Example 20 - Chapter 2 Class 9 Polynomials (Term 2)
Last updated at May 29, 2018 by
Transcript
Example 20 Expand (4a – 2b – 3c)2. (4a – 2b – 3c)2 = (4a + (–2b) + (–3c)) 2 Using (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx Putting x = 4a, y = -2b , z = -3c = (4a)2 + (–2b)2 + (–3c)2 + 2(4a)(–2b) + 2(–2b)(–3c) + 2(–3c)(4a) = 16a2 + 4b2 + 9c2 – 16ab + 12bc – 24ac
Examples
Example 1 (i)
Example 1 (ii)
Example 1 (iii) Important
Example 2 (i)
Example 2 (ii)
Example 2 (iii)
Example 3 Important
Example 4
Example 5 Important
Example 6
Example 7 Important
Example 8 Deleted for CBSE Board 2022 Exams
Example 9 Important Deleted for CBSE Board 2022 Exams
Example 10
Example 11 Important
Example 12 Important
Example 13 Important
Example 14
Example 15 Important
Example 16 (i)
Example 16 (ii)
Example 17
Example 18 (i) Important
Example 18 (ii)
Example 19
Example 20 You are here
Example 21 Important
Example 22 (i)
Example 22 (ii)
Example 23 (i) Important
Example 23 (ii)
Example 24 Important
Example 25 Important Deleted for CBSE Board 2022 Exams
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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.