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Section 1 - Algebra
Check In – Use these quick questions to highlight your strengths and weaknesses in Algebra.
Use the notes and examples on the following pages if you need any help with a particular topic.
You should know how to…
Check in…
multiply out brackets and collect like terms
3(2x + 4y) – 4(3x – 5y)
multiply 2 sets of brackets
(2x – 3)(3x + 1)
factorise an expression
3x2 + 12xy – 9x
factorise a quadratic expression
x2 + 4x - 12
solve an equation by factorising
Solve x2 + 4x – 12 = 0
solve an equation by using the quadratic
Solve 2x2 + 5x - 1
formula
solve a pair of simultaneous equations
Solve the simultaneous equations,
3y + 2x = 1
5y – 3x = -11
Solutions
6x + 12y – 12x + 20y
6x2 + 2x – 9x – 3
-6x + 32y or 32y – 6x
6x2 – 7x – 3
3x(x + 4y – 3)
(x + 6)(x - 2) or (x - 2)(x + 6)
x = -6 or x = 2
a = 2, b = 5, c = -1
x=2
y=-1
Section 1 - Algebra
Multiplying out brackets and collecting terms
These may look easy to start with, but watch out for careless mistakes that often cause
problems later on in more complex questions.
Example
Notice how the second
bracket is multiplied by –y
Remember to
collect like terms
Multiplying out 2 brackets occurs in many of the different chapters of C1 and C2 so make sure
you are confident at doing this. Don’t forget that (x + 3)2 would also need multiplying out in the
same way (x + 3)(x + 3).
OR you could use the grid multiplication method…
3x
-4
x
3x 2
- 4x
+2
+ 6x
-8
= 3x2 + 6x – 4x – 8
= 3x2 + 2x – 8
You also need to be able to deal with more complicated multiplications where there may be
more than 2 terms in each bracket. Each term in the first bracket needs to be multiplied by each
term in the second bracket.
OR again use a multiplication grid...
2
2x
-x
+4
x2
2x 4
- x3
+4x 2
+ 3x
+6x 3
-3x 2
+12x
-2
-4x 2
+2x
-8
= 2x 4 + 5x 3 – 3x 2 + 14x – 8
You may need to multiply out more than 2 brackets, in this case it is easiest to multiply 2
brackets together at a time, and then multiply the answer by another of the brackets.
Factorising
To factorise an expression, look for a common factor in each term which may be a letter, a
number or both. Remember you can always check your answer by multiplying out the bracket.
Each term can be divided
by 3, so 3 is a factor
It is essential that you can factorise quadratics quickly and confidently as this skill occurs
regularly in AS maths. Remember quadratic expressions are of the form ax2 + bx + c and these
will factorise into 2 brackets (
)(
). Start by finding numbers that multiply to make c. If a =
1 then you need the pair of factors that add to make b. If a is not 1, you will need to use trial and
error to check which pair is required.
Solution
i) Factor pairs of +3: 1 and 3, -1 and -3 (the pair needs to add up to +4, so we need 1 + 3 = 4)
(x + 1)( x + 3)
ii) Factor pairs of -12: 1 and -12, -1 and 12, 2 and -6, -2 and 6, 3 and -4, -3 and 4 (we need a pair that add
to -4, so we need 2 + -6 = -4)
(x + 2)( x – 6)
iii) Factor pairs of +6: 1 and 6, -1 and -6, 2 and 3, -2 and -3. To get 2x 2 we must have 2x in one of the
brackets, now try different pairs of factors, remember you need 2 times one of the factors and 1 times the
other factor to make -7. (we need 2 x -2 and 1 x -3 so we get -4 – 3 = -7)
(2x – 3)( x – 2)
Solving equations
Solving quadratic equations is important as it tells you where a quadratic graph crosses the xaxis. Some can be solved by factorising; others are solved by using the quadratic formula.
Notice how the equations must equal zero before you factorise or use the quadratic equation.
-4
If 2x + 3 = 0
2x = -3
x = -3/2
or
2
x+4=0
x = -4
If the quadratic equation will not factorise or it has large/difficult numbers, the quadratic equation
can be used to find the solutions.
Start by writing down the values of a, b and c being careful to include the negative signs. Then
substitute the values into the formulae, writing this step down as your method.
Example
Use the quadratic formula to solve
 (4)  (4) 2  4  2  1
2 2
4  16  8
x
2 2
x
Simultaneous equations
The simultaneous equations that you have worked with so far have involved 2 equations with 2
unknowns. One method of solving the simultaneous equations involves adding or subtracting
the 2 equations together (perhaps after multiplying them by a constant first) so that one of the
unknowns disappears. This method is called elimination.
Make sure every term is
multiplied by 2 on both
sides of the equation
Sometimes it is easier to use the method of substitution where one of the equations is
substituted into the other. This method tends to be used if one equation gives one of the
variables as the subject as shown in this example.
Section 2 – Number
You’ll be amazed how often you need to use fractions at AS level and it is often easier to use
fractions than decimals. Here are a few key tips for using fractions…
When solving equations…
7x+5=3
7 x = -2
x = -2/7 (don’t even try to write this as a decimal!)
Multiplying fractions: multiply the numerators, then multiply the denominators.
2 3 6
3
  SIMPLIFY 
5 8 40
20
Dividing fractions: leave the first fraction as it is, change the ÷ to x and turn the second fraction
upside down. Now repeat the process above for multiplying fractions.
4 2 4 7 28 14
4
   
 2
5 7 5 2 10 5
5
Fractions with squares and square roots: Squaring a number means multiplying it by itself, so
square the top and square the bottom. The same applies for square roots, square root the top
and square root the bottom.
2
2 2 4
2
    
3 3 9
3
1
1 1


4
4 2
Adding or subtracting fractions and whole numbers: To add or subtract fractions you need to
find equivalent fractions to the ones given that have the same denominators. The same can be
applied to whole numbers, make them into fractions by writing them over 1.
2 1 6 5 11
   
2 is a whole number
5 3 15 15 15
so is written as 2
1
2 2 2 6 2 4
1
2       or1
3 1 3 3 3 3
3
You will frequently need to substitute fractions into equations, you may need to find coordinates
for turning points on graphs or the intersection of lines.
Example
Given a line with equation y 
1
3
1
x  , find y when x 
4
5
2
Solution
Substituting the value of x into the equation gives…
1 1 3
y  
4 2 5
1 3
 
8 5
5 24


40 40
29

40
Surds
Surds can be simplified by using these two rules.
ab  a  b
a
a

b
b
Hint: To simplify surds look
for square numbers, note
how √4 changed to 2.
Example
Simplify the following:
i) √12
ii) √150
Solution
i ) 12  4  3  4  3  2  3  2 3
ii ) 150  25  6  25  6  5  6  5 6
Adding and subtracting surds is more like collecting like terms.
This is like multiplying the
second bracket by -1…
or, taking away each term
in the second bracket.
Multiplying out brackets containing surds should be done in the same way as explained above in
the algebra section. Use FOIL or a table to help you make sure every term in the first bracket is
multiplied by every term in the second bracket.
OR using
a grid…
1
2
2
+√3
+√3
-2√3
-4√3
-2√3√3
= 2 + √3 - 4√3 - 2√9
= 2 – 3√3 – (2x3)
= -4 – 3√3
Rationalising the denominator is the process of getting rid of the surds from the bottom of a
fraction. To do this you multiply by a fraction that is the same on the top and the bottom
(because then it is equivalent to multiplying by 1!). If the denominator of the fraction has 2
terms, multiply the top and bottom by the same numbers but use the opposite sign in between
them.
Notice how the
original denominator
was √2 – 1 so √2 + 1
is used to rationalise
the denominator.
(1-√3)(1+√3)
= 1 + √3 – √3 – √9
=1–3
Indices
You are probably quite confident on the first 3 rules of indices but make sure you practice the
indices questions that involve negatives and fractions.
6.
It is often easier to find
the root first before
doing the power.
Notice how the work on revising fractions can help you here when applying more than one of the
rules of indices…
Example
 34 
Simplify  2 
 
Solution
3 2

3
24
1
22

1
1
22
1

2
6
 2 12

2
3
Using rule 3, multiply
the indices.
Then after simplifying
use rules 4 and 6.
Section 3 – Graphs
The distance between 2 points and their midpoint
Finding the midpoint of 2 coordinates can be seen from a sketch or simply using the formula
shown here.
The distance between 2 points is the same as the length of the line between the 2 coordinates;
The equation of a straight line
The equation of a straight line is of the form y = mx + c where m is the gradient and c is the yintercept. You need to be confident at rearranging equations so they are in this form to be able
to give the gradient and y-intercept of a line.
 3
 0, 
 5 
as a coordinate
The gradient of the line can also be found from 2 points on the line using the change in the ycoordinate and the change in the x-coordinates. Remember if 2 lines are parallel they have the
same gradient.
It doesn’t matter which
coordinate you choose
for P and which is Q
Combining all of the information above can allow you to find the equation of a straight line by
using substitution.
Using the equation of a line: y = mx + c you can now substitute in the relevant values.
Using either P or Q from above you can obtain x and y values. Using P: x = 3, y = 8 and m = ¾
3
8  3 c
4
9
8 c
4
9
8  c
4
32 9 23
3
c
  or 5
4 4 4
4
Substitute the values into
y = mx + c and then rearrange
to get c. Don’t forget to re-write
the equation of the line!
So the equation of the line is:
y
3
3
x5
4
4
or
4 y  3 x  23
(Notice how each term is multiplied by 4 so the equation of the line is given with all the
coefficients as whole numbers, either equation is acceptable!)
Sketching Graphs
Linear graphs are straight lines and are of the form y = ax + b where a is the gradient and b is
the y-intercept.
Note the negative gradient means
the line slopes downwards. The
+1 indicates the line cuts through
the y-axis at (0,1).
Quadratic graphs are parabolas and are of the form y = ax2 + bx + c. If a is positive the graph is
shaped as below Graph 1, but if a is negative the graph is the other way up as in Graph 2.
Graph 1: y = x2 - x – 2
Graph 2: y = -2x2 + 3x + 1
Putting the function = 0 and solving the equation by factorising or using the quadratic formula
tells you where the graph cuts the x-axis.
Solving for Graph 1: x2 - x – 2 = 0
(x – 2)(x + 1) = 0
So x - 2 = 0 or x + 1 = 0
So x = 2 or x = -1 (Look how the parabola goes through these values on the graph above)
Putting x = 0 into the equation will tell you where the graph cuts the y-axis.
For Graph 1: y = 02 - 0 – 2 = -2 (the graph cuts the y – axis at -2, check this on the diagram)
Cubic Graphs are of the form y = ax3 + bx2 + cx + d. As above the general shape when a is
positive is shown in Graph 1, and when a is negative is shown in Graph 2.
Graph 1: y = x3 - 2x + 2
Graph 2: -x3 - 2x2 + x + 2
Putting the functions = 0 and solving the equations by factorising again tells you where the graph
cuts the x-axis (however in graph 1 the curve sits on the x-axis at 1 because of the repeated
root). Substituting x = 0 into the functions gives the y-intercept.
Questions
The best way to check you have understood all of this work and can remember the key skills
needed is to try some questions (and more importantly check you get the correct answers!)
Have a go at the following questions and keep note of any topics you need further help with.
Section 1 - Algebra
Multiplying out brackets and collecting terms
1. Simplify the following expressions:
(i)
2x + 3y – x + 5y + 4x
(ii)
5a – 2b + 3c – 2a + 5b
(iii) 4p + q – 6p – 5q + 5p + 4q
2. Multiply out the brackets and simplify where possible:
(i)
3(2x + 3y)
(ii)
4(3a – 2b) – 3(a + 2b)
(iii) p(2p – q) + 2q(p – 3q)
3. Multiply out these expressions.
(i)
(x + 1)(x – 3)
(iii) (x – 3)(x – 4)
(v)
(2x + 1)(4x – 1)
(vii) (3 + 2x)2
4.
(ii)
(iv)
(vi)
(viii)
(x + 2)(2x + 1)
(3x + 2)2
(1 – 2x)(1 + x)
(5x – 3)(2x + 5)
(i) Multiply (x³ + 4x² – 2x + 3) by (2x – 1)
(ii) Multiply (x² + 2x + 3) by (x² – x + 1)
Factorising
5. Factorise the following expressions:
(i)
10ab + 5ac
(ii)
2x² + 4xy – 8xz
(iii) 3s²t – 9s³t + 12s²t²
6. Factorise the following expressions as far as possible
(i)
3x²y – 6xy + 12xy²
(ii)
2a²b + 8a²b²
(iii) x² + 5x – 6
7. Factorise these quadratic expressions.
(i)
x² + 5x + 6
(ii)
(iii) x² – 6x + 8
(iv)
(v)
3x² + x – 2
(vi)
(vii) 4x² + 5x – 6
(viii)
x² + x – 12
2x² + 3x + 1
4x² – 8x + 3
6x² – x – 12
Solving equations
8. Solve the equations
(i)
5x + 1 = 2x – 11
(ii)
1 – 3y = 2(y – 8)
9. Solve the following equations:
(i)
2x – 3 = 8
(ii)
(iii) 3 – 2a = 3a – 1
(iv)
(v)
2(1 – z) + 3(z + 3) = 4z + 1
Start by collecting all of the
x terms onto one side of
the equation. (Multiply out
the brackets first!)
3y + 2 = y – 5
3(p – 3) = 2(2p + 1)
10. Solve these quadratic equations by factorising.
(i)
x² + 4x + 3 = 0
(ii)
x² + 5x – 6 = 0
(iii)
x² = 6x – 8
(iv)
x² = 7x + 18
(v)
2x² + 5x + 3 = 0
(vi)
2x² + x – 6 = 0
(vii) 4x² – 3x = 10
(viii) 6x² + 10 = 19x
11. Solve the following quadratic equations (where possible) using the quadratic formula.
(i)
x² + 2x – 2 = 0
(ii)
x² – 3x + 5 = 0
(iii) 2x² + x – 4 = 0
(iv)
2x² – 5x = 12
(v)
x² = 5x + 3
(vi)
3x² + x + 1 = 0
(vii) 4x² + 12x + 9 = 0
(viii) 4x² + 10x + 5 = 0
Simultaneous equations
12. Solve the following simultaneous equations:
(i)
2x + 5y = 11
(ii)
x = 6 – 2y
2x – y = 5
4x + 3y = 4
(iii)
3a – 2b = 4
5a + 4b = 3
(iv)
2p – 5q = 5
3p – 2q = –9
(v)
5x + 3y = 9
y = 3x – 4
(vi)
3a + 2b = 1
9a – 4b = 4
Make mixed numbers
into improper fractions
before multiplying or
dividing.
Section 2 – Number
Fractions
1. Calculate the following:
2 1
(i)

3 2
(v)
4
 
5
(ii)
1 2
2 
7 5
(iii)
1 2
2 
3 4
(vi)
36
100
(viii)
2
2
5
6
(iv)
3
1
1 2
10
4
(ix)
3
Surds
1. Write the following in terms of the simplest possible surd.
(a)
(b)
(c)
27
50
96
(d)
12  15
4. Rationalise the denominators of the following and simplify as far as possible
1
2 3
12
(a)
(b)
(c)
3
32
6
(d)
1 2
3 2
(e)
1 2
2 3
Indices
1. Find the values of
(a)
41/2
(d)
60
(b)
(e)
641 3
251 2
 1 2
 
9
(h)
 27  3


 125 
(a)
23  28  25
(b)
(32)3 x (37)-2
(d)
34
35  3 2
(e)
3 
25
93 2
(c)
( a 3 )5
( a 2 )5 2
2
1
(g)
(c)
(f)
2. Simplify
3. Simplify
(a)
(b)
(c)
323 2  85  25 2
35/2 + 31/2
2
3
1
3
2  4  53 4
2
5 3
1
 34
3
8
Section 3 – Graphs
The distance between 2 points and their midpoint
1. For the points A(3, 1) and B(7, 4) calculate
(i)
the gradient of AB
(ii)
the midpoint of AB
(iv)
the distance AB
2. For the points P(-2, 5) and Q(-6, -4) calculate
(i)
the midpoint of the line
(ii)
the distance of P from Q
Equation of a straight line
3. For the points A(-2, 9) and B(3, -1)
(i)
Calculate the gradient of AB
(ii)
Write the equation of the line through A and B.
4. For each of the lines below, give the gradient and y-intercept.
(i)
3y = 2x - 5
(ii)
4x – 2y = 7
(iii) 2y – 5x – 3 = 0
5. Write the equation of a line parallel to 5y = 15x + 4 that cuts the y-axis at (0, 2)
Sketching Graphs
6. Sketch the following lines.
(a)
y x 3
(b)
(d)
4y x 12
(e)
y 2x 1
3y x 6 0
(c)
(f)
x y 5
5y 15 2x
7. Sketch the following quadratic graphs, showing the points where the graph
cuts the coordinate axes in each case. (Hint: you factorised these above!)
(i)
y = x² + 4x + 3
(ii)
y = x² + 5x – 6
(iii) y = x² – 6x + 8
(iv)
y = 2x² + 5x + 3
(v)
y = 4x² – 3x – 10
(vi)
y = 6x² – 19x + 10
8. Sketch these curves, showing the points where the graph cuts the coordinate axes in each case. Some
of the functions have been factorised for you.
(i)
y = -x2 – 2x + 3 = (-x + 1)(x + 3)
(ii)
y = x3 – x2 – 2x - 8 = (x – 2)(x + 1)(x + 4)
(iii) y = (x + 1)(x – 3)2
(iv)
y = (x – 2)2
(v)
y = (-x + 2)(x + 3)(x – 4)