2
(a − b) 2 =a 2 − 2ab + b 2
( x − 7) 2
using the identity
=
a x=
,b 7
( x − 7) 2
( x − 7)( x − 7)
=
= x( x − 7) − 7( x − 7)
= x22 − 2 × x × 7 + 7 22
−2 ab
a
+b
= x 2 − 7 x − 7 x + 49
=x − 14 x + 49
2
=x 2 − 14 x + 49
(4 x − y ) 2
=
using the identity
a 4=
x, b y
= (4 x) 2 − 2 × 4 x × y + y 2
+2 ab
a2
= 16 x − 8 xy + y
2
+ b2
2
(4 x − y ) 2
= 4 x(4 x − y ) − y (4 x − y )
= 16 x 2 − 4 xy − 4 xy + y 2
= 16 x 2 − 8 xy + y 2
Although using identities can be useful, the most important skill is knowing how to expand logically
as this will give the skills to cope with any given situation.
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Take care:
A commonly made mistake is when expanding a perfect square is:
( x + 2) 2 ≠ x 2 + 4
This is incorrect because the middle term is omitted.
An example that combines the skills covered so far are:
Expand 3 x (2 x − 3)( x + 2)
With complex expansions like this, start with the perfect square, then multiply by the
binomial factor and then the factor 3x.
2
( x + 2) 2
=
( x + 2)( x + 2)
= x( x + 2) + 2( x + 2)
= x2 + 4x + 4
3 x(2 x − 3)( x 2 + 4 x + 4)
(2 x − 3)( x 2 + 4 x + 4)
Now the expression is:
= 2 x( x 2 + 4 x + 4) − 3( x 2 + 4 x + 4)
= 2 x3 + 8 x 2 + 8 x − 3 x 2 − 12 x − 12
= 2 x3 + 5 x 2 − 4 x − 12
3
2
Now the expression is: 3 x (2 x + 5 x − 4 x − 12)
3 x(2 x 3 + 5 x 2 − 4 x − 12)
= 3 x × 2 x 3 + 3 x × 5 x 2 − 3 x × 4 x − 3 x × 12
=6 x 4 + 15 x 3 − 12 x 2 − 36 x
There is one more identity worth knowing about. This identity is about the product of two binomial
factors in the form of
(a + b)(a − b)
sum
difference
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Expanding these two binomial factors gives:
(a + b)(a − b)
= a ( a − b) + b( a − b)
= a 2 − ab + ab − b 2
= a 2 − b2
For example:
( x − 3)( x + 3)
= x ( x + 3) − 3 ( x + 3)
( x − 3)( x + 3)
( a −b )
( a +b )
= x 2 2− 32 2
= x 2 + 3x − 3x − 9
a −b
= x2 − 9
= x2 − 9
( 5 + 2 x )( 5 − 2 x )
= 5(5 − 2x ) + 2x (5 − 2x )
(5 + 2 x)(5 − 2 x)
( a +b )
( a −b )
= 52 − (2 x) 2
=25 − 10 x + 10 x − 4 x 2
a 2 −b 2
= 25 − 4 x 2
= 25 − 4 x 2
More examples that combine the skills cover so far are:
Expand
x 2 ( 5 − 3 x )( 5 + 3 x )
( 5 − 3x )( 5 + 3x )
= 25 − 9 x 2
Now the expression is:
x 2 (25 − 9 x 2 )
= 25 x 2 − 9 x 4
Expand
(3 x − 1)3
this is the same as
(3 x − 1)(3 x − 1) 2
(3 x − 1) 2
= 9 x2 − 6 x + 1
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Now the expression is:
(3 x − 1)(9 x 2 − 6 x + 1)
= 3 x(9 x 2 − 6 x + 1) − 1(9 x 2 − 6 x + 1)
= 27 x3 − 18 x 2 + 3 x − 9 x 2 + 6 x − 1
= 27 x3 − 27 x 2 + 9 x − 1
Expand
2
1 
1
2
 2 x −   2 x +  ( 3x − 4 )
3
2 
2
Start with the more complex. There is a perfect square
(a + b)(a − b)
sum
( a − b) 2
and a
structure. These are equally complex so order for these two parts is
difference
unimportant.
( 3x − 4 )
2
= 9 x 2 − 24 x + 16
Now the expression is
2
1 
1
2
 2 x −   2 x +  ( 9 x − 24 x + 16 )
3
2 
2
1 
1

 2x −   2x + 
2 
2

1
= 4x2 −
4
Now the expression is
=
=
2 2 1
2
 4 x −  ( 9 x − 24 x + 16 )
3
4
2 2
1

4 x ( 9 x 2 − 24 x + 16 ) − ( 9 x 2 − 24 x + 16 ) 

3
4

2
9
24
16 
36 x 4 − 96 x 3 + 64 x 2 − x 2 +
x− 

3
4
4
4
2
247 2

4
3
x + 6x − 4 
 36 x − 96 x +
3
4

247 2
8
= 24 x 4 − 64 x3 +
x + 4x −
6
3
=
The video below covers some examples similar to those above. Click on the image below to start
downloading the video. Depending on your internet connection, this may take a minute or two.
Video ‘Multiplying (Using identities)’
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Polynomial division
In this section, a polynomial (dividend) will be divided by another polynomial (divisor) using a
method similar to long division. The resulting answer is not necessarily a polynomial. It may be
worth reviewing long division with numbers to help follow the example below.
Example: Dividing by a single term (monomial)
(5 x3 − 2 x 2 + 4 x) ÷ x
This is best written as
5 x3 − 2 x 2 + 4 x
x
3
5x 2 x2 4 x
=
−
+
x
x
x
2
= 5x − 2 x + 4
Example: Dividing by a single term (monomial)
( 2 x − 3) − 2 ( 3 − 2 x )
2x
2x − 3 − 6 + 4x
=
2x
6x − 9
=
2x
6x 9
=
−
2x 2x
9
= 3−
2x
Example: Dividing by a single term (monomial)
3 xy 2 + 5 x 2 y − 15 x 2 y 2
3x 2 y 2
3 xy 2
5 x 2 y 15 x 2 y 2
+
−
3x 2 y 2 3x 2 y 2 3x 2 y 2
1 5
= +
−5
x 3y
=
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When dividing by a binomial factor, the process is similar to performing long division with numbers.
Example:
8
82
821
6 4926 → 6 4926 → 6 4926 → 6 4926
48 ↓
48 ↓
48 ↓
1
12
12
12 ↓
12 ↓
06
0
6
0
Example:
(4 x 2 + 4 x + 1) ÷ (2 x − 3)
2 x − 3 4 x 2 + 4 x + 1 • think: what is 4x 2 ÷ 2 x ( =
2x)
This gives;
2x
2 x - 3 4 x 2 + 4 x + 1 • Now subtract the product of 2x & 2 x - 3 (= 4 x 2 - 6 x)
-
4x 2 - 6 x
( 6 x) =10 x
• Subtract 4x --
10x
Now, consider the next term in the dividend;
2x + 5
2 x - 3 4 x 2 + 4 x + 1 • bring down the next term
-
4x 2 - 6 x ↓
10 x + 1
- 10 x - 15
5)
• think: what is 10x ÷ 2 x ? (=
• multiply 5 and 2x - 3 ( = 10 x - 15)
• subtract 1 -( 15) =16
16
The answer is expressed as
(4 x 2 + 4 x + 1) ÷ (2 x − 3) = 2 x + 5 +
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16
2x − 3
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( x 3 + 2 x 2 − 5 x − 6) ÷ ( x + 1)
Example:
x + 1 x 3 + 2 x 2 − 5 x − 6 • think: what is x 3 ÷ x ( =
x2 )
This gives;
x2
x + 1 x 3 + 2 x 2 -5 x 6 • Now subtract the product of x 2 & x + 1 ( = x 3 + x 2 )
- x3 + x 2
• Subtract 2 x 2 - x 2 =
x2
x2
Now, consider the next term in the dividend;
x2 + x
x + 1 x 3 + 2 x 2 -5 x 6 • bring down the next term
- x3 + x 2
-
↓
• think: what is x 2 ÷ x ? (=
x)
x2 - 5x
• multiply x and x + 1 ( = x 2 + x)
x2 + x
- 6x
• subtract - 5 x - x =-6 x
Now, consider the next term in the dividend;
x2 + x - 6
x + 1 x 3 + 2 x 2 -5 x 6 • bring down the next term
- x3 + x 2
↓
↓ • think: what is - 6 x ÷ x ? (= -6)
x2 - 5x
↓ • multiply -6 and x + 1 ( - 6 x - 6)
- x2 + x
-
↓ • subtract
- 6x - 6
-6x 6
0
There is no remainder, so
( x 3 + 2 x 2 − 5 x − 6) ÷ ( x + 1) = x 2 + x − 6
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With numbers; we can say that 8 is a factor of 96 because 96 ÷ 8 = 12 with no remainder. In the
example above; ( x 3 + 2 x 2 − 5 x − 6) ÷ ( x + 1) gave the result x 2 + x − 6 with no remainder, therefore
it is possible to say that ( x + 1) is a factor of ( x 3 + 2 x 2 − 5 x − 6) .
The video below covers some examples similar to those above. Click on the image below to start
downloading the video. Depending on your internet connection, this may take a minute or two.
Video ‘Dividing (Polynomial division)’
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Activity
1. Expand the following, collect like terms if possible.
(a)
3 ( 9 − 2x )
(b)
(c)
4x − 3( x − 4)
(d)
1
4x2 − x − 9)
(
2
2 ( x2 + x + 4) + 3( x + 6) − ( x2 − 2x + 9)
(f)
4 ( x + 4 y ) − 5 ( x − 2 y + 1) + 2 ( x − 3)
(e)
75 − 4 ( 2 x 2 + x − 6 )
2. Expand the following, collect like terms if possible.
(a)
(c)
(e)
a ( 3a − 4b + 1)

5
p2  2 p + 4 + 
p

1
−4 x ( 9 − x ) − ( 4 − x )
2
(b)
4ac ( b + 2c − 7 )
(d)
x ( x 2 + 4 x − 1) − 3 ( x 2 + 1)
(f)
4
( 6 x3 + 5 x 2 − x + 4 )
3x
3. Expand the following, collect like terms if possible.
(a)
(b)
( 2a − 1)( 4b − 3)
(c)
(e)
(g)
(i)
( x + 3)( x − 2 )
2
( x − 4)
( x − 2 )( x + 2 )
3
( x − 2)
(d)
(f)
(h)
(j)
( 3x − 2 )( y + 4 )
( x − 4 )( 2 x + 1)
( 2 x − 3)( 7 − x )
( 2 x + 1)( x − 3)( x + 2 )
4 ( 2 x − 1) ( x 2 + 7 x + 5 )
4. Perform the following divisions.
(a)
(b)
(c)
( x − 4 x + 11) ÷ ( x + 4 )
( x + 4 x + x − 6 ) ÷ ( x − 1)
( 2 x − 5 x − 14 x + 8) ÷ ( 2 x − 1)
2
3
2
3
2
The answers to the activity questions are found in the pdf named:
‘Answers to Activity Questions (PCA)’
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