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Example 15 - Chapter 2 Class 9 Polynomials
Last updated at May 29, 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
Example 4
Example 5 Important
Example 6 Important
Example 7 Important
Example 8 Important
Example 9
Example 10 Important
Example 11 (i)
Example 11 (ii)
Example 12
Example 13 (i) Important
Example 13 (ii)
Example 14
Example 15 You are here
Example 16 Important
Example 17 (i)
Example 17 (ii)
Example 18 (i) Important
Example 18 (ii)
Example 19 Important
Example 20 Important
Question 1 Deleted for CBSE Board 2024 Exams
Question 2 Important Deleted for CBSE Board 2024 Exams
Question 3 Deleted for CBSE Board 2024 Exams
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 5 Deleted for CBSE Board 2024 Exams
Chapter 2 Class 9 Polynomials
Serial order wise
- Ex 2.1
- Ex 2.2
- Ex 2.3
- Ex 2.4
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Examples
- Remainder Theorem
Transcript
Example 15 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
