Example 13 - 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
Example 4
Example 5 Important
Example 6
Example 7 Important
Example 8 Deleted for CBSE Board 2022 Exams
Example 9 Important Deleted for CBSE Board 2022 Exams
Example 10
Example 11 Important
Example 12 Important
Example 13 Important You are here
Example 14
Example 15 Important
Example 16 (i)
Example 16 (ii)
Example 17
Example 18 (i) Important
Example 18 (ii)
Example 19
Example 20
Example 21 Important
Example 22 (i)
Example 22 (ii)
Example 23 (i) Important
Example 23 (ii)
Example 24 Important
Example 25 Important Deleted for CBSE Board 2022 Exams
Transcript
Example 13 (Splitting the middle term) Factorise 6x2 + 17x + 5 by splitting the middle term, and by using the Factor Theorem. 6x2 + 17x + 5 We factorize using the splitting the middle term method = 6x2 + 2x + 15x + 5 = 2x(3x + 1) + 5(3x + 1) = (3x + 1) (2x + 5) Example 13 (Factor Theorem) Factorise 6x2 + 17x + 5 by splitting the middle term, and by using the Factor Theorem. Step 1: We check if x2 is multiplied by 1. 6x2 + 17x + 5 = 6 ("x2 + " 17/6 " x + " 5/6) = 6 p(x) Step 2: We factorize p(x) Let "x2 + " 17/6 " x + " 5/6 = (x – a) (x – b) So, ab = Constant term = 5/6 We find factors of 5/6 5/6 = 5/6 × 1 5/6 = 5/3 × 1/2 5/6 = 5/2 × 1/3 So, factors of 5/6 are 1, 5/3 , 5/2, 1/3 , 1/2 We take value of x as 1, 5/3 , 5/2, 1/3 , 1/2 –1, (−5)/3 , (−5)/2, (−1)/3 , (−1)/2 and check value of p(x) i.e. "x2 + " 17/6 " x + " 5/6 Hence, a = (−5)/2 , b = (−1)/3 Now, "x2 + " 17/6 " x + " 5/6 = (x – a) (x – b) = ("x – " ((−5)/2))("x – " ((−1)/3)) = ("x + " 5/2) ("x + " 1/3) So, 6x2 + 17x + 5 = 6 ("x2 + " 17/6 " x + " 5/6) = 6 ("x + " 5/2) ("x + " 1/3) = 6 ((2𝑥 + 5)/2) ((3𝑥 + 1)/3) = (2𝑥 + 5) (3𝑥 + 1)
