Solve all your doubts with Teachoo Black (new monthly pack available now!)
Example 11 - Chapter 2 Class 9 Polynomials (Term 2)
Last updated at Aug. 25, 2021 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 Deleted for CBSE Board 2023 Exams
Example 4
Example 5 Important
Example 6
Example 7 Important Deleted for CBSE Board 2023 Exams
Example 8
Example 9 Important
Example 10
Example 11 Important You are here
Example 12 Important Deleted for CBSE Board 2023 Exams
Example 13 Important
Example 14
Example 15 Important Deleted for CBSE Board 2023 Exams
Example 16 (i)
Example 16 (ii)
Example 17
Example 18 (i) Important Deleted for CBSE Board 2023 Exams
Example 18 (ii)
Example 19
Example 20
Example 21 Important Deleted for CBSE Board 2023 Exams
Example 22 (i)
Example 22 (ii)
Example 23 (i) Important Deleted for CBSE Board 2023 Exams
Example 23 (ii)
Example 24 Important Deleted for CBSE Board 2023 Exams
Example 25 Important
Transcript
Example 11 Examine whether x + 2 is a factor of x3 + 3x2 + 5x + 6 and of 2x + 4. First checking whether x + 2 is a factor of x3 + 3x2 + 5x + 6 Finding remainder when x3 + 3x2 + 5x + 6 is divided by x + 2 Step 1: Put Divisor = 0 x + 2 = 0 x = 2 Step 2: Let p(x) = x3 + 3x2 + 5x + 6 Putting x = 2 p( 2) = ( 2)3 + 3( 2)2 + 5( 2) + 6 = 8 + 12 10 + 6 = 0 Remainder = p( 2) = 0 Since remainder is zero x + 2 is a factor of x3 + 3x2 + 5x + 6 Checking whether x + 2 is a factor of 2x + 4 Finding remainder when 2x + 4 is divided by x + 2 Step 1: Put Divisor = 0 x + 2 = 0 x = 2 Step 2: Let p(x) = 2x + 4 Putting x = 2 p( 2) = 2( 2) + 4 = 4 + 4 = 0 Remainder = p( 2) = 0 Since remainder is zero x + 2 is a factor of 2x + 4
