Algebra 2
Mathematics Worksheet
This is one of a series of worksheets designed to help you increase your confidence
in handling Mathematics. This worksheet contains both theory and exercises which
cover:1.
2.
3.
4.
Expanding brackets
Simple factorisation
Quadratic factorisation
Completing the square
There are often different ways of doing things in Mathematics and the methods
suggested in the worksheets may not be the ones you were taught. If you are
successful and happy with the methods you use it may not be necessary for you to
change them. If you have problems or need help in any part of the work then there
are a number of ways you can get help.
For students at the University of Hull
 Ask your lecturers.
 You can contact a Mathematics Tutor from the Skills Team on the email shown
below.
 Access more Maths Skills Guides and resources at the website below.
 Look at one of the many textbooks in the library.
Web: http://libguides.hull.ac.uk/skills
Email: [email protected]
1. Expanding Brackets
Brackets are used for convenience in grouping terms together. When removing
brackets each term within the bracket must be multiplied by the quantity outside the
bracket. For example 2 x  y   2 x  2 y .
We can see this by using the idea from Algebra 1 where x and y were the lengths of
two pieces of wood (in whatever units you like!).
x
y
x-y
x
x
x-y
y
y
x-y
2(x - y)
2x-y
We can see from the diagram that 2( x  y ) is not the same as 2x  y .
In the same way 7( x  2 y )  ( 7)  x  ( 7)  ( 2 y )  7 x  14 y
When simplifying expressions containing brackets, first multiply out the brackets and
then collect like terms.
Worked Examples
(a) Simplify 3  y  2   5  3 y  2 x   2  6 y  3 
3  y  2  5 3 y  2 x   2  6 y  3
 3 y  6  15 y  10 x  12 y  6  10 x
With practice you may be able to do the process in one step by collecting terms as
you multiply. For instance, in the above, collecting y -terms 3 y  15 y  12 y and then
the numbers 6  6 and finally the x -term 10x .
(b) Simplify x( y  2 z )  x(3 y  4 z )
x( y  2 z)  x(3 y  4 z)
 xy  2 xz  3 xy  x  4 z 
 xy  2 xz  3 xy  4 xz  2 xy  6 xz
this is better written as 6 xz  2 xy otherwise the – sign can get lost!
Multiplying brackets together can be done in a number of ways, here are four:1.
( x  y )(a  b)
 x ( a  b)  y ( a  b)
multiply first bracket out
then remaining brackets
 xa  xb  ya  yb
(remember to be careful if you have a mixture of signs)
page 1
2. The ‘eye-brows method
(x
+
y)
(a
+
b)
Collecting terms gives
( x  y )( a  b )
 xa  ya  xb  yb
3. We can use a diagram and think in terms of area
x
y
From the diagram we see that the total
a
xa
ya
area is given by ( x  y )( a  b )
b
yb
xb
This is equivalent to adding all the bits
together giving
( x  y )( a  b)  xa  ya  xb  yb
4. Multiply, using a table and ignoring lengths
a
xa
y
ya
b
xb
yb
x
( x  y )( a  b)  xa  ya  xb  yb
this method is especially useful when some of the terms are negative or the brackets
contain more than 2 terms. (see example d below)
From the above, putting a  x and b  y we have
y
x
yx
x
x2
( x  y )( x  y )  x 2  yx  xy  y 2
y
xy
 x 2  2 xy  y 2
y2
Similarly
x  y 2  x 2  2 xy  y 2
x  y 2  x 2  2 xy  y 2
x  y x  y   x2 - y 2
These very important expressions occur often and should be learnt!
Examples
With practice you will be able to cut out some intermediate stages.
Expand and simplify the following
(a) ( x  3)(2 x  5)
page 2
( x  3)(2 x  5)
Method 1
 x(2 x  5)  3(2 x  5)
Method 2
2x
 2 x 2  5 x  6 x  15
5
 2 x 2  x  15
x
-3
2x 2
5x
6x
-15
In both cases ( x  3)(2 x  5)  2 x 2  x  15
(b) ( x 2  2 xy)( x  2 y )
Method 1
( x 2  2 xy)( x  2 y )  x 2 ( x  2 y )  2 xy( x  2 y )
 x 3  2 x 2 y  2 x 2 y  4 xy 2
Method 2
x2
x3
2 x 2 y
2 x2 y
4 xy 2
x
2y
 x 3  4 xy 2
2 xy
In both cases ( x 2  2 xy)( x  2 y )  x 3  4 xy 2
(c) (2 x  3) 2
Comparing this with the expression for ( x  y) 2 we have
2 x  32  2 x 2  22 x 3  32  4 x 2  12 x  9
This could also have been done using either of the methods above.
(d) 2 x  4 y  33 x  2 y  5
using method 2 (much easier than method 1 in most cases)
2x
4 y
3
2
12xy
3x
9x
6x
2 y
4 xy
8y 2
6 y
-5
10 x
20 y
-15
2 x  4 y  33x  2 y  5 =
6 x 2  8 y 2  16 xy  x  14 y  15
e) ( x  2)3
using
we have
and hence
 x  y   x 2  2 xy  y 2
2
 x  2  x 2  4 x  4
3
 x  3    x  2  x 2  4 x  4   x3  6 x 2  12 x  8
2
page 3
Exercise 1
Expand and simplify where possible
2. x2a  xy 
1. 3 x 2  2 y
4. ( x  7)( x  4)
3. x  5x  3
6. 2 x  13x  2
5. x  32
8. 3x  43x  4
7. 2 x  52  3x 


9. 3x  12   x  42
10. 3( x  2 y ) 2
11. 3x  2 y 2
12.
13. 3x  2 y 2  2 y  3x 2
14.
15. 3x  7 2  7x  32
16.
x  12 x  2
18.
17.
x  13
x  y  z 2  x  y  z 2
32 x  12  23x  12
x  14
2. Simple Factorisation
Factorisation is the reverse process to expanding brackets
Expanding 3( x  y) gives 3x  3 y so factorising 3x  3 y gives 3( x  y) .
To do this we need to find common factors in the given terms.
Examples
(a) Factorise 3x  12 y
Both terms can be divided by 3 (i.e. they have a factor of 3),
so 3x  12 y = 3( x  4 y ) .
(b) Factorise 7 x 2  14 xy
Both terms can be divided by 7x (i.e. they have a factor of 7x ),
so 7 x 2  14 xy = 7 x( x  2 y ) .
(c) Factorise mn 3  3n 2 p  2m 4 p 3
There is no factor common to all three terms hence the expression cannot be
factorised.
(d) Factorise 14mn3  35n 2  21m5n 4
To factorise this expression we need to find a factor common to all 3 terms
14, 35 and 21 have a factor of 7;
mn3 , n 2 and m5n4 have a factor of n 2 ,
so 14mn3  35n 2  21m5n 4 = 7n 2 (2mn  5  3m5n 2 )
Exercise 2 - Factorise fully (where possible)
2. 3a  4ab
1. 2a  4b
4. 3ab  4ac  5ad
5. ut  5t 2
7. 36d 3  12d 2
10. 5a 4b3  10a3c  b 4c 2
8. a 4bc  a3b 2c
11. 6ab 2  9a3b3  3b5c 2
3. 8 gt  12ht
6. 2ab 2  6ab  a
9. 2r 2  2rh
12. 4 2r 3  2rh 2  8r 2
page 4
3. Quadratic Factorisation
As ( x  2)(2 x  3)  2 x 2  x  6 so factorising 2 x 2  x  6 gives ( x  2)(2 x  3)
We can find clues to help in factorisation by considering the patterns of signs:
Expanding
( x  2)(2 x  3)  2 x 2  7 x  6
Factorising
2 x 2  7 x  6  ( x  2)(2 x  3)
( x  2)(2 x  3)  2 x 2  7 x  6
2 x 2  7 x  6  ( x  2)(2 x  3)
The signs
when factorising
.. + .. + ..  (.. + ..)(.. + ..)
.. – .. + ..  (.. – ..)(.. – ..)
If the constant term is positive then the brackets will both have the same sign.
Expanding
( x  2)(2 x  3)  2 x 2  x  6
Factorising
2 x 2  x  6  ( x  2)(2 x  3)
( x  2)(2 x  3)  2 x 2  x  6
2 x 2  x  6  ( x  2)(2 x  3)
when factorising
The signs
.. – .. – .. or .. + .. – .. 
(.. – ..)(.. + ..) or (.. + ..)(.. – ..)
If the constant term is negative then the brackets will have different signs.
So, when factorising the quadratic ax 2  bx  c (with a  0 ) then
if b and c are both positive the brackets will both have plus signs in
if b is negative and c is positive the brackets will both have minus signs in
and if c is negative one bracket will have a plus and the other a minus sign.
Examples
(The method used here is one of many.)
(a) Factorise x 2  4 x  3
Ax
B
Cx
To get the x 2 term both A and C must be 1
x2
D
3 From the signs both B and D are positive so to get +3
we need B = 1 and D = 3
page 5
This leads to:
x
x
x2
3x
3
1
x
3
This gives x and 3x in the other two boxes
a total x -term of x  3 x  4 x as required.
Solution Factors of x 2  4 x  3 are (x + 1)(x + 3).
Or you could have started with x 2  4 x  3   Ax  B   Cx  D  and equated
coefficients.
The solution is always the same.
(b) Factorise 2 x 2  7 x  3
Ax
B
2
Cx
To get the 2 x 2 term A = 1 and C = 2
2x
D
3 From the signs both B and D are negative so
to get +3, B = -1 and D = -3 or vice-versa
This leads to EITHER
x
-1
2x
2 x
2 x2
3 x
-3
3
( x -term 2 x  3x  5 x incorrect)
OR
x
-3
2x
6 x
2 x2
x
-1
3
( x -term 6 x  x  7 x correct)
Solution Factors of 2 x 2  7 x  3 are ( x  3)(2 x  1) .
(again with practice this will become easier)
(c) Factorise 2 x 2  5 x  3
Ax
B
Cx 2 x 2
To get 2 x 2 we need A = 1 & C = 2
D
-3 To get -3 we need B = 1, & D = -3 or vice-versa
or B = 3 & D = -1 or vice-versa
This gives 4 possibilities
x

1
2
2x
2x
2x
-3
-3
-3x
( x -term total 2 x  3x   x )
x

3
2
2x
6x
2x
x
-1
-3
( x -term total 6 x  x  5 x )

2x
1
( x -term total
-3
6 x
2x
x
-3
6 x  x  5 x )
x
2
x

-1
2
2x
2 x
2x
3x
3
-3
( x -term total 2 x  3x  x )
From  we haveSolution factors of 2 x 2  5 x  3 are ( x  3)(2 x  1)
In practice, once the correct pair have been found it is not necessary to work out any
more except, perhaps as a check.
page 6
(d) Factorise 4 x 2  9
Ax
B
Cx
To get 4 x 2 we need A = 1 & C = 4 or vice-versa or A = 2 & C = 2
4 x2
D
-9 To get -9: B = 1 & D = -9 or B = 3 & D = -3
or vice-versa in both cases
This gives a lot of possibilities but symmetry suggests we try
2x
3
2 x 4 x2
6 x This gives x -term total 6 x  6 x  0 which is what we want
-3 6 x
-9
Solution factors of 4 x 2  9 are (2 x  3)(2 x  3)
This is an example of the difference of two squares. Earlier we had
( x  y)( x  y)  x 2  y 2 , similarly, 4 x 2  9   2 x   32   2 x  3  2 x  3
2
(e) Factorise px  5 p  qx  5q
Pairing the terms up to put the two p terms and the two q terms together gives
px  5 p  qx  5q  p( x  5)  q( x  5)
pA  qA
This last expression is similar to
which can be factorised as
( p  q) A
giving px  5 p  qx  5q  ( p  q)( x  5)
Solution factors of px  5 p  qx  5q are ( p  q)( x  5) or ( x  5)( p  q)
(f) Factorise 6 x 2  7 x  3
Ax
B
Cx
To get 6 x 2 : A = 1 & C = 6 or A = 2 & B = 3
6 x2
D
-3
To get -3: B = 1 & D = -3 or B = -1 & D = 3 or vice-versa in
both cases
This gives 8 possibilities but you cannot have a 6 x and a 3 in the same bracket, nor
a 3 x and a 3 (why? - answer later) so this reduces the possibilities to 4:
x

3
2
6x
18 x
6x
x
-1
-3
( x -term total 18 x  x  17 x )

2x
3
3x
9x
6 x2
2 x
-1
-3
( x -term total 9 x  2 x  7 x )
x

-3
2
6x
18 x
6x
x
1
-3
( x -term total 18 x  x  17 x )

2x
-3
3x
6 x 2 9 x
2x
1
-3
( x -term total 9 x  2 x  7 x )
From  we have Solution factors of 6 x 2  7 x  3 are (3x - 1)(2x + 3)
page 7
To answer the earlier question consider (3 x  3)(2 x  1) .
This expands to give 6 x 2  3 x  3  3( x 2  x  1) which has a factor of 3
and 3 x  3  3( x  1) also has a factor of 3
From this we can see that, unless the quadratic has a numerical factor, the terms in
the brackets cannot have a common numerical factor.
This is why we could ignore (6 x  3), (3 x  3) as possible factors of 6 x 2  7 x  3 .
Exercise 3
Factorise as far as possible
1. x 2  3 x  2
2. x 2  6 x  9
3.
x2  4 x  4
4.
x2  2 x  8
5.
x 2  6 x  16
6.
2 x 2  3x  1
7.
6x2  x  2
8.
9 x2  6x  1
9.
2d 2  3d  1
10.
4 x 2  16 y 2
11.
4a 2  16a
12.
9 x2  1
13.
6 x2  5x  6
14.
( x  1)2  4
15.
4 x2  4 x  3
16.
(3x  2)2  ( x  3)2
17. m 2n 2  4mn  21
4. Completing the Square
Completing the square is the technique which is the basis for obtaining the formula
for the solution of a quadratic equation. It can also be used in solving quadratic
equations and finding maximum and minimum values of quadratics. It involves
finding what needs to be added to expressions such as x 2  8 x and y 2  10 y to get
them into the form ( x  a)2 and ( y  b)2 respectively.
(a) Consider x 2  8 x ; we want to find a number a so that
( x  a)2  x 2  8 x  ... this gives ( x  a)2  x 2  2ax  a 2  x 2  8 x  ...
where  means equal for all values of x . This means that the coefficient of x on
both sides must be the same, giving 2a  8  a  4 and a 2  16 . We need to add
16 to complete the square.
x 2  8 x  16  ( x  4)2
or
x 2  8 x  ( x  4)2  16
Once this process is understood it becomes easy to complete the square as long as
the coefficient of the squared term is 1.
(b) Consider y 2  10 y :
Coefficient of y is -10; halve it giving –5. This suggests that we need to add
(5)2  25
giving ( y - 5) 2  y 2 - 10 y + 25
and y 2  10 y   y  52  25
page 8
c) Consider p 2  9 p  3 : We need to complete the square on p 2  9 p - coefficient
 2  20 14 to p
of p is 9; halve it giving 4 12 . This suggests we need to add 4 12
giving
and
2
 9p
p  4 12 2  p 2 + 9 p + 20 14
p 2  9 p 3  p  4 12 2  20 14   3  p  4 12 2  23 14
d) Consider 2 x 2 + 12 x  5 = 2( x 2 + 6 x)  5 :
In the bracket the coefficient of x is 6; halve it giving 3. This suggests that we add
32  9 (in the bracket)
giving
and

 

2 x 2 + 12x 5 = 2x 2 + 6 x 5  2( x + 3)2  18 5
2( x + 3) 2  2 x 2 + 6 x + 9  2 x 2 + 6 x  18 ,
 2( x + 3) 2  13

e) Consider (the nasty) 7 p 2  12 p  6  7 p 2 
12
7

p  6:
In the bracket the coefficient of p is 12 ; halve it giving 6 . This suggests that we
7
add 36
49
7
(in the bracket)
giving
and

7 p  76
2  7p 2  127 p  3649   7p 2  127 p  367
 7p 2  12 p   6  7  p 



 7p 
6
7
6
7


2
2


36 
49 
36
7

6
 6  7p 
6
7

2

6
7
Exercise 4
In each of the following add the term which will complete the square for the given
expressions. Then write the result down as the square of an expression
1. c 2  4c
2. b 2  14b
3. y 2  5 y
4. d 2  6 d
5. k 2  7k
6. m 2  11m
9. a 2  1 a
10. 3 y 2  12 y
2
7. n 2  n
11. 2 p 2  4 p
8. k 2  1 13 k
12. 3a 2  7a
Complete the square in the following
2
13. p  4 p  1
2
14. 3 y  12 y  2
2
15. 2 p  4 p  7
2
16. 5a  3a  4
page 9
ANSWERS
Exercise 1
1. 3 x 2  6 y
2. 2ax  x 2 y
3. x 2  8 x  15
4. x 2  3 x  28
5. x 2  6 x  9
6. 6 x 2  x  2
7. 6 x 2  11x  10
8. 9 x 2  16
9. 8 x 2  14 x  15
10. 3 x 2  12 xy  12 y 2
11. 9 x 2  36 xy  36 y 2
12. x3  3x 2  3x  1
13. 18 x 2  8 y 2
14.  4 xy  4 xz
15. 2 x 2  14
16.  6 x 2  1
17. x3  3x  2
18. x 4  4 x3  6 x 2  4 x  1
Exercise 2
1. 2  a  2b 
2.
a  3  4b 
3. 4t 2 g  3h 
6. a 2b 2  6b  1
9. 2r r  h 
4.
a  3b  4c  5d 
5.
t  u  5t 
7.
12d 2  3d  1
8.
a 3bc  a  b 
10. no simpler form
11.
Exercise 3
1.  x  2  x  1
4.
7.
10.
13.
16.
2.
 x  4  x  2 
5.
 3x  2  2 x  1
4  x  2 y  x  2 y 
 3x  2  2 x  3
 4 x  5 2 x  1
8.
11.
14.
17.


3b2 2a  3a 3b  b3c 2


12. 2r 2 2r 2  h 2  4r

 x  32
 x  8 x  2 
3.
 3x  12
4a  a  4 
 x  1 x  3
 mn  7  mn  3
9.
6.
12.
15.

 x  2 2
 2 x  1 x  1
 2d  1 d  1
 3x  1 3x  1
 2 x  1 2 x  3
Exercise 4
1. 4; (c  2)2
2. 49; (b  7)2
3. 25 ; ( y  5 )2
5. 49 ; (k  7 )2
6. 121 ; (m  11)2
7.
9.
4
1 ;
16
(a 
2
1 )2
4
13.  p  22 - 3
4
2
2
10. 12; 3( y  2)
14. 3 y  2  14
2
4
1 ;
4
(n 
2
1 )2
2
4. 9; (d  3)2
8. 4 ; (k  2 )2
9
49 ;
12
3
3(a  7 )2
11. 2; 2( p  1)2
12.
15. 2 p  12  5
2
16. 5 a  3  89

10

6
20
We would appreciate your comments on this worksheet,
especially if you’ve found any errors, so that we can improve it for
future use. Please contact the Maths tutor by email at
[email protected]
Updated 22nd June 2012
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alternative format on request. Telephone 01482 466199
© 2009
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