1. Chapter 2 Class 9 Polynomials
2. Serial order wise

Transcript

Ex 2.4, 1 Determine which of the following polynomials has (x + 1) a factor: (i) x3 + x2 + x + 1 Finding remainder when x3 + x2 + x + 1 is divided by x + 1 Step 1: Put Divisor = 0 x + 1 = 0 x = –1 Step 2: Let p(x) = x3 + x2 + x + 1 Putting x = –1 p(–1) = (–1)3 + (–1)2 + (–1) + 1 = –1 + 1 – 1 + 1 = 0 Thus, Remainder = p(–1) = 0 Since remainder is zero, x + 1 is a factor of x3 + x2 + x + 1 Ex 2.4, 1 Determine which of the following polynomials has (x + 1) a factor: (ii) x4 + x3 + x2 + x + 1 Finding remainder when x4 + x3 + x2 + x + 1 is divided by x + 1 Step 1: Put Divisor = 0 x + 1 = 0 x = –1 Step 2: Let p(x) = x4 + x3 + x2 + x + 1 Putting x = –1 p(–1) = (-1)4 + (−1)3 + (−1)2 + (−1) + 1 = 1 – 1 + 1 – 1 + 1 = 1 Thus, Remainder = p(–1) = 1 Since remainder is not zero, x + 1 is not a factor of x4 + x3 + x2 + x + 1 Ex 2.4, 1 Determine which of the following polynomials has (x + 1) a factor: (iii) x4 + 3x3 + 3x2 + x + 1 Finding remainder when x4 + 3x3 + 3x2 + x + 1 is divided by x + 1 Step 1: Put Divisor = 0 x + 1 = 0 x = –1 Step 2: Let p(x) = x4 + 3x3 + 3x2 + x + 1 Putting x = –1 p(–1) = (–1)4 + 3(−1)3 + 3(−1)2 + (−1) + 1 = 1 – 3 + 3 – 1 + 1 = 1 Thus, Remainder = p(–1) = 1 Since remainder is not zero, x + 1 is not a factor of x4 + 3x3 + 3x2 + x + 1 Ex 2.4, 1 Determine which of the following polynomials has (x + 1) a factor: (iv) x3 – x2 – (2 + √2) x + √2 Finding remainder when x3 – x2 – (2 + √2) x + √2 is divided by x + 1 Step 1: Put Divisor = 0 x + 1 = 0 x = –1 Step 2: Let p(x) = x3 – x2 – (2 + √2) x + √2 Putting x = –1 p(–1) = (–1)3 – (–1)2 – (2 + √2) (–1) + √2 = –1 – (1) + (2 + √2)  + √2 = –1 – 1 + 2 + √2 + √2 = –2 + 2 + √2 + √2 = 2√2 Thus, Remainder = p(–1) = 2√2 Since remainder is not zero, x + 1 is not a factor of x3 – x2 – (2 + √2) x + √2

Serial order wise

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can contact him here.