Topic 4: Expanding brackets
There are two major ideas in Algebra called Expanding and Factorising. 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
3(a + b) = 3a + 3b
Expanding
Sum
Product
3a + 3b = 3(a + b)
Factorising
To expand brackets, everything in the brackets must be multiplied by the term outside the
brackets. To help in this process, one of two methods is used. Remember that when no operation
is shown, multiplication is implied.
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Example:
Method 1:
Method 2:
3(a + b)
=3 × (a + b)
= 3× a + 3× b
= 3a + 3b
3(a + b)
= 3× a + 3× b
= 3a + 3b
7(a + 4)
=7 × (a + 4)
= 7×a + 7×4
= 7 a + 28
7(a + 4)
= 7×a + 7×4
= 7 a + 28
a (b − 7)
=a × (b − 7)
= a×b − a×7
= ab − 7 a
a (b − 7)
= a×b − a×7
= ab − 7 a
3(4 − 3n)
=3 × (4 − 3n)
= 3 × 4 − 3 × 3n
= 12 − 9n
3(4 − 3n)
= 3 × 4 − 3 × 3n
= 12 − 9n
−
7 a (a − 5b)
= − 7 × a × (a − 5b)
= − 7 × a × a − − 7 × a × 5b
= − 7 a 2 + 35ab
3 p ( − 5 p + 4q )
= − 3 × p × ( − 5 p + 4q )
= − 3 × p × − 5 p + − 3 × p × 4q
= 15 p 2 − 12 pq
−
7 a (a − 5b)
−
7 × a × a − − 7 × a × 5b
−
7 a 2 + 35ab
=
=
−
−
3 p ( − 5 p + 4q )
3 × p × − 5 p + − 3 × p × 4q
= 15 p 2 − 12 pq
=
−
Expanding brackets and collecting like terms
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Expanding
3(2a + 5b) + 4(a − 1)
= 6a + 15b + 4a − 4
Collecting like terms
= 10a + 15b − 4
Expanding
3a (b − 3) − 5(2ab + 7)
= 3ab − 9a − 10ab − 35
Collecting like terms
= − 7 ab − 9a − 35
Expanding
2a (b − 3c) + 5ab + 6ac − 12
= 2ab − 6ac + 5ab + 6ac − 12
Collecting like terms
= 7 ab − 12
Video ‘Expanding brackets’
Activity
1.
Expand the following expressions:
(a)
4(a − 8)
(b)
3(10 − r )
(c)
a (h + 3)
(d)
b(3b − 5)
(e)
2 s (t − 1)
(f)
n( n − m)
(g)
6(3 − 2a )
(h)
3m(n − 4)
(i)
2d (3d − e)
(j)
(k)
4a (a + b − 3c)
(l)
3( − s + 2t )
(n)
5ij ( j − 5)
(m)
−
a ( m − n)
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−
4(4 − 3m)
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2.
Simplify the following by expanding the brackets and collecting like terms
(a)
3(2a + b) + 7a
(b)
5(a − 3b) − a + 2
(c)
6 x − 2( x − 3 y )
(d)
5 xy + 6 x( y − 3) + 2 x
(e)
4(2 x − y ) + 3( x − 2 y )
(f)
4d (e + 4) + 5de − 10d + 9
2 x( f + 2 g ) + 3(5 g − f )
(h)
2(2 x − 3 y + z ) − 3 x(2 − y )
(j)
(g)
(i)
−
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−
2h(2i − j ) − 3i (2h + j )
7 x( x − 4) − x(7 − 3x)
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