Factors of Algebraic Expressions - Part II

3. Factors of binomials  
Since we have learnt how to factorize a monomial, factorization of a binomial becomes very simple. We know that a binomial is a sum of two monomials. Factorize the monomials independently and then add them. Common factors from both the monomials can be written on one side and the remaining sum of the two monomials is written in a bracket. As we solve more examples, the concept will become clear.

Example 1 : Factorize 2a³bc  +   4ab

Factors of 2a³bc  : 2 * a * a * a * b * c

Factors of 4ab   : 2 * 2 * a * b

2a³bc  +   4ab  = 2 * a * a * a * b * c    +    2 * 2 * a * b

Common factors in both the factorized monomials is 2ab

Thus   2a³bc  +   4ab  =  2ab (a²c  + 2)

Example 2  :   Factorize abx ñ aby

Factors of abx   : a * b * x

Factors of aby  :  a  *  b * y

Thus abx ñ aby  = a * b * x  - a  *  b * y

The common factor is ab.

Thus factorization of  abx ñ aby  = ab(x-y)

4. Factors of a polynomial by grouping the terms
When one has to factorize the polynomials, it is better to first write the polynomial in the descending order of the powers of the variables, then  group them as binomials. Then factorize each binomial. The final answer would be clear from the examples given below.

Example 1:  Factorize  4x³ + 2x + 2x² +1

4x³ + 2x + 2x² +1  =  4x³ + 2x² + 2x +1

                            =  (4x³ + 2x²) + (2x +1)

In the first binomial (4x³ + 2x²), the common factor is 2x² .

Thus (4x³ + 2x²)  =  2x²   *  ( 2x  + 1)

The second binomial (2x+ 1 ) cannot be factorized as there are no common terms between the two monomials other than 1.

Thus   4x³ + 2x + 2x² +1  =  2x²  *  ( 2x  + 1)  + (2x  + 1)

Now (2x + 1) becomes a common factor.

Thus 4x³ + 2x + 2x² +1  =  ( 2x  + 1)  (2x²  + 1)

        Therefore factorization of  (4x³ + 2x + 2x² +1 )  is   ( 2x  + 1)  (2x²  + 1)
To check if your answer is correct, cross multiply the two brackets and verify.

Example 2 : Factorize 5p² ñ 6pq  + q².

In this example we have to be a little clever and see that the middle term can be written as ñ 5pq ñ  1pq.

 5p² ñ 6pq  + q² =  5p²  ñ 5pq  ñ  1pq   + q²

Now group the binomials (5p²  ñ 5pq )  ñ  ( 1pq ñ q² )

(5p²  ñ  5pq )  =  5p  ( p ñ q )

(1pq  ñ q² )   =  q ( p ñ q )

Thus   5p² ñ 6pq  + q² =  5p (  p  ñ q )  ñ q ( p ñ q )

Thus  5p² ñ 6pq  + q² =  ( p ñ q )  ( 5p  ñ q )

The factorization of the trinomial 5p² ñ 6pq  + q²  is  (p ñ q )  ( 5p ñ q )

Cross multiply the two factors (p ñ q )  ( 5p  ñ q ) and check that you will get back the original trinomial 5p² ñ 6pq  + q².

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