Points to Remember- Factorisation

Table of Contents
1. What are Factors?
2. Factors of Algebraic expressions
3. Methods of Factorization
4. Factorization Using Common Factors
5. Factorisation By Regrouping Terms
6. Factorisation using identities
7. Division of Algebraic Expressions
8. Some Solved Examples for You:
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What are Factors?

Factors are numbers, letters, or expressions that, when multiplied together, create a specific number or expression. Factorization simplifies complex expressions into easier parts for examination or use.

Factors of Algebraic expressions

Note : 1 is a factor of every term but is usually not explicitly shown unless necessary.

Methods of Factorization

1. Factorization Using Common Factors

To factorize an algebraic expression, the highest common factors are determined.

Example 1:Factorise algebraic expression -2y² + 8y 

Solution: We have:   -2y² + 8y 
On factorising the terms, we get : 
-2y²= -1 × 2 × y × y
8y = 2 × 2 × 2 × y
The highest common factors of these terms are 2y
-2y²+8y = [ (- 1× 2 × y) × y ] + [ (2 × 2 × 2) × y]
=2 × y × (-y + 4)
Therefore, -2y² + 8y = 2y(-y + 4)

Example 2: Factorise 18x² - 14x³ + 10x⁴

Solution: We have  

$18x^2\; =\; 2\; \times \; 3\; \times \; 3\; \times \; x\; \times \; x$18x²= 2 × 3 × 3 × x × x
14x³=2 × 7 × x × x × x
10x⁴= 2 × 5 × x × x × x × x

Obviously, the common factors of these terms are 2, and two times x.

∴ 18x² - 14x³ + 10x⁴

=[2 × 3 × 3 × x × x] - [2 × 7 × x × x × x] + [2 × 5 × x × x × x × x] 

=[(2 × x × x) × 3 × 3]- [(2 × x × x ) × 7 × x] + [( 2 × x × x ) × 5 × x × x]

=(2 × x × x)[3 × 3 - 7 × x + 5 × x × x]

=2x²[ 9 - 7x + 5x²]

Thus, 18x² - 14x³ + 10x⁴ = 2x²[9 - 7x + 5x²]

2. Factorisation By Regrouping Terms

In some algebraic expressions, it is not possible that every term has a common factor. Therefore, to factorise those algebraic expressions, terms having common factors are grouped together. 

Example 1: Factorise 9x + 18y + 6xy + 27

Solution:Here, we have a common factor 3 in all the terms.

∴ 9x + 18y + 6xy + 27 = 3[3x + 6y + 2xy + 9]

We find that 3x + 6y = 3(x + 2y) and 2xy + 9 = 1(2xy + 9)

i.e. a common factor in both groups does not exist,

Thus, 3x + 6y + 2xy + 9 cannot be factorized.

On regrouping the terms, we have

3x + 6y + 2xy + 9 = 3x + 9 + 2xy + 6y

= 3(x + 3) + 2y(x + 3)

= (x + 3)(3 + 2y)

Now, 3[3x + 6y + 2xy + 9] = 3[(x + 3)(3 + 2y)]

Thus, 9x + 18y + 6xy + 27 = 3(x + 3)(2y + 3)

Example 2: Factorise $8xy\; -\; 6y\; +\; 4\; -\; 12x$8xy-6y+4-12x

Solution: Step 1: Check if there is a common factor among all terms. There is none.

Step 2: Think of grouping. Notice that the first two terms have a common factor:

8xy - 6y=  2y (4x - 3)

For the last two terms:

4 - 12x = - 4(3x - 1)

Step 3: Putting the groups together:

8xy - 6y + 4 - 12x
= 2y(4x - 3) - 4(4x - 3)

Factor out the common term (4x - 3):

=(4x - 3) (2y - 4)

The factors of 8xy - 6y + 4 - 12x are: (4x - 3) and (2y - 4)

3. Factorisation using identities

Some expressions can easily be factorized using these identities:

Example 1:Factorise 4x² - 20x + 25.

Solution: 4x² - 20x + 25
= (2x)² - 2 × 2x × 5 + (5)²
= (2x - 5)²  --[Using the identity a² - 2ab + b² = (a - b)²]

Example 2: Factorise $64x^2\; -\; 25$64x² - 25

Solution: There are two terms; both are squares and the second is negative. The expression is of the form (a²-b²). 
= 64x²- 25 = (8x)² - (5)²
Using identity a² - b² = (a + b) (a - b)
= (8x - 5) (8x + 5)

Example 2:Verify that : (11pq + 4q)² - (11pq - 4q)² = 176pq²

Solution: Solving L. H. S. = (11pq + 4q)² - (11pq - 4q)² 

[ Using identities (a + b)² = (a² + 2ab + b²) and (a - b)² = (a² - 2ab + b²) ]

= 121p²q² + 88pq² + 16q² - (121p²q² - 88pq² + 16q²)

= 121p²q² + 88pq² + 16q² - 121p²q² + 88pq² - 16q²

= 88pq² + 88pq²

= 176pq²

L. H. S. = 176pq²

Therefore, L. H. S. = R. H. S. (verified)

Division of Algebraic Expressions

1. Division of a Monomial by Another Monomial:

• Write the irreducible factors of both the monomials.
• Cancel out the common factors between the numerator and denominator.
• The remaining terms form the answer.

Example: Divide 12y² by 6y

Solution: Dividing 12y² by 6y 

2. Division of a Polynomial by a Monomial:

• Write the irreducible form of both the polynomial and monomial.
• Factorize the polynomial, if possible, to identify common terms.
• Cancel out the common factor between the terms.
• The remaining terms represent the final answer.

Example 1: Divide $x^6\; +\; 7x^5\; -\; 5x^4$x⁶+7x⁵-5x⁴ by x²

Sol: Dividing  x⁶+7x⁵-5x⁴ by x²
= x⁶+ 7x⁵- 5x⁴ ÷ x²
Now, we need to divide each term of the polynomial by the monomial and simplify:$=\; \backslash frac\{x^6\}\{x^2\}\; +\; \backslash frac\{7x^5\}\{x^2\}\; -\; \backslash frac\{5x^4\}\{x^2\}$

^{x⁶x² + 7x⁵x² - 5x⁴x²}

= x⁴ + 7x³ - 5x²

3. Division of A Polynomial by another  Polynomial:

• In the case of polynomials, reduce the terms and factorize using identities or by finding common terms. 
• Cancel out common factors to find the result.

Example1: Solve $(3x^3\; -\; 27x)\; \backslash div\; 3x(x\; -\; 3)$(3x³ - 27x) ÷ 3x( x - 3)

Sol: 

Factorize the polynomial:

3x³ - 27x = 3x(x² - 9)

Using the identity (a² - b² = (a + b)(a - b):

3x(x² - 9) = 3x(x + 3)(x - 3)

Divide by:

3x(x + 3)(x - 3)3x(x - 3)

Cancel the common factors 3x and (x - 3):

= x + 3

$\backslash frac\{54y^3\}\{9y\}\; =\; 6y^2$

Some Solved Examples for You:

Q1: Factors of x²+ xy + 8x + 8y are : 
A) (x + y)(x + 8)
B) (2x + y)(x + 8)
C) (x + 2y)(x + 8)
D) (x + 8)(x + y)

Solution: correct option is (A)
x² + xy + 8x + 8y
Factorizing each term 
 = (x × x) + (x × y) + (8 × x)+ (8 × y)
On taking common from terms 
=x(x + y) + 8(x+y)
=(x + y)(x + 8)

Q 2: Simplify 5(2x+1)(3x+5)÷(2x+1)
A) 10x + 5
B) 5(3x + 5)
C) 5(2x + 1)(3x + 5)
D) 3x + 5

Solution: correct option is (B)

5(2x + 1)(3x + 5 ) ÷ (2x + 1) =  5(2x + 1)(3x + 5)(2x + 1) 

= 5(3x + 5)