Last updated at May 29, 2018 by Teachoo

Transcript

Example 21 Factorize 4x2 + y2 + z2 – 4xy – 2yz + 4xz. 4x2 + y2 + z2 – 4xy – 2yz + 4xz = 22 x2 + y2 + z2 – 4xy – 2yz + 4xz = (2x)2 + y2 + z2 – 4xy – 2yz + 4xz = (2x)2 + (–y)2 + z2 – 4xy – 2yz + 4xz = (2x)2 + (–y)2 + (z)2 + 2(2x)(–y) + 2(–y)(z) + 2(2x)(z) Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac where a = 2x, b= –y , c = z = (2x + (–y) + z)2 = (2x – y + z)2 = (2x – y + z)(2x – y + z)

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
- Identity VIII

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.