In order to  understand  factors  you
must  first  understand  the  use  of
brackets. Take for example:
     4 x 2 + 4 x 5 = 8 + 20 = 28

This can be written  as  4(2+5) = 28.
    The brackets ( ) enclose a number
or  numbers each of which are multip-
lied   by  the  number   outside  the
brackets  in  this  case  4  as shown
above.
^z
Note:
It is  not  necessary  to  write  the
multiplication symbol x between 4 and
the brackets eg 4 x (2+5) = 28.
    This  is  understood  by  context
just as you  do not  write + in front
of a positive number.

    When you  are  dealing  with  the
same   quantities inside the  bracket
you add ,subtract ,multiply or divide
^z
the  quantities  inside  the brackets
FIRST !  and  then  the  result   is
multiplied by the number outside  the
brackets,ie always solve the brackets
first.

      Eg: 4(9 x 3 + 4)  =
          4( 27 + 4)    =
          4(   31  )    = 124

      Eg: 2(15  3 + 6) =
          2(   5   + 6) =
          2(     11   ) = 22
^z
5(27 + 12- 3)
3(7 - 1 - 13)
4(7 - 3 + 8)
4(12)
^t
Rewrite 5 x 27 + 5 x 12 -5 x 3 using
bracket notation
blank

^t
Rewrite 3 x 7 - 3 x 1 -3 x 13  using
bracket notation
blank

^t
Rewrite 4 x 7 - 3 x 4 + 4 x 8  using
bracket notation
blank

^t
which can be shortened to give
blank
^z
12(7 + 5 - 5)
84
2(7 - 2 - 8)
-6
14(7 - 4 + 6)
126
^t
Rewrite 12 x 7 + 5 x 12 -5 x 12 using
bracket notation  blank
^t
which equals blank

^t
Rewrite  2 x 7 - 2 x 2 -2 x 8  using
bracket notation  blank
^t
which equals blank

^t
Rewrite 14 x 7 - 14 x 4 + 14 x 2 x 3
using bracket notation blank
^t
which equals blank
^z
4(7 + 6 + 4)
7(y + 10 + 2)
5a( 9)
3(9b + 8)
^t
4 x 7 + 4 x 6 + 4 x 4 can be
written as blank

^t
7y +7 x 10 + 14 can be written as
blank

^t
Notice you can split 14 up into 7x2
for the purposes of factorisation.

5a + 25a + 15a   = blank

^t
27b + 3 + 7 x 3  = blank

^z
416
63
9
-108
100
10
93
^t
4(9 x 12  +  3  -  1 x 7) =   blank

^t
3(12  4   +   2 x 9 )    =   blank

^t
3(3)                      =   blank

^t
6(4 + 5   x   12 - 14)    =   blank

^t
5(5 x 5   -   15  3)     =   blank

^t
5 x 2(4  +  6 x 2 - 15)   =   blank

^t
3(7 x 7  -  48  3  -  2) =   blank
^z
4
20
56
9
115
3
-8
^t
2(28  2  -  3 x 2  -  6) =   blank

^t
5(24  3  -  7  +  6  2) =   blank

^t
7(3  +  12  -  56  8)    =   blank

^t
3(4  +  5  -  12 x 2)     =   blank

^t
5(4 x 3  +  21  3  +  4) =   blank

^t
(15  -  6  -  9  3)     =   blank

^t
(4  -  5 x 4  -  14)  =   blank
^z
Brackets may be nested eg:

       4(3(3 + 6)) =
       4(3(  9  )) =
       4(   27   ) = 108

In this case you work on the  numbers
in  the innermost  brackets first and
then work your way outwards.
      eg: (3 + 6) = 9
         3(  9  ) = 27
         4( 27 )  = 108

^z
innermost
5
3
7( 5 x 3 )
105
^t
Let's try an example of the use of
nested brackets.
What does  7( (15  3)(7 - 4) ) =

We start from the blank
brackets and work our way outwards.

^t
(15  3) = blank     and
^t
(7 - 4)  = blank

^t
Thus the expression can be written
in a simpler form as blank

^t
Therefore 7( (15  3)(7 - 4)) = blank

^z
4 - x
160 - 40x
^t
Here's another one
      4( (2 + 8)(4 - x) ) =

Again working our way outwards we try
and simplify the expression.
         (2 + 8) = 10
       10(blank ) = 40 -10x

^t
     4( 40- 10x ) = blank

Therefore 4( (2 + 8)(4 - x) ) =
                    160 - 40x
^z
(2 x 3)
8
6( 8( 4 - x ) )
6( 32 - 8x )
192 -48x
^t
Now try these :
      6( (48  (2 x 3) )(4 - x) ) =

The innermost expression is blank

^t
48  6 = blank

^t
Therefore the expression can be
rewritten as blank

^t
Which simplifies to give blank

^t
Therefore 6( (48  (2 x 3) )(4 - x) )
          = blank
^z
(4 + 8)  (2 x 3)
12  6
6( ( 2 )(4 - x)
6( 8 - 2x )
48 -12x
^t
 6( ( (4 + 8)  (2 x 3) )(4 - x) ) =

The innermost expressions are
blank

^t
These simplify to give blank

^t
Therefore the expression can be
rewritten as blank

^t
Which simplifies to give blank

^t
Therefore
6( ( (4 + 8)  (2 x 3) )(4 - x) )
         = blank
^z
(9  3) & (4x + 3)
15x + 9
2x(15x + 9)
18x
^t
 2x( x( 9  3 ) + 3( 4x + 3 ) )

The innermost expressions are
blank

^t
Therefore the expression inside the
first set of brackets is blank

^t
Therefore the expression can be
rewritten as blank

^t
Therefore
2x( x( 9  3 ) + 3( 4x + 3 ) )
         = 30x + blank
^z

Notice  that  every time  you  open a
bracket you must close it.

    Eg:
       4(3x(3 +6)) =
        ^  ^    ^^
              Ŀ
              Ŀ      
           second set       
                              
             first set    
^z
A  factor  of a  given  number  is  a
number which divides evenly into  the
given number.

  4  is a factor of  12  because it
  divides evenly into 12.

  8 is a factor of 64 because it
  divides evenly into 64 to give
  you 8

^z
A factor  of an algebraic expression
on  the other  hand need  not divide
evenly  into it  but  by multiplying
the factors  of an  expression toge-
ther  we can obtain the  expression.

       3(4 + 2x) = 12 + 6x
      factors > expression
              <

How can we derive  the  factors of an
expression ?
^z
    If the  entire  expression can be
divided   by  a  number   or  simpler
expression  this number or expression
is called a common factor.
For example:
            12 + 6x is an  expression
which can be divided by  the number 3
(12 + 6x)    3  =  (4 + 2x).
    It  can  also  be  divided by the
simpler expression  (4 + 2x)
(12 + 6x)    (4 + 2x) = 3.
    Therefore (4 + 2x)  and 3 are the
factors  of  the  expression 12 + 6x.
^z
Naturally this  may not seem  obvious
or apparent  to you  at first , so we
will establish a systematic technique
which will allow us  to find  factors
for even the most complex of express-
ions.
    We will use a  systematic step by
step approach called an  algorithm to
solve  for  the  factors of an expre-
ssion.
^z
          FACTOR ALGORITHM
  Ŀ
  1 Look at smallest coefficient 
    of the expression if there   
    is a common factor then it   
    cannot be greater than this  
    number.                      
  
                 
                 V
^z
                 ?
  Ŀ
  2 Does  this  number  divide   
    evenly  into  every  other   
    coefficient of the expression
    ie is it a factor ?          
  
                ?         
^z
                          
        No                Yes
Ŀ Ŀ
3 Can you think  3               
of a number that                 
will divide         This is the  
evenly into this   common factor 
number ie a                      
factor?                          
 
        ?  
           Ŀ
                         
^z
                          
       No                 Yes
Ŀ Ŀ
4 The expression 4 Is  this  the 
does not have a  largest  number 
common factor we that will divide
will see how to  into  all  the  
factorise it     coefficients  of
later.           the expression ?
 
                         ?  
             
                          
^z

                          
       No                 Yes

Ŀ Ŀ
5                5               
The largest num-                 
ber  that  will   This number is 
divide in is the    the common   
common factor         factor     
                                 
 
^z
divides
Multiplying
evenly
^t
A factor is a number or expression
which blank   evenly into  another
number or expression.

^t
blank       the factors of an
expression together gives you the
expression.

^t
A common factor of an expression is
one which divides blank  into the
expression.

^z
2
common
Dividing
x - 4
factors
^t
Lets use the factor algorithm to
find the factors of the following
expressions :    2x -8

The smallest coefficient that
divides evenly into the expression
is blank.

^t
Therefore 2 is the blank  factor.

^t
blank    the expression by the
common factor gives us
^t
blank .
^t
Therefore the blank   of the
expression are 2 and x - 4
^z
3
3x
3x(x + 3y -5z)
3x
(x + 3y -5z)
^t
Find the factors of 3x + 9xy - 15xz

The smallest coefficient of the
expression is blank.

^t
However not only will 3 divide
evenly into the expression but so
will blank

^t
Therefore 3x is a common factor.
So we can  rewrite the expression
as blank

^t
Thus the factors of 3x + 9xy -15xz
are blank
^t
and blank
^z
a
+ c
j
(-1  -3p)
5a
(1  -  3b  +  4c)
^t
What are the factors of ab + ac

              blank and
^t
              b blank

^t
jp - j - 4jp
              blank and
^t
              blank

^t
5a -15ab +20ac
              blank and
^t
              blank
^z
r
-3s + 5
ab
(5a + 10b + 1)
(x + y)
(5 + z)
(5a - 4b)
(4x + 3y)
^t
rs + 5r -3rs
              blank and
^t
              (s blank  )

^t
5ab + 10ab + ab
              blank and
^t
              blank

^t
5(x + y) + z(x + y)
              blank   and
^t
              blank

^t
4x(5a -4b) + 3y(5a -4b)
              blank     and
^t
              blank
^z
Sometimes an expression does not have
a common factor.In this case we split
the  expression up  into two  or more
parts and factorise them individually.
For example :
             xy + xz + ay + az
We split this up into
   xy + xz             ay + az

has the common     has the common
factor x           factor a
  x(y + z)             a(y + z)
^z
x(y + z) + a(y + z) = xy + xz + ay +az

(y + z) is now a common factor of the
expression written in this form.

(x + a)(y + z) = xy + xz + ay + az

Thus   (x + a)   and   (y + z)  are
the factors of the expression.
^z
Find the factors of
                    ab - ac - db + dc
First split  it  into  two parts  and
extract  the  common factor from both
parts.
       a(b - c)  and  d(-b + c)
Since   (b - c) = -(-b + c)
These factors can be written as :
       a(b - c)  and -d(b - c)
Therefore  (a - d) & (b - c)  are the
factors of the expression.
^z
ab + ac
db + dc
a(b + c)
d(b + c)
(a + d)
(b + c)
^t
Let's try a few examples of these
types of problems

            ab + ac + db + dc

Splitting it into two parts we get
blank     and
^t
blank

^t
Factorising these parts individually
we get blank     and
^t
blank

^t
Therefore the factors are blank
^t
and blank
^z
4a + 4
ab + a
4(a + 1)
b(a + 1)
(4 + b)
(a + 1)
^t

            4a + 4 + ab + b

Splitting it into two parts we get
blank     and
^t
blank

^t
Factorising these parts individually
we get blank     and
^t
blank

^t
Therefore the factors are blank
^t
and blank
^z
5 + 15y
-x -3xy
5(1 + 3y)
-x(1 + 3y)
(5 - x)
(1 + 3y)
^t

            5 - x + 15y - 3xy

Splitting it into two parts we get
blank     and
^t
blank

^t
Factorising these parts individually
we get blank     and
^t
blank

^t
Therefore the factors are blank
^t
and blank
^z
-3dc + 3bd
ab - ac
3d(-c + b)
a(b - c)
(3d + a)
(b - c)
^t

          ab - 3dc - ac + 3bd

Splitting it into two parts we get
blank      and
^t
blank

^t
Factorising these parts individually
we get blank      and
^t
blank

^t
Therefore the factors are blank
^t
and blank
^z
Next we come to a very important  use
for factors:
 -  Solving Quadratic Equations  -

An expression of the form
   ax + bx + c
{ Where a is not equal to 0 }
is called a quadratic expression.

    Quadratic  expressions   may   be
written in  many  different  forms  to
allow you to factorise them.
^z
There are  three  steps to  solving a
quadratic expression.

    Step 1: Multiply  the   constant
    { c } by the coefficient  of the
    x term and call this number the
    guide number.

    Step 2: Find the factors  of the
    guide  number  which  when added
    together give the coefficient of
    the x term.
^z
    Step 3: Rewrite  the expression
    using   these  factors  as  the
    coefficients of x.

    Now   factorise  the   expression
    using using the methods  you have
    already seen.
^z
As this  section will be dealt  with
in  another  program  we  will  just
illustrate one example here :
     x + x - 8 = 4

The constant here  is not -8 we must
first  bring  4  over  to  the other
side  of the equation  to  determine
the constant :
     x + x - 12 = 0
     The  constant = -12  the  guide
number = -12.
     The factors of the guide number
which  when  added together give the
^z
coefficient of  the x term are 4 and
-3.
     There is no easy way of getting
these its just a matter of trial and
error and experience.
     Lets just examine  one  way  of
doing it.
     -12 has the factors
      4 , -3         3  , -4
      2 , -6        -6  ,  2
      1 , -12       -12 ,  1
Now which pair of factors when added
together give the coefficient of the
x term  1 ?
     Thats right  4 and -3.
^z

Rewriting the equation

         x + 4x - 3x -12 = 0
      x(x + 4)  -3(x + 4) = 0

Therefore the factors are :

           (x + 4)(x - 3) = 0
^z


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