1. Chapter 2 Class 9 Polynomials (Term 2)
  2. Serial order wise
  3. Ex 2.5

Ex 2.5, 1 (iii) - Chapter 2 Class 9 Polynomials (Term 2)

Last updated at Aug. 25, 2021 by

Ex 2.5, 1 - Chapter 2 Class 9 Polynomials - Part 3

  1. Chapter 2 Class 9 Polynomials (Term 2)
  2. Serial order wise

    3. Ex 2.5

    ExamplesΒ β†’

Transcript

Ex 2.5, 1 Use suitable identities to find the following products: (iii) (3x + 4) (3x - 5) (3x + 4) (3x – 5) = 3 (π‘₯+ 4/3) 3(π‘₯βˆ’ 5/3) = (9) (π‘₯+ 4/3) (π‘₯βˆ’ 5/3) Using (x + a) (x + b) = x2 + (a + b) x + ab, Where a = 4/3 , b = (βˆ’5)/3 = 9 [π‘₯2+(4/3+(βˆ’5)/3)π‘₯+(4/3)(βˆ’ 5/3)] = 9 [π‘₯2 +((4 βˆ’ 5))/3 π‘₯ βˆ’20/9] = 9 [π‘₯2 βˆ’ 1/3 π‘₯ βˆ’20/9] = 9(π‘₯2) βˆ’9(1/3)π‘₯ βˆ’9(20/9) = 9x2 – 3x – 20

Chapter 2 Class 9 Polynomials (Term 2)
Serial order wise

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