Examples
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 Important
Example 7 Important
Example 8 Important You are here
Example 9
Example 10 Important
Example 11 (i)
Example 11 (ii)
Example 12
Example 13 (i) Important
Example 13 (ii)
Example 14
Example 15
Example 16 Important
Example 17 (i)
Example 17 (ii)
Example 18 (i) Important
Example 18 (ii)
Example 19 Important
Example 20 Important
Question 1 Deleted for CBSE Board 2025 Exams
Question 2 Important Deleted for CBSE Board 2025 Exams
Question 3 Deleted for CBSE Board 2025 Exams
Question 4 Important Deleted for CBSE Board 2025 Exams
Question 5 Deleted for CBSE Board 2025 Exams
Examples
Last updated at April 16, 2024 by Teachoo
Example 8 (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 8 (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)