LEARN MATH STEP BY STEP THROUGH VERY EASY PROCESS

FACTORS

__ALGEBRAIC 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 factorization 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) 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) 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, so the common factors of 4X & 12Y is = 4**

** 4X – 12Y**

** 4 X
------------- = 4 ( X – 3Y )**

** 4**

**Example.2) Factorization 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 would be = 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 would be = 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 would be = 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 required HCF or the 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 )**