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Example 10 - 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
Example 10 Factorise x3 23x2 + 142x 120. Let p(x) = x3 23x2 + 142x 120 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) = ( ( ))/(( 1)) g(x) is obtained after dividing p(x) by x 1 So, g(x) = x2 22x + 120 So, p(x) = (x 1) g(x) = (x 1) (x2 22x + 120) We factorize g(x) i.e. x2 22x + 120 x2 22x + 120 We factorize using the splitting the middle term method = x2 12x 10x + 120 = x(x 12) 10(x 12) = (x 12) (x 10) So, p(x) = (x 1)(x 10)(x 12)
