The factorization is an important chapter with wide applications in Algebra and students must be well conversant with the different types of problems to solve any problem in factorization. To start with, the factorization is the process of simplifying an algebraic expression by identifying the individual factors that ultimately combine to form the given expression. Many times, the factorization is very helpful to simplify and solve the problem with identifying and eliminating the factors that are common to both numerator and denominator.

For example, consider 117 ÷ 72. Here the numerator is 117. Now, 117= 3.3.13; and the denominator is 72=3.3.2.2.2
 So 117 ÷ 72=
 
\frac{117}{72}=\frac{3.3.13}{3.3.2.2.2}=\frac{13}{8}

Greatest common factor of 117 and 72 is 9. Similarly we can find Greatest common factor(GCF) of algebraic expressions.

Examples:

1.Find the GCF of 6a2b3 and 9a3b2

2. Find the GCF of 15a2b3c3, 25a3b2c2 and 10a4b5c6

Factorizing Algebraic expressions

Factorization by taking out the common factor

Factorize:

  1. 8xy – 3xyz= xy (8-3z)
  2. 4a2b-6ab2 +8ab = 2ab(2a-3b+4)
  3. 2p ( 3x – 5y) -5q (3x – 5y) = (3x – 5y) ( 2p – 5q)
  4. m(m + n)2 – 3m2n(m + n) = (m + n)(m(m + n) – 3m2n) = (m + n)(m2+mn+3m2n) = m(m + n)(m+n+3mn)

Factorization by grouping

Examples:
1. a2 - b + ab - a = a2 +ab - a - b
                   = a(a + b) - 1(a + b)
                   = (a + b) (a - 1)
2. am + an + bm + bn = a(m + n) + bm + bn
                     = a(m+n)+b(m+n)
                     = (m+n)(a+b)  
3. x4 -x3 -xy + x + y + 1= (x4 -x3)-xy + y + x - 1  
                         = x3(x-1)-y(x-1)+1(x-1)
                         = (x-1)(x3-y+1)

Factoring difference of two squares

Factoring difference of squares, we use the identity

a2 – b2 = (a + b) (a – b)

Examples:

  1. x2 – 121 = (x + 11) (x – 11)
  2. 25x2 – 16 y2 = (5x)2 – (4y)2 = (5x + 4y) (5x – 4y)
  3. 9 – 4(2a-3b)2 = (3)2 – (2(2a-3b))2 = (3-2(2a-3b))(3+2(2a-3b)) = (3-4a+6b) (3+4a-6b)
  4. 36(x + y)3 – 25(x + y) = (x + y)((6(x + y)2 – 52 = (x + y) (6(x + y)+5)(6(x + y)-5) = (x+ y) (6x + 6y + 5) (6x + 6y – 5)

Factorization of Trinomials of the form x2 + (a + b)x + ab

  1. Factorize x2 +8x – 9
In this trinomial a+b = 8 and ab = -9, we need to find factors of -9, sum = 8, we have 9 + (-1)= 8. 
We split the middle term 8x= 9x + (-1)x
x2 +8x - 9 = x2 + 9x + (-1)x - 9
           = x(x + 9) -1(x + 9)
           = (x + 9) (x - 1)
2. Factorize x2 - 5x + 6
x2 - 5x + 6 = x2 + (-3 + -2)x + (-3)(-2)
            = x2 - 5x +6

Factorizing ax2 + bx + c

Splitting the middle term: sum of factors of a.c = b. 
Example 
1. Factorize 2x2 + x -3 
   2(-3) = -6, Splitting the middle term = +3x -2x = x
   2x2 + x - 3 = 2x2 + 3x -2x -3
               = x(2x + 3) -1(2x + 3)
               = (2x + 3) (x - 1)

2. Factorize 3x2 + 7x +2 

  3x2 + 7x +2 = 3x2 + 6x + x +2
              = 3x( x + 2) + 1(x + 2)
              = (x + 2) (3x + 1)

3. Factorize 2y2 -5y -7 

  2y2 -5y -7 = 2y2 - 7y + 2y - 7
             = y(2y-7) +1(2y - 7)
             = (2y - 7) ( y + 1)

Factorize the following expressions

  1. 8p²q + 4pq²- 20pq -10q²
  2. 4p²-12p+9
  3. 16a5b²-36a³b4
  4. p²(p+2) –p(p+2) – 12(p+2)