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Example 20 - 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
- Identity V
- Identity VI & VII
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Identity VIII
Transcript
Example 20 Factorise : 8x3 + y3 + 27z3 – 18xyz 8x3 + y3 + 27z3 – 18xyz = (2x) 3 + y3 + (3z) 3 – 18xyz = (2x) 3 + y3 + (3z) 3 – 3(2x)(y)(3z) Using a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca) Putting a = 2x, b = y , c = 3z = (2x + y + 3z)[(2x)2 + y2 + (3z) 2 – (2x)(y) – (y)(3z) – (2x)(3z)] = (2x + y + 3z) (4x2 + y2 + 9z2 – 2xy – 3yz – 6xz)
