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Lesson 3.1 factorisation

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A V Prakasam

The document discusses different methods for factorizing quadratic expressions. It explains that the traditional method relies on multi-step processes, while the Vedic method makes factorization simpler using two sub-sutras called "Anurupyena" and "Adyamadyenantyamantyena", which mean "proportionately" and "the first by the first and the last by the last". This Vedic method splits the middle term of a quadratic expression into two parts proportionately so that the factors can be determined directly from the coefficients and terms of the expression.

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Factorisation of Simple Quadratics Factorisation of Simple Quadratcis Example 1: Factorise x2+7x+10 When the coefficient of x2 is 1, factorisation is easy even according to the current system. Split the middle coefficient into two parts so that their sum is the middle coefficient and their product is the absolute term. In this example split 7 into 5 and 2 sothat 5+2=7 and 5X2=10. Then the facotrs can be written as (x+2)(X+5) But in the case of quadratics whose coefficent of x2 is not unit, the students depend upon a multi-step method. For example: Example 2: Factorise 2x2 +5x+2 2x2 +5x+2 = 2x2+4x+x+2 = 2x(x+2)+1(x+2) = (2x+1)(x+2) The vedic system, however, makes the factorisation symple through two small sub-sutras – Anurupyena and Adyamadyenantyamantyena, which mean ‘proportionately’ and ‘the first by the first and the last by the last’. This will be explained in the next slide.
Factorisation of Simple Quadratics Consider the previous example – 2x2+5x+2 (i) Split the middle quotient, 5x, into two parts so that the coefficient of the first part is the same as the ratio of the second part to the last quotient. In other words split 5 into 4 and 1 so that the ration2:4 is the same as the ratio1:2. now this ratio 1:2 is one factor which is x+s (ii) The second factor is obtained by dividing the first quotient of the quadratic, i.e. 2x 2, by the first quotient of the factor already found i.e. x, giving 2x and the last quotient of the quadratic i.e. 2 by the last quotient of the factor already found i.e. by 2 giving 1. The second factory is then 2x+1 The following additional examples would be found useful; Examples: 1. 2x2-5x-3 = (x+3)(2x-1) 2. 2x2+7x+5 = (x+1)(2x+5) 3. 2x2+9x+10 = (x+2)(2x+5) 4. 2x2-5x-3 = (x-3)(2x+1) 5. 9x2-15x+4 = (3x-1)(3x-4)

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Lesson 3.1 factorisation

  • 1. Factorisation of Simple Quadratics Factorisation of Simple Quadratcis Example 1: Factorise x2+7x+10 When the coefficient of x2 is 1, factorisation is easy even according to the current system. Split the middle coefficient into two parts so that their sum is the middle coefficient and their product is the absolute term. In this example split 7 into 5 and 2 sothat 5+2=7 and 5X2=10. Then the facotrs can be written as (x+2)(X+5) But in the case of quadratics whose coefficent of x2 is not unit, the students depend upon a multi-step method. For example: Example 2: Factorise 2x2 +5x+2 2x2 +5x+2 = 2x2+4x+x+2 = 2x(x+2)+1(x+2) = (2x+1)(x+2) The vedic system, however, makes the factorisation symple through two small sub-sutras – Anurupyena and Adyamadyenantyamantyena, which mean ‘proportionately’ and ‘the first by the first and the last by the last’. This will be explained in the next slide.
  • 2. Factorisation of Simple Quadratics Consider the previous example – 2x2+5x+2 (i) Split the middle quotient, 5x, into two parts so that the coefficient of the first part is the same as the ratio of the second part to the last quotient. In other words split 5 into 4 and 1 so that the ration2:4 is the same as the ratio1:2. now this ratio 1:2 is one factor which is x+s (ii) The second factor is obtained by dividing the first quotient of the quadratic, i.e. 2x 2, by the first quotient of the factor already found i.e. x, giving 2x and the last quotient of the quadratic i.e. 2 by the last quotient of the factor already found i.e. by 2 giving 1. The second factory is then 2x+1 The following additional examples would be found useful; Examples: 1. 2x2-5x-3 = (x+3)(2x-1) 2. 2x2+7x+5 = (x+1)(2x+5) 3. 2x2+9x+10 = (x+2)(2x+5) 4. 2x2-5x-3 = (x-3)(2x+1) 5. 9x2-15x+4 = (3x-1)(3x-4)