algebraic factorisation

Any number can be expressed in the form of its factors, for example, 12 = 4 × 3. Similarly, an algebraic expression can also be expressed in the form of its factors. Let us take an example, 4x² + 12xy. This equation has two terms 4x² and 12xy.

We can express 

4x² as 4 ⋅ x ⋅ x and

12xy as 12 ⋅ x ⋅ y or 3 ⋅ 4 ⋅ x ⋅ y.

Notice that in both the terms 4x is a common factor, therefore, we can also write the expression as \(4x(x + 3y)\). Expand \(4x(x + 3y)\) and you will get back the same expression. We just factorized our first algebraic expression!

An Algebraic expression can sometimes be represented in the form of a product of two or more algebraic expressions. Every algebraic expression in the product is called a factor of the given expressions. For example, 4x and x + 3y are factors of expression 4x² + 12xy. Finding factors of a given expression is called Algebraic Factorization. 

Let us learn how to factorize under various cases:

When an expression has a Monomial as a common factor of all its terms

Identify the largest monomial which is a factor of each term of the expression.

Example:

1. Factorize 3x²y+ 9xy² + 12xyz
3xy is the largest monomial common to the three terms 3x²y, 9xy², and 12xyz
Therefore, this term can be expressed as

 \( 3xy(x) + 3xy(3y) + 3xy(4z) \)

⇒ \(3xy(x + 3y + 4z)\)

2. Factorize x³y²z + x²y + 2xy²
xy is the largest monomial common to the three terms x³y²z, x²y, 2xy²
Therefore, \(x^3y^2z + x^2y + 2xy^2 \) = \(xy(x^2yz + x + 2y)\)

When an expression has a compound factor common to all its terms

Example: \(3(x + 2y) + 5x(x + 2y)^2 \)

Therefore, it can be written as 

\((x + 2y) (3 + 5x^2+ 10xy)\)

Factoring by Grouping

Step 1: Arrange the terms of the given expression in groups in such a way that all the groups have a common factor.

Step 2: Factorize each group.

Step 3: Take out the factor which is common to each group.

Example:
\(5ax^2 + 3axy −5bxy − 3by^2 \\ ax(5x + 3y)− by(5x + 3y) \\ (5x + 3y) (ax − by)\)

When an expression fits into an algebraic formula

Try to use an algebraic formula to factorize an algebraic expression. 

Example:

1. \(49m^2b^4 − 4m^2b^6 \)
   \(= m^2b^4(49 − 4b^2)\)
   Using the formula x²−y² = (x+y)⋅(x−y), 49 − 4b² can be written as (7+2b)⋅(7−2b) that is 7²− (2b)²
   \( = m^2b^4[7^2 − (2b)^2] = m^2b^4(7 − 2b)(7 + 2b) \)
    
2. \(32x^2+32x + 8 = 2( 16x^2+ 16x + 4)\) 
   Using formula (x+y)² = x² + 2xy + y² we can replace 16x²+16x+4 as (4x+2)²
   \( = 2[(4x)^2+ 2⋅4x⋅2 + 2^2] = 2(4x + 2)^2 \)

Factorization of a Second - Degree Trinomial

Can a second-degree or a quadratic polynomial be factorized? The answer is "yes"
A quadratic polynomial is expressed as ax² + bx + c , where a, b and c are not equal to zero.

Let's discuss two cases

1. a = 1
2. \(a \neq 1\)

Case 1: If a = 1
let represent \(ax^2  + bx + c  = (x + l) (x + m)\), where l and m are integers.

\(ax^2  + bx + c  = x^2 + lx + mx + lm                       = x^2 + (l + m )x + lm\)

To factorize an expression of type ax² + bx + c, look for two integers l and m such that their sum is b and product is c.

Example: x²  + 6x + 8

find two integers l and m whose sum is 6 and product is 8.

As 4 + 2 = 6 and 4 × 2 = 8, therefore

\(x^2  + 6x + 8 = x^2 + (4 + 2)x + (4 × 2) \)

x² + 4x + 2x + 8 or x⋅(x+4) + 2⋅(x+4) 

=(x+4)(x+2)

Case 2: If \(a \neq 1\) in ax² + bx + c

find two integers l and m such that

l × m = ac and l + m = b

Example: 3x² − 10x + 8
Find two integers such that l × m = 24 and l + m = −10

Two integers that fulfill these two criteria are −6, −4:  −6 × −4 = 24 and \( −6 + (−4) = −10\)

Therefore, 3x² − 10x + 8 = 3x² − 6x − 4x + 8
= \(3x(x − 2) − 4(x − 2) = (3x − 4) (x − 2)\)