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Example 15 - Factorise x3 - 23x2 + 142x - 120 - Class 9 - Examples

  1. Chapter 2 Class 9 Polynomials
  2. Serial order wise
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Example 15 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) = (𝑝(𝑥))/((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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