Example 10 - Chapter 2 Class 9 Polynomials

Last updated at August 11, 2026 by

Example 10 - Factorise x3 - 23x2 + 142x - 120 - Class 9 - Examples - Examples

part 2 - Example 10 - Examples - Serial order wise - Chapter 2 Class 9 Polynomials

part 3 - Example 10 - Examples - Serial order wise - Chapter 2 Class 9 Polynomials

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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)

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