FACTOR - This is a whole number that divides a number leaving no remainder;
For example :
10/2 = 5 , here 2 divides 10 giving a quotient of 5 and no remainder so 2 is a FACTOR of 10
FACTORISATION - The process of expressing a value as the product of its factors.
6/2 = 3 ,here both 2 and 3 are factors of 6 also 2 X 3 = 6 ,
so 2 X 3 or 2(3) or 3(2) is a factorisation of 6.
When a number x is divided by one of its factors y the product of the divisor y and the quotient is equal to the number x.
For example :
20/5 = 4 ,here the divisor is 5 and the quotient is 4 ,so 5 X 4 = 20
and 5 X 4 is a factorisation of 20.
A common factor of two numbers X and Y is a number that is factor of both X and Y ,i.e a number that divides both X and Y.
For Example :
6/2 = 3 and 8/2 = 4 ,here 2 divides both 6 and 8, thus 2 is a common factor of 6 and 8.
Note that when we say that A divides B we mean that A divides B without leaving a remainder..
Highest Common Factor (HCF):
The highest common factor of two numbers X and Y is the greatest number that divides
both X and Y without leaving a remainder .
suppose we want to find the HCF of 8 and 20
2 divides 8 and 2 divides 20 ,so 2 is a common factor of 8 and 20.
4 divides 8 and 4 divides 20, so 4 is also a common factor of 8 and 20
so 4 and 2 are both common factors of 8 and 20 ,
4 is greater than 2 and since there is no number bigger than 4 that
divides both 8 and 20 , 4 is the HCF (highest common factor) of 8 and 20.
Solving the quadratic equation relies on bringing all its terms on one side
then expressing the quadratic expression (so formed with all the terms on one side of the equation) in product of factors including the Highest common factor,
then setting each factor equal to 0 ,and solving the equations so formed.
If ab = 0
then , a = 0 and b = 0
further if , x(x+1) = 0
then x = 0 and x+1 =0
then x+2=0 and x+1=0
TO solve Quadratic equation do the following
Step (1) make sure that all non zero terms are one one side
if they are not put them all on one side paying attention to their signs.
Step (2)factorize the non zero side of equation (which should be a quadratic expression)
Step (3) Set factors equal to zero ,and solve the consequent equations
Solve x2 + 3x + 2 = 0
since all the non-zero terms are on one side we proceed to step (2)
now we have to factorise the quadratic expression
x2 + 3x + 2
since ,1 x2 = 1 times x2 = x2
the numerical coefficient of the quadratic term is 1
in the equation above the constant term is 2 and numerical coefficient of the quadratic term is 1
now multiply the constant term by the numerical coefficient of the quadratic term
i.e 2 x 1 = 2
in the equation above the linear term is 3x and the numerical coefficient of 3x is 3.
Now we look for two numbers whose product is 2 and whose sum is 3.
since 1 x 2 = 2
and 1 + 2 = 3
1 and 2 will work
now multiply both 1 and 2 by x (the linear variable)
that is 1 times x = x and 2 times x = 2x
since x + 2x = 3x replace 3x with x+2x in the above equation
so , x2 + 3x + 2 = 0
becomes , x2 + x + 2x + 2 = 0
now put the terms on the left hand side into groups of two
as such ( x2 + x ) + (2x + 2) = 0
now factorise each binomial group on the left hand side of equation ,factorising the HCF out of each group
since the HCF of x2 and x is x and since the HCF of 2x and 2 is 2
we have , x[(x2/x) + (x/x)] + 2[(2x/2) + (2/2)] = 0
since x2/x = x , x/x =1 , 2x/2 = x , 2/2 =1
x(x + 1)+ 2(x + 1) = 0
now factoring the two terms x(x+1) and 2(x+1) , we see that x+1 is the HCF
(x+1)[x(x+1)/(x+1) + 2(x+1)/(x+1)] = 0
now since x(x+1)/(x+1) = x , and 2(x+1)/(x+1) = 2
(x+1)(x+2) = 0
setting both factors equal to zero
x + 1 = 0 and x + 2 = 0
solving the equations immediately above ,we have
x + 1 = 0
x = 0 - 1
x = -1
x + 2 = 0
x = 0 - 2
x = -2
Hence , the solution or roots of the above quadratic equation
x2 + 3x + 2 = 0
are x = -1 and x = -2
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