Algebra
Multiplying out quadratic expressions
(3x + 5)(4x  7)
Use FOIL (first, outer, inner, last)
The first terms are 3x*4x = 12x^{2}
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 12x^{2} 
21x + 20 x 35 which is then simplified to 12x^{2}
 x  35
Factorisation (4 types) There are four
different ways of factorising expressions
1 Take out a common factor
x^{2} + 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
x^{2} + 4x  5
In this case, x^{2}
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 ax^{2} + bx
+c. 'p' and 'q' must add to b and multiply to c.
3 Harder quadratic
factorisation
20x^{2}  7x  6
You will notice that in method 2, 'a' in ax^{2}
+ bx +c was equal to 1. Method three is
used when the number in front of x^{2}
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 ax^{2}
+ 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
ax^{2}  c where a and c are
both square numbers)
4x^{2}  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
4x^{2}  9 and the outer and inner are
6x + 6x and therefore cancel.
Completing the square
This is a slightly different form of factorisation
x^{2} + 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 x^{2} + 10x + 25 so
(x+5)(x+5) is correct. The reason why it works out is because c in
ax^{2} + bx +
c was exactly the square of half of b
What if c is not exactly the
square of half of b?
x^{2} + 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 x^{2} + 10x + 25. But the
original question said x^{2} + 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.
x^{2} + 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. 3x^{2} + 6x = 7
2re write in the form ax^{2}
+ bx + c = 0
3x^{2} + 6x  7 = 0
3 a= 3, b=
6, c= 7
4 Plug these into the
quadratic formula x=[b +√(b^{2}
 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}/_{[n1]} = 5
Multiply both sides by (n+2) and then by
(n1) to get rid of the denominators
2(n1) + 3(n+2) = 2(n+2)(n+1)
*Remember to multiply the right hand side by (n+2) and (n+1) as
well*
