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Factorisation



Factors of Natural Number

Factors of Natural number are the number which can divide the Natural Number
Consider 16
We can write
16= 2×8= 4×4=2×2×2×2=1×16
So 1,2,4,8,16 all are factors of Natural Number 16. Prime Factors are the factors which are prime number.
A number can be expressed as product of prime factors and that form is called prime form
16=2×2×2×2

Factors of algebraic expression

An algebraic expression can also be expressed as Products of factors
Example
6xy
6xy= 6×(xy) = 6×(x) ×(y)= 2×3×(xy)= 2×3×(x) ×(y)
We can see that 2,3,x,y cannot be further expressed as products of factor.  So they are called the prime factors. We called them irreducible factors in terms of algebraic expression
 

Factorisation of algebraic expression

When we factorise an algebraic expression, we write it as a product of factors. Thesefactors may be numbers, algebraic variables or algebraic expressions
The expression 6x (x - 2). It can be written as a product of factors. 2,3, x and (x - 2)
6x (x - 2).  =2×3× x× (x - 2)
The factors 2,3, x and (x +2) are irreducible factors of 6x (x + 2).
 

Method of Factorisation

Common factor method

1) We can look at each of the term in the algebraic expression, factorize each term into irreducible factors
2) Then find common factors to factorize the expression
 
Example
1) 2x+4
Here we can see 2 is the common factor in the terms
=2(x+2)
2) 11x2y +5x
First reduce it to irreducible factors form
=11×(x) ×(x) ×y+5×(x)
Now we can see that x is the common factor
=x(11xy+5)
 

Factorisation by regrouping terms

1) First we see common factor across all the terms in the algebraic expression
2) we look at grouping the terms and check if we find binomial factor from both the groups.
3) Take the common Binomial factor out
Example
1) 2xy + 3x + 2y + 3
First reduce it to irreducible factors form. There is no common factor across all the terms
= 2 × x × y + 3 × x + 2 × y + 3
Now we think about grouping the terms. We can see that we have common factor between first two terms
= x × (2y + 3) + 1 × (2y + 3)
= (2y + 3) (x + 1)
 
2) 6xy – 4y + 6 – 9x
There are no common factors across all the terms, so we think of grouping and rearranging the terms
=6xy – 4y  – 9x+6
=2y(3x-2) -3(3x-2)
=(2y-3)(3x-2)
 

Factorisation using identities

We can use the below identities to factorise algebraic expression
(a + b)2 = a2 + 2ab + b2
(a - b)2 = a2 - 2ab + b2
(a + b) (a – b) = a2– b2
 
Example
1) 4x212x + 9
We can write this as
(2x)2 -2×2x×3 + 32
Which resembles the second identity
Here a=2x and b=3
So
4x212x + 9 =(2x-3)2
2) 16x2 -81
We can write this as
(4x)2 -92
Which resembles the third identity
Here a=4x and b=9
So
16x2 -81=(4x-9)(4x+9)
 
Watch this tutorial for more example on Factorisation




Factorisation of the form (x+a)(x+b)

Given x2+ px + q,
1) we find two factors a and b of q (i.e., the constant term) such that
ab = q and a + b = p
2) Now expression can be written
x2+ (a + b) x + ab
or x2 + ax + bx + ab
or x(x + a) + b(x + a)
or (x + a) (x + b) which are the required factors.
Example
1)x2– 7x+ 12
Comparing to x2+ px + q
We have p=-7 and q=12
Now 12 =(- 3) × (-4) and (-3) + (-4) =- 7
 =x2 – 3x – 4x + 12
= x (x –3) – 4 (x –3) = (x –3) (x – 4)
 
2))y2+ 11y+ 30
Comparing to x2+ px + q
We have p=11 and q=30
Now 30 =( 5) × (6) and (5) + (6) =11
 =y2 + 5x + 6x + 30
= y (y +5) +6 (y +5) = (y +5) (y + 6)
 

Division of algebraic expression

Division of algebraic expression is performed by Factorisation of both the numerator and denominator and then cancelling the common factors.
General Steps of Division
1) Identify the Numerator and denominator
2) Factorise both the Numerator and denominator by the technique of Factorisation using common factor, regrouping, identities and splitting
3) Identify the common factor between numerator and denominator
4) Cancel the common factors and finalize the result
 
Example
48 (x2yz + xy2z + xyz2) /4xyz
=48xyz (x + y + z)/4xyz
= 4 ×12×xyz (x + y + z) /4xyz
= 12 (x + y + z)
 
Here Dividend=48 (x2yz + xy2z + xyz2)
Divisor=4xyz
Quotient=12 (x + y + z)
 
So, we have
Dividend = Divisor × Quotient.
In general, however, the relation is
Dividend = Divisor × Quotient + Remainder
When reminder is not zero
 

Solved Example on Division

 
1)  34a2bc2  /  51 abc
Solution
=17×2×a×a×b×c×c / 17×3×a×b×c
Cancelling common factors
= 2ac/3
 
2) 36(x2– 7x+ 12)/2(x-4)        
Solution
Here we need to factorize both the numerator and denominator
Let’s take a look at Numerator first
36(x2– 7x+ 12)
Here we can factorize this using splitting method
=36(x2 – 3x – 4x + 12)
=36(x-3)(x-4)
So division will become
36(x2– 7x+ 12)/2(x-4)
=36(x-3)(x-4)/2(x-4)
=18(x-3)
 
 
Watch this tutorial for more example on Division

Common Mistake done in algebraic expression and equation

 
Mistake 1
Coefficient 1 of a term is usually not shown. So student often ignores that. But while adding like terms, we should include it in the sum.
Wrong Approach
x+2x+3x
=5x
We ignored the coefficient 1
Right Approach
x+2x+3x
= (1) x+2x+3x
=6x
 
Mistake 2
When you multiply the expression enclosed in a bracket by a constant (or a variable) outside, you usually applied the multiplication to first term only. This is wrong calculation. Each term of the expression has to be multiplied by the constant (or the variable).
Wrong Approach
2(2x+3)
=4x+3
We ignored the second term
Right Approach
2(2x+3)
=4x+6
 
Mistake 3
when you square a monomial, you often ignored the   numerical coefficient. This is wrong approach. The   numerical coefficient and each factor has to be squared.
Wrong Approach
(2x)2
=2x2
We ignored the numerical coefficient
Right Approach
(2x)2
=4x2
 
Mistake 4
when you square a binomial, you often do not apply the proper formula. This is wrong approach. The right formula should be used
Wrong Approach
(2x+3)2
=4x2 +9
We did not use the proper formula
Right Approach
(2x+3)2
=4x2 +9+12x
 
 
 
 

 

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