Factorization by Splitting middle term: Part IV

Factorization by splitting the middle term:- Part IV

There are some expression   which can be factorized by splitting middle term; let us consider this expression

Px2 + qx + r and  Px2 + qxy + ry2

Both are neither a square nor in the term of (x + a)(x + b) = x2 + (a + b)x + ab. Such expressions are factorized by splitting their two middle terms.

viz the middle terms qx of the expression px² + qx + r is split into two terms ax and bx such that their sum is qx and their product is equal to the product of the first and the third term. Let us consider on following example:-

Exp 1: 6x² – 5xy – 6y²

1st step: Get the product of first and third terms

i.e  6x² x 6y² = 36x²y²

2nd step: Split middle terms such as it product would be equal to product of first and third terms.

Here, middle term =  – 5xy

So, we can write it as 4xy – 9xy                     (36x²y²)

Therefore,

6x² – 5xy – 6y² =6x² +4xy – 9xy – 6y²

                         

                            =2x(3x + 2y)(2x – 3y)

Exp 2: 7x² + 23xy + 6y²

Product of the first and the third terms = 42x²Y²

Therefore, the middle term 23xy can be write as

21xy + 2xy

Hence  7x² + 23xy + 6y² = 7x² +21xy + 2xy +6y²

                                  

                                       =7x(x + 3y) + 2y(x + 3y)

                                       = (x +3y)(7x +2y)

Exp 2: 14a² + 11ab – 15b²

Product of the first and the third terms = – 210a²b²

Therefore it can be write as 21ab – 10ab

Hence,  14a² + 11ab – 15b² = 14a²  – 21ab – 10ab – 15b²

                                                 =7a(2a – 3b) – 5b(2a – 3b)

                                                 =(2a – 3b)(7a – 5b)

Exp 4:- x² - 9x + 20 = x²  – 4x – 5x + 20

                                 = x(x – 4) + 5(4 – x)

By taking 5 common from (20 – 5x), we are not getting the expression that is common in both terms. Therefore, we need to take – 5 common from (20 – 5x).

The second step in the above solution will be correctly written as

x²  – 9x + 20 = x(x – 4) – 5(x – 4)

                     = (x – 4)(x – 5)  

Note: it is sometimes necessary to change the order of the terms for getting the expression that is common in terms of the given expression.

Factorization Using Identities:- Part III

There is some more example of Factorizationusing Identities:- Part II. Let us see some example before moving further:-

Examp:-(e.) 75a³ b² – 108ab⁴  

Solution: The given expression is not in the form of any identity. However, its two terms have common factors. Let us rewrite the expression, we have

(3ab² × 25a²) – (3ab² × 36b²) = 3ab² (25a² – 36b²)

                

                                                    = 3ab² {(5a)² – (6b)²}

             

                                                     = 3ab² (5a + 6b)(5a – 6b)

By using identity,            a² – b² = (a +b)(a – b)

Examp:-(f.) 9 – x⁶ + 2x³y³ – y⁶ = 9 – (x⁶ – 2x³y³ + y⁶)

Solution:-

                                                   = 9 - {(x³)² – 2(x)³(y)³ + (y³)²

                                                   = (3)² – (x³ – y³)²

[On substituting x³ = a, y³ = b in the identity a² -2ab + b² = (a – b)², we have

(x³)² – 2(x³)(y³) + (y³)² = (x³ – y³)²]

                                      = {3 + (x³ – y³)}{3 – (x³ – y³)}

Using, a² – b² = (a + b)(a – b)

                                       =(3 + x³ –y³)(3 – x³ + y³)

                  

Factorization by using the identity (x + a) (x +b) = a² + (a + b)x + ab: – ×

An expression of the type x2 +mx +n  is not a perfect square, therefore, cannot be factorized by identities (i) and (ii) used in the preceding topic. This expression also does not fit the identity type (a 2 – b2) either.

However, we can use identity (iv)  (x + a) (x +b) = a² + (a + b)x + ab.

To factorize this type of expression, we need to find two factor a and b of the constant term n such that ab = n and a + b = m (the coefficient of the middle term).

Then, the given expression is factorized in the following way:-

x² +mx + n = x² + (a + b) x + ab

                   = x² + ax + bx + ab = x (x + a) + b(x + a)

                  = (x + a)(x + b)

This is required factorization.

Exmp:-  a. x² + 14x + 45

Solution:

The constant term 45 can be factorized as 45 = 15 × 3 or 9 × 5,

We’ll consider only those factors of 45, whose sum is equal to 14 (the coefficient of middle term).

Hence, factor 15 + 3 is not required as 15 + 3 ≠ 14. We have to take other pair of factors;

 i.e  9 +5 = 14

Therefore, the required factor of 45 is 9 and 5.

Putting these factors in the given expression we have,

x² + 14x + 45 = x² + (9 + 5)x + (9)(5)

                       

                        = x² + 9x + 5x + (9)(5)

                        = x² + 14x +45              

Exmp (b):- x² – 22x -48

Sol:- The constant term 48 can be factorized as

                 48 = 8 × 6 or 12 × 4 or 16 × 3 or 24 × 2

Since, 2 – 24 = -22, 2 and – 24 are the required factors of – 48.

So,  x² – 22x – 48 = x² + (2 – 24)x + (2) (– 24)

                              = (x + 2)(x -24)

By using identities,  (x +a )(x + b) = x² + (a + b)x + ab   

Exmp ©:- x² -13 x + 36

The constant term 36 can be factorized as

36 = 6 × 6 or 9 × 4 or 12 × 3 or 18 × 2

Since, (– 9) + (– 4) = 1 – 13, - 9  and – 4 are required factors of 36.

Hence, x² – 13x + 36 = x² + (- 4 – 9)x + (-9)(-4)

                                     =x2 – 4x – 9x + (- 9) (-4)

                                     = x(x – 4) – 9(x – 4)

                                      = (x – 4)(x – 9), required factorization.

Negative or Positive Factors of the Constant Term:-

In the given expression x² +mx + n, where ab = n and a + b = m, we should follow some rules  regarding positive or negative factors of the constant terms.  The rule is following: -

1. If m is positive and n is also positive, then both the factors a and b of n will be positive.

2. If m is positive and n is negative, then both the factors a and b of n will have opposite signs. It means one factor will be positive and the other will be negative.

3. If m is negative and n is positive, then the factors a and b of n will be negative sign.

4. If m and n are both negative, then the factors a and b of n will have opposite sings.  

Continue..........