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

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

Transcript

Ex 2.5, 1 Use suitable identities to find the following products: (iii) (3x + 4) (3x - 5) (3x + 4) (3x โ€“ 5) = 3 (๐‘ฅ+ 4/3) 3(๐‘ฅโˆ’ 5/3) = (9) (๐‘ฅ+ 4/3) (๐‘ฅโˆ’ 5/3) Using (x + a) (x + b) = x2 + (a + b) x + ab, Where a = 4/3 , b = (โˆ’5)/3 = 9 [๐‘ฅ2+(4/3+(โˆ’5)/3)๐‘ฅ+(4/3)(โˆ’ 5/3)] = 9 [๐‘ฅ2 +((4 โˆ’ 5))/3 ๐‘ฅ โˆ’20/9] = 9 [๐‘ฅ2 โˆ’ 1/3 ๐‘ฅ โˆ’20/9] = 9(๐‘ฅ2) โˆ’9(1/3)๐‘ฅ โˆ’9(20/9) = 9x2 โ€“ 3x โ€“ 20

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