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Algebra Geometry Indices
Irrational numbers Circle theorems Simultaneous equations
Equations of Circles Proportion  Rearranging equations
Solving quadratics using graphs Histograms  Sampling
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Trigonometric functions Sine and Cosine rules Transformations
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Multiplying out quadratic expressions

(3x + 5)(4x - 7)

Use FOIL (first, outer, inner, last)

The first terms are 3x*4x  =  12x2

The outer terms are 3x*[-7]  =  -21x

The inner terms are 5*4x  =  20x

The last terms are 5*[-7]  =  -35

Hence the answer is 12x2 - 21x + 20 x -35    which is then simplified to 12x2 - x - 35

 

Factorisation (4 types) -There are four different ways of factorising expressions

1-    Take out a common factor

                    x2 + 3x + x

                    =x (x + 3 + 1)

                    =x (x + 4)

In the example above, each of the three parts that are added together are divisible by "x", hence we divide each by "x", put the whole thing in brackets and put "x" in front of the bracket.

2-    Quadratic factorisation

                    x2 + 4x - 5

In this case, x2 and 4x are both divisible by x but 5 is not hence we can't use method 1. We have to factorise in the form (x + p)(x + q). 'p' and 'q' are two numbers that together add to 4 and multiply to -5.

Quadratics that you have to factorise will always be in the form ax2 + bx +c. 'p' and 'q' must add to b and multiply to c.

3-    Harder quadratic factorisation

                    20x2 - 7x - 6

You will notice that in method 2, 'a' in ax2 + bx +c was equal to 1. Method three is used when the number in front of x2  is greater than 1. The problem with this is that we don't know whether to write (20x + p)(x + q) or (10x +p)(10x + q) etc...

                    1- Always write (20x + p)(20x + q) / 20

                    2- p and q must add to b and multiply to c*a in ax2 + bx +c.

              Qu) But in method 2 they had to multiply to c, not c*a (???)

              Ans) Actually, they have to add to b and multiply to c*a in both, except in method 2, 'a' was 1 so 'c*a' was the same as 'c'

                    3- Write in p and q...        (20x + [-15])(20x + [8]) / 20  ...and cancel to get (4x - 3)(20x + 8) / 4 and then (4x - 3)(5x + 2)

4- Difference of two squares (expressions in the form ax2 - c where a and c are both square numbers)

                    4x2 - 9

The square root both parts of it to get 2x and 3

                    write (2x + 3)(2x - 3)

This means that when you multiply it out using FOIL (first, outer, inner, last) the first and last terms give us  4x2 - 9 and the outer and inner are -6x + 6x and therefore cancel.

 

Completing the square

This is a slightly different form of factorisation

x2 + 10x + 25

25 is "the square"

10 is "twice the square root of the square"

1- always write (x+p)(x+p) where p is half of "twice the square root of the square" , so in this case we write (x+5)(x+5)

2- multiply it out and see what you get. In this case we get x2 + 10x + 25 so (x+5)(x+5) is correct. The reason why it works out is because c in ax2 + bx + c was exactly the square of half of b

What if c is not exactly the square of half of b?

x2 + 10x + 22

1- do exactly the same...  Write (x+p)(x+p) where p is half of b, so in this case we write (x+5)(x+5)

2- multiply it out and see what you get. In this case we get x2 + 10x + 25. But the original question said x2 + 10x + 22 so our answer is +3 too big therefore we have to change it to (x+5)(x+5) - 3.

NB) The usual way of writing these answers is (x+5)2 - 3 instead of (x+5)(x+5) - 3.

Completing the square can be used to solve quadratic equations.

x2 + 10x + 22 = 0

(x+5)2 - 3 = 0

(x+5)2 = 3

x+5 = +-√3

x = -5 +-√3

x = - 3.27 or x = -6.73             (all answers must be to 3 s.f. always!)

 

The Quadratic formula- alternative to completing the square

1- write down the equation you have to solve            e.g. 3x2 + 6x = 7

2-re write in the form ax2 + bx + c = 0                                3x2 + 6x - 7 = 0

3-    a= 3, b= 6, c= -7

4-    Plug these into the quadratic formula    x=[-b +-√(b2 - 4ac)] / 2a

5- The answer for this example turns out to be x = 0.826 or x = -2.83

Exam tip) Any time the question says "give your answer to a suitable degree of accuracy" you know that the answer will be a very long decimal and so the original expression does not factorise so you have to use the formula!

 

Fractions in equations

E.g. 1

[x+2]/6 +3 = [4x]/5               

Multiply both sides by 6 and then by 5 to get rid of the denominators

5x + 10 + 90 = 24x

E.g. 2

2/[n+2] +3/[n-1] = 5

Multiply both sides by (n+2) and then by (n-1) to get rid of the denominators

2(n-1) + 3(n+2) = 2(n+2)(n+1)

*Remember to multiply the right hand side by (n+2) and (n+1) as well*

 

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Topics on this site include: Algebra, Geometry, Indices, Irrational numbers, Circle theorem, Simultaneous equations, Equations of Circles, Proportionality, Rearranging Equations, Graphical solution of equations (using graphs to solve equations), Histograms, Sampling, Probability, Similar Solids, Proof, Accuracy and Error, Trigonometric functions, Sin (Sine) and cosine rule, Transformations and Vectors. Although this website is based on the AQA syllabus spec. A, many topics for other boards such as EdExcell and the IGCSE overlap with topics here. We hope this site is one of the best on the web, however mathsmad cannot guarantee the accuracy or completeness of the information on this website. All pages of this site are the copyright of the owner and may not be reproduced without written permission from the owner. This statment supercedes any other information that may be provided on this website in the case of contradiction.
 

GCSE maths revision notes