Class 9 Mathematics
Chapter 2: Polynomials
Exercise 2.3 – Stepwise Solutions
Important Formula Used
Factor Theorem: If p(a) = 0, then (x − a) is a factor of p(x)
Q1. Determine which of the following polynomials has (x + 1) as a factor
Method: Put x = −1
(i) p(x) = x³ + x² + x + 1
p(−1) = −1 + 1 − 1 + 1 = 0
Answer: (x + 1) is a factor
(ii) p(x) = x⁴ + x³ + x² + x + 1
p(−1) = 1 − 1 + 1 − 1 + 1 = 1 ≠ 0
Answer: Not a factor
(iii) p(x) = x⁴ + 3x³ + 3x² + x + 1
p(−1) = 1 − 3 + 3 − 1 + 1 = 1 ≠ 0
Answer: Not a factor
(iv) p(x) = x³ − x² − (2 + √2)x + √2
p(−1) = −1 − 1 + (2 + √2) + √2 = 2√2 ≠ 0
Answer: Not a factor
Q2. Use Factor Theorem
(i) p(x) = 2x³ + x² − 2x − 1, g(x) = x + 1
p(−1) = −2 + 1 + 2 − 1 = 0
Answer: (x + 1) is a factor
(ii) p(x) = x³ + 3x² + 3x + 1, g(x) = x + 2
p(−2) = −8 + 12 − 6 + 1 = −1 ≠ 0
Answer: Not a factor
(iii) p(x) = x³ − 4x² + x + 6, g(x) = x − 3
p(3) = 27 − 36 + 3 + 6 = 0
Answer: (x − 3) is a factor
Q3. Find the value of k if (x − 1) is a factor
Method: Put x = 1
(i) p(x) = x² + x + k
1 + 1 + k = 0 → k = −2
(ii) p(x) = 2x² + kx + √2
2 + k + √2 = 0 → k = −2 − √2
(iii) p(x) = kx² − √2x + 1
k − √2 + 1 = 0 → k = √2 − 1
(iv) p(x) = kx² − 3x + k
k − 3 + k = 0 → 2k = 3 → k = 3/2
Q4. Factorise
(i) 12x² − 7x + 1
= (3x − 1)(4x − 1)
(ii) 2x² + 7x + 3
= (2x + 1)(x + 3)
(iii) 6x² + 5x − 6
= (3x − 2)(2x + 3)
(iv) 3x² − x − 4
= (3x − 4)(x + 1)
Q5. Factorise
(i) x³ − 2x² − x + 2
= (x − 2)(x² − 1)
= (x − 2)(x − 1)(x + 1)
(ii) x³ − 3x² − 9x − 5
= (x + 1)(x² − 4x − 5)
= (x + 1)(x − 5)(x + 1)
(iii) x³ + 13x² + 32x + 20
= (x + 1)(x² + 12x + 20)
= (x + 1)(x + 10)(x + 2)
(iv) 2y³ + y² − 2y − 1
= (2y + 1)(y² − 1)
= (2y + 1)(y − 1)(y + 1)
Final Answers Summary
- Q1: Only (i) has factor (x + 1)
- Q2: (i) and (iii) are factors
- Q3: k = −2, −2 − √2, √2 − 1, 3/2
- Q4 & Q5: Factorised forms shown above