In this article, we will learn about factorization using various topicIdentities or various formulas for finding the factors of algebraic expressions. 

(The symbol “^” here signifies the power of a number or raise to )

(a + b)^2 = a^2 + 2ab + b^2

Whenever the numbers having a perfect square are given, and an addition sign is present in the question, we will use this topicIdentity of (a + b )^2 to solve an algebraic expression. 

       Solve: 4x^2 + 20xz + 25z^2

⦁   = (2x)^2  + 2 (2x) (5z) + (5z)^2. This is of the form (a + b) ^ 2 = a^2 + 2ab + b^2

                 = (2x + 5z) ^2

(a - b) ^ 2 = a^2 - 2ab +b^2

Whenever the numbers having a perfect square are given and subtraction sign is present in the question, at that time we use this topicIdentity to solve the algebraic expression. 

             Solve: 49a^2b^2 - 112abz + 64z^2

               = (7ab) ^2 - 2(7ab) (8z) + (8z)^2. This is of the form (a - b) ^ 2 = a^2 - 2ab +b^2

               = (7ab - 8z) ^2 

(a - b) (a+ b) = (a^2 - b^2) 

Suppose algebraic expression with addition sign (contained within the bracket) such as (2h + k) is multiplied with algebraic expression with subtraction sign (contained within the bracket) such as  (2h - k), then we will use the below technique. 

 (2h - k) (2h + k) 

              = (2h)^2  -  (k)^2

              = 4h^2 - k^2

Note: Here we don’t use the above formula, as we have the same 2h and k in both brackets.

(x+ a) (x+ b) = x^2 + xb + xa + ab 

                                = x^2 + (a + b ) x + ab

Solve: z^2 + 7z + 12

Here let us begin by checking the constant or the last number 12, it is not a perfect square. 

Thus, we have to first multiply the coefficient of the first term, here 1 by the last term 12. 

So, we get, 12 x 1 =12.

 The next step is now to pick two numbers, which are factors of 12. You will have to select them in such a way that when the factors are multiplied, you shall get 12 and when the factors are added, you will get the answer as 7.

Let us choose the factors as 4 and 3, which are the factors of 12.

Here, 4 x 3 = 12 and 4 + 3 =7. SO, we can use factors 4 and 3 in the factorization method.

Now, z^2 + 7z + 12 can be rewritten as:

          z^2 + 4z + 3z + 12 

          ————-    ————

Now we have to find the common factors between these two pairs of numbers. That is, take the variable “z” as a common variable out in the first bracket and three as a common number from the second bracket.

          =  z *( z + 4) + 3 * ( z +  4).

  Now, (z+ 4) is common here, so we can rewrite this as: 

           (z+4) (z+3) 

Hence, factors of z^2 + 7z + 12 = (z+4) (z+3)  

Following are the examples which will show how we will use these topicIdentities for finding factors of algebraic expressions. We have to observe the algebraic expression and we have to use the topicIdentities or the formulas accordingly to find the factors easily. 

Problem sum 1:  Factorize t^4 - 121

Solution: 

= t^4 – 121.

 We note t^4= (t^2)^2    And 121 = (11)^2.

 Observing this algebraic expression resembles the third topicIdentity or the third formula, we get: 

(t^2)^2  - (11)^2 

= (t^2 - 11) (t^2 + 11)

Therefore, the factors are (t^2 - 11) (t^2 + 11) 

Problem sum 2: Factorize x^2 - 4xv + 4v^2 - y^2 

Solution: 

= x^2 - 4xv + 4v^2 - y^2 

————————    ——-

The first three terms given are in the form of the second topicIdentity, and after solving it we have to use the third topicIdentity to find all the factors of the given algebraic expression.

= (x^2) - 2(2x) (v) + (2v) ^2   - y^2

We can rearrange the above algebraic equation in the following manner to compare it with other factorization topicIdentities.

{(x^2) - 2(2x) (v) - y^2} + (2v) ^2  

Here {(x^2) - 2(2x) (v) - y^2} is similar to (a - b) ^ 2 = a^2 - 2ab +b^2

= (x - 2v)^2 - y^2

Now, let us suppose that (x - 2v) = m 

= m^2 - y^2 

Here m^2 - y^2 is similar to the topicIdentity (a^2 - b^2) = (a - b) (a+ b).

Thus, m^2 - y^2 = (m- y ) (m + y )  ……..(1)

Now, putting the value of m in the equation, we get.

= (x - 2v - y ) ( x - 2v + y ) 

Therefore factors of the algebraic expression is ( x - 2v -y ) ( x - 2v + y ) 

Problem Sum 3: Factorize 12a^2 - 11a - 15 

Solution: 

Here 15 is not a perfect square. So, we will multiply the coefficient of the first term and last term. 12 X 15 = 180.

Now, we have to find two factors of 180 such that when we multiply those two numbers we must get 180, and if we add, we should get -11.

So 20 X 9 = 180 and also -20 + 9 = -11.

= 12a^2 -20a + 9a - 15

——————  ———-

Now, we have to find the common factors between these two pairs of algebraic terms.

= 4a (3a- 5) + 3 (3a - 5)

Here (3a - 5) is common Factor 

= (3a - 5) (4a + 3) 

So, factors for 12a^2 -11a - 15 are (3a -5) & (4a + 3)