FACTORS
An algebraic expression may be expressed as the product of two or more terms or expression, called factors of the original expression –
Example.-1) X⁴ + 2X²- 4X = X ( Xᶟ + 2X – 4),
In the above equation X & ( Xᶟ + 2X – 4 ) are the factors of X⁴ + 2X² - 4X
Example.-2) X² - Y² = ( X + Y ) ( X – Y )
In the above equation ( X + Y ) and ( X – Y ) are the factors of X² - Y²
Method of Factorizing an Expression –
One can use a number of ways to express an algebraic expression as the product of two or more factors, we have to consider two of them.
a) by finding the common factor in each term.
b) by writing the expression as a difference of square.
Finding the common factor –
Step.1) First we have to find the highest common factors (HCF) of the terms of the polynomial.
a) first we have to find the HCF of the numerical coefficients of the term
b) take the highest power of each literal common to the terms
c) the product of the factors in the steps a & b is the HCF.
Step.2) we have to divide each term in the expression by this common factor.
Step.3) then we have to place the quotient within brackets and the common factors outside the brackets.
Example.1) Please do factorize 4X – 12Y
Ans.) If we factorize both the number then we find - 4X = 2 x 2 x X and 12Y = 2 x 2 x 3 x Y, as we know that, the common factors of 4X & 12Y is = 4
4X – 12Y
4 X --------------- = 4 ( X – 3Y )
4
Example.2) Please do factorization of 3x + 5x + 7x
Ans.) Here, the common factor in all the terms is x
So, 3x + 5x + 7x = x ( 3 + 5 + 7 )
Example.3) Factorize 6a⁴bc² - 12aᶟbc² + 4a²b²c⁴
Ans.) In the above equations, there are three numbers 5a⁴bc², 7aᶟbc², 4a²b²c⁴
5a⁴bc² = 5 X a X a X a X a X b X c X c
7aᶟbc² = 7 X a X a X a X b X c x c
4a²b²c⁴ = 4 X a X a X b X b X c X c X c X c
The required HCF of the given numerical coefficients = the HCF of 6, 12 & 4 is = 2
The highest power of ‘a’ common to all three terms = a X a = a²
The highest power of ‘b’ common to all three terms = b
The highest power of ‘c’ common to all three terms = c X c = c²
So, the HCF or common factor of the term of the given equation is = 2 X a² X b X c² = 2a²bc²
So, factors are = 2a²bc² ( 3a² - 6abc² + 2bc² )
Example.4) Factorize 12a⁵b⁸c⁴ + 6a⁴b⁵cᶟ + 24 a⁶b⁴cᶟ
Ans.) In the above equation, there are three numbers 12a⁵b⁸c⁴, 6a⁴b⁵cᶟ, 24 a⁶b⁴cᶟ
12a⁵b⁸c⁴ = 2 X 2 X 3 X a X a X a X a X a X b X b X b X b X b X b X b X b X c X c X c X c X c
6a⁴b⁵cᶟ = 2 X 3 X a X a X a X a X b X b X b X b X b X c X c X c
24 a⁶b⁴cᶟ = 2 X 2 X 2 X 3 X a X a X a X a X a X a X b X b X b X b X c X c X c
The required HCF of the given numerical coefficients = the HCF of 12, 6 & 24 is = 6
The highest power of ‘a’ common to all three terms = a X a X a X a = a⁴
The highest power of ‘b’ common to all three terms = b X b X b X b = b⁴
The highest power of ‘c’ common to all three terms = c X c X c = cᶟ
So, the HCF or common factor of the term of the given equation is = 2 X a² X b X c² = 2a²bc²
So, factors are = 2a²bc² ( 3a²- 6abc²+ 2bc² )
Example.5) Factorize 8 ( a – b )⁴ + 16 ( a – b )⁶
Ans.) In the above equation, there are two numbers
8 ( a – b )⁴ = 2 X 2 X 2 X ( a – b ) X ( a – b ) X ( a – b ) X ( a – b )
16 ( a – b )⁶ = 2 X 2 X 2 X 2 X ( a – b ) X ( a – b ) X ( a – b ) X ( a – b ) X ( a – b ) X ( a – b )
The required HCF of the given numerical coefficients = the HCF of 8 & 16 is = 8
The highest power of ‘( a – b )’ common to all three terms
= ( a – b ) X ( a – b ) X ( a – b ) X ( a – b ) = ( a – b )⁴
So, the HCF or common factor of the term of the given equation is
= 8 ( a – b )⁴ { 1 + 2 ( a – b )}
= 8 ( a – b )⁴ ( 1 + 2a – 2b )
Difference of Two Square –
Polynomials which are in the form of the difference of two squares can be factorized by using the following relations
=> a² -b² = ( a + b ) ( a – b )
Example.1) Factorized 49a² - 36b²
Ans.) 49a² - 36b² = ( 7a + 6b ) ( 7a – 6b )
Example.2) Factorize a² - ( b + c )²
Ans.) a² - ( b + c )² = { a – ( b + c )} { a + ( b + c )}
= ( a – b – c ) ( a + b + c )
Example-3) Factorize a⁴ - b⁴
Ans.) a⁴ - b⁴ = ( a² + b² ) ( a² - b² )
= ( a² + b² ) ( a + b ) ( a – b )
Example.4) Factorize (0.50 )⁴ - (0.40)⁴
Ans.) (0.50 )⁴ - (0.40)⁴
= {(0.50 )² + (0.40)²} {(0.50 )² - (0.40)²}
= {(0.50 )² + (0.40)²} ( 0.50 + 0.40 ) ( 0.50 – 0.40 )
= ( 0.25 + 0.16 ) ( 0.90 ) ( 0.10 )
= ( 0.41 ) X ( 0.09)
= 0.0369 (Ans.)