Looking for accurate SEBA Class 9 Maths Exercise 2.3 (Polynomials) textual solutions for the 2026–27 academic year? Access complete step-by-step solutions covering the Factor Theorem, determining whether (x + 1) or (x – a) is a factor of a polynomial, finding the value of k, and factorizing quadratic and cubic polynomials to score top marks in your annual exams.
EXERCISE 2.3
1. Determine which of the following polynomials has (x + 1) a factor :
(i) x³ + x² + x + 1
soln: To check if (x + 1) is a factor, put x + 1 = 0 => x = -1.
(i) p(x) = x³ + x² + x + 1
p(-1) = (-1)³ + (-1)² + (-1) + 1
p(-1) = -1 + 1 – 1 + 1 = 0
Since p(-1) = 0, (x + 1) is a factor.
(ii) x⁴ + x³ + x² + x + 1
soln: p(-1) = (-1)⁴ + (-1)³ + (-1)² + (-1) + 1
p(-1) = 1 – 1 + 1 – 1 + 1 = 1 ≠ 0
Since p(-1) ≠ 0, (x + 1) is not a factor.
(iii) x⁴ + 3x³ + 3x² + x + 1
soln: p(-1) = (-1)⁴ + 3(-1)³ + 3(-1)² + (-1) + 1
p(-1) = 1 – 3 + 3 – 1 + 1 = 1 ≠ 0
Since p(-1) ≠ 0, (x + 1) is not a factor.
(iv) x³ – x² – (2 + √2)x + √2
soln: p(-1) = (-1)³ – (-1)² – (2 + √2)(-1) + √2
p(-1) = -1 – 1 + 2 + √2 + √2
p(-1) = 2√2 ≠ 0
Since p(-1) ≠ 0, (x + 1) is not a factor.
2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:
(i) p(x) = 2x³ + x² – 2x – 1, g(x) = x + 1
soln: Put x + 1 = 0 => x = -1
p(-1) = 2(-1)³ + (-1)² – 2(-1) – 1
p(-1) = -2 + 1 + 2 – 1 = 0
By Factor Theorem, g(x) is a factor of p(x).
(ii) p(x) = x³ + 3x² + 3x + 1, g(x) = x + 2
soln: Put x + 2 = 0 => x = -2
p(-2) = (-2)³ + 3(-2)² + 3(-2) + 1
p(-2) = -8 + 12 – 6 + 1 = -1 ≠ 0
By Factor Theorem, g(x) is not a factor of p(x).
(iii) p(x) = x³ – 4x² + x + 6, g(x) = x – 3
soln: Put x – 3 = 0 => x = 3
p(3) = (3)³ – 4(3)² + 3 + 6
p(3) = 27 – 36 + 3 + 6 = 0
By Factor Theorem, g(x) is a factor of p(x).
SEBA Class 9 Maths Exercise 2.3 Solutions
3. Find the value of k, if x – 1 is a factor of p(x) in each of the following cases:
(i) p(x) = x² + x + k
soln: Since (x – 1) is a factor of p(x), p(1) = 0.
(i) p(x) = x² + x + k
p(1) = (1)² + 1 + k = 0
1 + 1 + k = 0
2 + k = 0 => k = -2
(ii) p(x) = 2x² + kx + √2
soln: p(1) = 2(1)² + k(1) + √2 = 0
2 + k + √2 = 0 => k = -(2 + √2)
(iii) p(x) = kx² – √2x + 1
soln: p(1) = k(1)² – √2(1) + 1 = 0
k – √2 + 1 = 0 => k = √2 – 1
(iv) p(x) = kx² – 3x + k
soln: p(1) = k(1)² – 3(1) + k = 0
k – 3 + k = 0
2k – 3 = 0 => k = 3/2
4. Factorise :
(i) 12x² – 7x + 1
= 12x² – 4x – 3x + 1
= 4x(3x – 1) – 1(3x – 1)
= (3x – 1)(4x – 1)
(ii) 2x² + 7x + 3
= 2x² + 6x + x + 3
= 2x(x + 3) + 1(x + 3)
= (x + 3)(2x + 1)
(iii) 6x² + 5x – 6
= 6x² + 9x – 4x – 6
= 3x(2x + 3) – 2(2x + 3)
= (2x + 3)(3x – 2)
(iv) 3x² – x – 4
= 3x² – 4x + 3x – 4
= x(3x – 4) + 1(3x – 4)
= (3x – 4)(x + 1)
(v) 2x² + x – 45
= 2x² + 10x – 9x – 45
= 2x(x + 5) – 9(x + 5)
= (x + 5)(2x – 9)
(vi) y² + 18y + 65
= y² + 13y + 5y + 65
= y(y + 13) + 5(y + 13)
= (y + 13)(y + 5)
(vii) p² + 14p + 13
= p² + 13p + p + 13
= p(p + 13) + 1(p + 13)
= (p + 13)(p + 1)
(viii) -18 + 11x – x²
= -(x² – 11x + 18)
= -(x² – 9x – 2x + 18)
= -[x(x – 9) – 2(x – 9)]
= -[(x – 9)(x – 2)]
= (9 – x)(x – 2)
(ix) 8a² – 22ab + 15b²
= 8a² – 12ab – 10ab + 15b²
= 4a(2a – 3b) – 5b(2a – 3b)
= (2a – 3b)(4a – 5b)
Class 9 Maths Chapter 2 Exercise 2.3 Question Answer SEBA 2026-27
5. Factorise :
(i) x³ – 2x² – x + 2
= x²(x – 2) – 1(x – 2)
= (x – 2)(x² – 1)
= (x – 2)(x – 1)(x + 1)
(ii) x³ – 3x² – 9x – 5
soln: For x = -1: (-1)³ – 3(-1)² – 9(-1) – 5 = -1 – 3 + 9 – 5 = 0
So (x + 1) is a factor.
x³ – 3x² – 9x – 5 = x²(x + 1) – 4x(x + 1) – 5(x + 1)
= (x + 1)(x² – 4x – 5)
= (x + 1)(x² – 5x + x – 5)
= (x + 1)[x(x – 5) + 1(x – 5)]
= (x + 1)(x + 1)(x – 5)
(iii) x³ + 13x² + 32x + 20
soln: For x = -1: (-1)³ + 13(-1)² + 32(-1) + 20 = -1 + 13 – 32 + 20 = 0
So (x + 1) is a factor.
x³ + 13x² + 32x + 20 = x²(x + 1) + 12x(x + 1) + 20(x + 1)
= (x + 1)(x² + 12x + 20)
= (x + 1)(x² + 10x + 2x + 20)
= (x + 1)[x(x + 10) + 2(x + 10)]
= (x + 1)(x + 2)(x + 10)
(iv) 2y³ + y² – 2y – 1
= y²(2y + 1) – 1(2y + 1)
= (2y + 1)(y² – 1)
= (2y + 1)(y – 1)(y + 1)
(v) x³ + x² – x – 1
= x²(x + 1) – 1(x + 1)
= (x + 1)(x² – 1)
= (x + 1)(x – 1)(x + 1)
(vi) x³ + x² + x + 1
= x²(x + 1) + 1(x + 1)
= (x + 1)(x² + 1)
(vii) x³ + 2x² – x – 2
= x²(x + 2) – 1(x + 2)
= (x + 2)(x² – 1)
= (x + 2)(x – 1)(x + 1)
(viii) x³ + 3x² – 7x – 6
soln: For x = 2: (2)³ + 3(2)² – 7(2) – 6 = 8 + 12 – 14 – 6 = 0
So (x – 2) is a factor.
x³ + 3x² – 7x – 6 = x²(x – 2) + 5x(x – 2) + 3(x – 2)
= (x – 2)(x² + 5x + 3)
(ix) 3x³ + 5x² – 16x – 2
soln: For x = 2: 3(2)³ + 5(2)² – 16(2) – 2 = 24 + 20 – 32 – 2 = 10 ≠ 0
For x = -1: 3(-1)³ + 5(-1)² – 16(-1) – 2 = -3 + 5 + 16 – 2 = 16 ≠ 0
For x = -3: 3(-3)³ + 5(-3)² – 16(-3) – 2 = -81 + 45 + 48 – 2 = 10 ≠ 0
(Note: This polynomial does not have rational roots, retaining original factorized form)
= 3x³ + 5x² – 16x – 2
Factor Theorem Class 9 SEBA Ex 2.3 Solutions
6. If x + a is a common factor of the polynomials x² + px + q and x² + mx + n then show that a = (q – n)/(p – m)
soln: Let f(x) = x² + px + q and g(x) = x² + mx + n.
Since (x + a) is a factor, x = -a makes both equal to zero:
f(-a) = (-a)² + p(-a) + q = 0 => a² – pa + q = 0 — (1)
g(-a) = (-a)² + m(-a) + n = 0 => a² – ma + n = 0 — (2)
Equating (1) and (2):
a² – pa + q = a² – ma + n
-pa + q = -ma + n
ma – pa = q – n
a(m – p) = q – n
a(p – m) = -(q – n)
a = (q – n)/(p – m)
Hence proved.
7. For factorization of 2x² + x – 45 the steps of factorization are given below randomly
(i) 2x(x – 5) + 9(x – 5)
(ii) 2x² – (10 – 9)x – 45
(iii) (x – 5)(2x + 9)
(iv) 2x² – 10x + 9x – 45
Choose the correct order of the steps —
(a) (ii) -> (i) -> (iii) -> (iv)
(b) (i) -> (iv) -> (ii) -> (iii)
(c) (ii) -> (iv) -> (i) -> (iii)
(d) (i) -> (ii) -> (iv) -> (iii)
soln: 2x² + x – 45
= 2x² + (10 – 9)x – 45 (Step ii)
= 2x² + 10x – 9x – 45 (Step iv)
= 2x(x + 5) – 9(x + 5)
Sequence: (ii) -> (iv) -> (i) -> (iii)
Answer: (c) (ii) -> (iv) -> (i) -> (iii)
SEBA Class 9 Maths Chapter 2 Exercise 2.3 Textbook Answers
8. The following question contains Assertion (A) and Reason (R) .
Choose the correct choice.
Assertion (A) : If a ≠ 0 and ax² + bx + a is exactly divisible by (x – a), then a² + b + 1 = 0
Reason (R) : If (x – a) is a factor of a polynomial f(x), then f(a) = 0
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
soln: For Assertion (A):
Let f(x) = ax² + bx + a. Since (x – a) is a factor, f(a) = 0.
a(a)² + b(a) + a = 0
a³ + ab + a = 0
Dividing by a (a ≠ 0):
a² + b + 1 = 0
(Assertion A is true).
Reason (R) states the Factor Theorem, which is true and correctly explains Assertion (A).
Answer: (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
9. If (x + 1) and (x – 1) are the factors of P(x) = ax³ + bx² + cx + d then
(a) a + d = 0
(b) b + d = 0
(c) b + c = 0
(d) a + b = 0
soln: Given P(x) = ax³ + bx² + cx + d
Since (x + 1) is a factor:
P(-1) = a(-1)³ + b(-1)² + c(-1) + d = 0
-a + b – c + d = 0 — (1)
Since (x – 1) is a factor:
P(1) = a(1)³ + b(1)² + c(1) + d = 0
a + b + c + d = 0 — (2)
Adding (1) and (2):
2b + 2d = 0
2(b + d) = 0 => b + d = 0
Answer: (b) b + d = 0
Class 9 Maths Exercise 2.3 Factorize Quadratic Cubic Polynomials
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