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Ex 2.3, 5 (i) - Chapter 2 Class 9 Polynomials
Last updated at April 16, 2024 by Teachoo
Factorizing cubic equation
Chapter 2 Class 9 Polynomials
Concept wise
- Definition
- Degree & Coefficient of a polynomial
- Value of a polynomial at a given point
- Verifying Zeroes of a polynomial
- Finding Zeroes of a polynomial
- Remainder Theoram
- Check if factor
- Factorisation by middle term
- Factorisation by factor formula
-
Factorizing cubic equation
- Identity I - IV
- Identity V
- Identity VI & VII
- Identity VIII
Transcript
Ex 2.3, 5 Factorise: (i) x3 − 2x2 − x + 2 Let p(x) = x3 – 2x2 – x + 2 Checking p(x) = 0 So, at x = 1, p(x) = 0 Hence, x – 1 is a factor of p(x) Now, p(x) = (x – 1) g(x) ⇒ g(x) = (𝑝(𝑥))/((𝑥 − 1)) ∴ g(x) is obtained after dividing p(x) by x – 1 So, g(x) = x2 – x – 2 So, p(x) = (x – 1) g(x) = (x – 1) (x2 – x – 2) We factorize g(x) i.e. x2 – x – 2 x2 – x – 2 We factorize using the splitting the middle term method = x2 – 2x + x – 2 = x(x – 2) + 1 (x – 2) = (x + 1) (x – 2) So, p(x) = (x – 1)(x + 1)(x – 2)
