Ex 2.4, 1 (iii) - Chapter 2 Class 9 Polynomials
Last updated at Dec. 16, 2024 by Teachoo
Ex 2.4
Ex 2.4, 1 (i)
You are here
You are here
Chapter 2 Class 9 Polynomials
Serial order wise
- Ex 2.1
- Ex 2.2
- Ex 2.3
-
Ex 2.4
- Examples
- Remainder Theorem
You saved atleast 2 minutes by viewing the ad-free version of this page. Thank you for being a part of Teachoo Black.
Transcript
Ex 2.4, 1 Use suitable identities to find the following products: (iii) (3x + 4) (3x - 5) (3x + 4) (3x – 5) = 3 (𝑥+ 4/3) 3(𝑥− 5/3) = (9) (𝑥+ 4/3) (𝑥− 5/3) Using (x + a) (x + b) = x2 + (a + b) x + ab, Where a = 4/3 , b = (−5)/3 = 9 [𝑥2+(4/3+(−5)/3)𝑥+(4/3)(− 5/3)] = 9 [𝑥2 +((4 − 5))/3 𝑥 −20/9] = 9 [𝑥2 − 1/3 𝑥 −20/9] = 9(𝑥2) −9(1/3)𝑥 −9(20/9) = 9x2 – 3x – 20
