Finding Situational Factors

The process of multiplying two polynomial in the simplest form is the definition of situation of factorization. The product of functions containg the x in the term for example x+2 and x+5.This function will be multiplied we get polynomial in product.

Example, x + 3 and x – 3 are factors of x2 – 9 this is the situation of factor.

Here x2 – 9 is a second degree polynomial whereas x + 3 and x – 3 are first degree polynomials.

Thus factorization is useful in simplifying expressions.We see the steps to solve situational factors definition using formulaes and common factors Steps for Finding Situational Factors Using Definition

Step 1: (Finding a common factor) When the terms of an algebraic expression A have a common factor B, we divide each term of A by B and get an expression C. Now, A is factored as B × C in the situation of factors.

Ex : 1 Factorize 6x4y3 – 4x2y2 + 10xy3.

Solution : We observe that 2xy2 is a common factor .

Therfore 6x4y3 – 4x2y2 + 10xy3 is divided by 2xy2

2xy2{(`(6x^4y^3)/(2xy^2) - (4x^2y^2)/(2xy^2) - (10xy^3)/(2xy^2)` )}

= 2xy2(3x3y – 2x + 5y).

Step 2: (Grouping the terms) When the terms of an algebraic expression do not have a common factor, the terms may be grouped in an appropriate manner and a common factor is determined in the situation of factors.

Ex : 2 Factorize x2 – 2xy – x + 2y.

Solution: The terms of the expression do not have a common factor. However, we observe that the terms can be grouped as follows:

x2 – 2xy – x + 2y = (x2 – 2xy) – (x – 2y) = x(x – 2y) + (–1) (x – 2y)

= (x – 2y) [x + (–1)] = (x – 2y) (x – 1).

Ex : 3 Factorize 6x5y2 + 6x4y3 + 9x2y4 + 9xy5.

Solution : Applying both step 1 and step 2, we have explain in the situation of factors.

6x5y2 + 6x4y3 + 9x2y4 + 9xy5 = 3xy2(2x4 + 2x3y + 3xy2 + 3y3)

= 3xy2 [(2x4 + 3xy2) + (2x3y + 3y3)] = 3xy2 [x(2x3 + 3y2) + y(2x3 + 3y2)]

= 3xy2 (2x3 + 3y2) (x + y). Definitions of Algebraic Fomula Used to Find Situational Factors

We use factorization formulae. These factorization formulae are obtained from the product formulae. We have learnt that the product formulae are in the situation of factors

(i) (X + Y)2 = X 2 + 2XY + Y 2

(ii) (X – Y)2 = X 2 – 2XY + Y 2

(iii) (X + Y)(X–Y) = X2 – Y 2

(iv) ( X + Y) (X 2 – XY + Y 2) = X 3 + Y 3

(v) (X – Y)(X 2 + XY + Y 2) = X 3 – Y 3

(vi) (X + Y)3 = X 3 + Y 3 + 3X 2Y + 3XY 2 = X 3 + Y 3 + 3X Y (X +Y )

Ex: 4 Factorization using X 2 + 2XY + Y 2 = (X + Y)2

Solution: The given expression can be written as follows:

9x2+ 12xy + 9y2 = (3x)2 + 2(2x)(3y) + (3y)2

Setting X = 3x, Y = 3y, the R.H.S. is X 2 + 2XY + Y 2 and so it is factored as (X + Y)2. Hence we get 9x2 + 12xy + 9y2 = (2x + 3y)2.

Ex : 5 Factorization using X 2 – 2XY + Y 2 = (X – Y)2

Solution : The given equation can be written as follows by situation of factors:

p2 – 18pq + 81q2 = p2 – 2(p)(9q) + (9q)2

Setting X = p and Y = 9q, the R.H.S. is X 2 – 2XY + Y 2 and so it is factorised as (X – Y)2. Hence we get p2– 18pq + 81q2 = (p – 9q)2.

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