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Factorization - Class 8 : Notes


Factors:
These are the number when multiplied give another number.
Example: 2 and 3 are factors of 6, since 2 × 3 gives 6.
In other words, pair of all the numbers whose product results in one single number are said to be the factors of that number.
Example: The numbers 1, 2, 3, 4, 6, 12 are the factors of 12.

1. Prime Factor Form:
When number is written as a product of prime factors then it is said to be its prime factor form.
Example: Prime factor form of 70 is 2, 5, and 7.

2. Algebraic Expressions Factors:
The irreducible form of algebraic expressions is known as its factors.
Example 1: The factors of 5xy will be 5, x and y.
Example 2: The factors of 3x(x + 2) will be 3, x and (x + 2).


Factorization:
It is the process of decomposition of algebraic expression into product of other objects or factors which gives the original expression. The following process are:

1. Method of common factors:
Step 1: Decompose every term into irreducible factors.
Step 2: Find all the common terms amongst all the obtained irreducible factors.
Step 3: The product of common terms and the left out terms will give the desired factor form.

Example 1: Factorise 5x + 20.
Solution:
Step 1: Decompose every term into irreducible factors.
                    Here, 5x = 5 × x and 20 = 2 × 2 × 5.
Step 2: Find all the common terms amongst all the obtained irreducible factors.
                    Here, the only common term is 5.
Step 3: The product of common terms and the left out terms will give the desired factor form.
                    Thus, the desired factor form will be 5 (x + 4).

Example 2: Factorise 8x2 + 6x + 4.
Solution8x2 = 2 × 2 × 2 × x × x
               6x = 3 × 2 × x
                 4 = 2 × 2
Thus, 8x2 + 6x + 4 = 2 (4x2 + 3x + 2). This is the desired factor form.

2. Factorisation by regrouping
For given expression, if there is no term common to all the terms then we take out common amongst some terms and then obtain desired factor form.

Example: Factorise xy + x + y + 1.
SolutionWe will take (xy + x) as one group and (y + 1) as another.
                Factor form of (xy + x) = (x × y) + (x × 1)
                                                      = x (y + 1)
                Factor form of (y + 1) = (y × 1) + (1 × 1)
                                                    = (y + 1)
               On combining, we get,
                                      xy + x + y + 1 = x (y + 1) + (y + 1)

               Now we can take common (y + 1) from both the term, we get,
                                                             = (x + 1) (y + 1)


3. Factorization using identities:
There are many standard identities. Some of them are given below:
(i) (a + b)2 = a2 + 2ab + b2
(ii) (a - b)2 = a2 - 2ab + b2
(iii) (a + b) (a - b) = a2 - b2

Example 1: Factorise x2 + 10x + 25.
SolutionWe will use the identity (a + b) 2 = a2 + 2ab + b2 here.
               Therefore, x2 + 10x + 25 = x2 + 2 × 5 × x + 52
                                                       = (x + 5)2

Example 2: Factorise 4y 2 – 12y + 9
Solution: We will use the identity  (a - b)2 = a2 - 2ab + b2
                Since  4y 2 = (2y) 2 , 9 = 32 and 12y = 2 × 3 × (2y)
               Therefore, 4y 2 – 12y + 9 = (2y) 2 – 2 × 3 × (2y) + (3)2
                                                         = ( 2y – 3)2

Example 3: Factorise 49p 2 – 36
Solution:  There are two terms; both are squares and the second is negative.
               The expression is of the form (a 2 – b2 ). We will use the identity (a + b) (a - b) = a2 - b2
                                      49p 2 – 36 = (7p) 2 – ( 6 )2
                                                       = (7p – 6 ) ( 7p + 6)

4. Factors of the form (x + a) (x + b):
In this type, factorise given expression such that (x + a) (x + b) = x2 + (a + b) x +ab.

Example: Factorise x2 + 3x + 2.
SolutionOn comparing with the identity (x + a) (x + b) = x2 + (a + b) x +ab; we get (a + b) = 3 and ab = 2.
                On solving, we get a = 1 and b = 2.
Substituting these values into the identity, we can expression as (x + 1) (x + 2).


Division of Algebraic Expressions:
(a) Division of monomial by a monomial:
Let us take example to understand this:

Example: Divide 14x2 by 7x.
SolutionWriting both the given terms into irreducable form, we get,
                 14x2 = 2 × 7 × x × x
                    7x = 7 × x
Now, diving in usual manner, we can write
                14x2 / 7x = (2 × 7 × x × x) / (7 × x) = 2x

(b) Division of polynomial by a monomial:
Let us take example to understand this:

Example: Divide 4x3 + 2x2 + 2x by 2x.
SolutionWe can write,
               (4x3 + 2x2 + 2x) / 2x
We can take the 2x common from each the term, and divide by 2x, then we get-
= 2x (2x2 + x + 1 ) / 2x
= (2x2 + x + 1)


(c) Division of polynomial by a polynomial:
Let us take example to understand this:

Example: Divide 9x2 + 3x by (3x + 1).
SolutionWe can write,
               (9x2 + 3x) / (3x + 3)
We can take the 3x common from each the term, and divide by (3x+1), then we get
= 3x × (3x + 1) / (3x + 1)
               = 3x



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