Get live Maths 1-on-1 Classs - Class 6 to 12
Example 15 - Chapter 2 Class 9 Polynomials (Term 2)
Last updated at March 28, 2023 by Teachoo
Examples
Example 1 (i)
Example 1 (ii)
Example 1 (iii) Important
Example 2 (i)
Example 2 (ii)
Example 2 (iii)
Example 3 Important Deleted for CBSE Board 2023 Exams
Example 4
Example 5 Important
Example 6
Example 7 Important Deleted for CBSE Board 2023 Exams
Example 8
Example 9 Important
Example 10
Example 11 Important
Example 12 Important Deleted for CBSE Board 2023 Exams
Example 13 Important
Example 14
Example 15 Important Deleted for CBSE Board 2023 Exams You are here
Example 16 (i)
Example 16 (ii)
Example 17
Example 18 (i) Important Deleted for CBSE Board 2023 Exams
Example 18 (ii)
Example 19
Example 20
Example 21 Important Deleted for CBSE Board 2023 Exams
Example 22 (i)
Example 22 (ii)
Example 23 (i) Important Deleted for CBSE Board 2023 Exams
Example 23 (ii)
Example 24 Important Deleted for CBSE Board 2023 Exams
Example 25 Important
Transcript
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)
