Factorization | Factorization by grouping

Factorization | Factorization by grouping

The process of finding out factors of an algebraic expression is known as factorization. It is also called the resolution of algebraic expression into its factors.

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Cases of factorization:

Case 1: Factorization of an expression which have got a common factor in all terms:

In this case we can resolve the algebraic expression into two factors, one of which is simple and the other is compound. For example: ax+bx = x(a+b). x is simple factor and a+b is compound factor.

Similarly,

8x²y – 8xy² + 2y³

= 2y(4x² – 4xy + y²)

 

Case 2: Factorization by grouping:

If there are more than two terms in an expression and all of the terms have no common factor except 1, in this case, terms containing like common factors are grouped together. After grouping, the expression is factorized.

For example: In p² – 15q – 5p + 3pq, the first two terms have no common except 1 and last two terms have p as common. So, if we make group of 1^st and 3^rd together and last and 2^nd together we can get (p – 5) as common factor.

So,

            P² – 15q – 5p + 3pq

            = p² – 5p + 3pq – 15q

            = p(p – 5) + 3q(p – 5)

            = (p – 5)(p + 3q)

Workout Examples

Example 1: Factorize the following expression:

a)    ab + ac

b)    -2x² – 4xy

c)    2a²bc – 4ab²c + 8abc²

d)    p(x + y) + q(x + y)

e)    (a + b)(a – b) – 2(a – b)                        

Solution: a) ab + ac

                = a(b + c)

 

               b) -2x² – 4xy

                = -2x(x + 2y)

 

               c) 2a²bc – 4ab²c + 8abc²

                = 2abc(a – 2b + 4c)

 

               d) p(x + y) + q(x + y)

                = (x + y)(p + q)

 

               e) (a + b)(a – b) – 2(a – b)

                = (a – b)(a + b – 2)

 

Example 2: Factorize the following expression:

a)    a² – ab – ac + bc

b)    a² – b(a – c) – ca

c)    ma + mb + na + nb

d)    3a – 3b + ab – b²

e)    ax – bx – ay + by                                    

Solution: a) a² – ab – ac + bc

                = a(a – b) – c(a – b)

                = (a – b)(a – c)

 

               b) a² – b(a – c) - ca

                = a² – ba + bc - ca

                = a² – ba – ca + bc

                = a(a – b) – c(a – b)

                = (a – b)(a – c)

 

               c) ma + mb + na + nb

                = m(a + b) + n(a + b)

                = (a + b)(m + n)

 

               d) 3a – 3b + ab – b²

                = 3(a – b) + b(a – b)

                = (a – b)(3 + b)

 

               e) ax – bx – ay + by

                = x(a – b) – y(a – b)

                = (a – b)(x – y)

Example 3: Factorize the following expression:

a)    xy + x +y + 1

b)    a² + 3b + 3a + ab

c)    p² – 15q – 5p + 3pq

d)    ac² – 2a – bc² + 2b

e)    a³ – a² – ab + a + b - 1                           

Solution: a) xy + x + y +1

                = x(y + 1) + 1(y + 1)

                = (y + 1)(x +1)

 

               b) a² + 3b + 3a + ab

                = a² + ab + 3a + 3b

                = a(a + b) + 3(a + b)

                = (a + b)(a + 3)

 

               c) p² – 15q – 5p + 3pq

                = p² – 5p + 3pq – 15q

                = p(p – 5) + 3q(p – 5)

                = (p – 5)(p + 3q)

 

               d) ac² – 2a – bc² + 2b

                = ac² – bc² – 2a + 2b

                = c²(a – b) – 2(a – b)

                = (a – b)(c² – 2)

 

               e) a³ – a² – ab + a + b - 1

                = a³ – a² – ab + b +a – 1

                = a²(a – 1) – b(a – 1) + 1(a – 1)

                = (a – 1)(a² – b + 1)

You can comment your questions or problems regarding the factorization of algebraic expressions here.