LEARN MATH STEP BY STEP THROUGH VERY EASY PROCESS

FRACTION

__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.)**