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Example 15 - Chapter 2 Class 9 Polynomials
Last updated at April 16, 2024 by Teachoo
Chapter 2 Class 9 Polynomials
Concept wise
- Definition
- Degree & Coefficient of a polynomial
- Value of a polynomial at a given point
- Verifying Zeroes of a polynomial
- Finding Zeroes of a polynomial
- Remainder Theoram
- Check if factor
- Factorisation by middle term
- Factorisation by factor formula
- Factorizing cubic equation
- Identity I - IV
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Identity V
- Identity VI & VII
- Identity VIII
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
