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Example 15 - Factorise x3 - 23x2 + 142x - 120 - Class 9 - Factorizing cubic equation

Example 15 - Chapter 2 Class 9 Polynomials - Part 2
Example 15 - Chapter 2 Class 9 Polynomials - Part 3

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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) = ( ( ))/(( 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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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.