NCERT Solutions for Class 9 Maths Chapter 2 Polynomials (Ex 2.4)

Exercise 2.4 of Class 9 Maths Chapter 2 Polynomials is all about algebraic identities. An identity is a formula that is true for every value you put into it. Once a student learns these eight identities by heart, long multiplication work becomes very short. For example, instead of multiplying 103 by 107 the normal way, the identity gives the answer in one line. This exercise has 16 questions, and each one is solved on this page in easy steps.

These NCERT (National Council of Educational Research and Training) Solutions follow the latest textbook for the current session. Every question shows which identity is being used and why, so students learn the method and not just the final answer. Teachers can use the same steps on the board, and parents can use them to check homework. A free PDF of the full exercise is also given, so students can practise offline before a test.

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials: Exercise 2.4

Q1. Determine which of the following polynomials has (x + 1) a factor:
(i) x³ + x² + x + 1
(ii) x⁴ + x³ + x² + x + 1
(iii) x⁴ + 3x³ + 3x² + x + 1
(iv) x³ – x² – (2 + √2)x + √2

Answer:
The zero of (x + 1) is –1, because x + 1 = 0 gives x = –1.
By the factor theorem, (x + 1) is a factor of p(x) only if p(–1) = 0.

(i) Let p(x) = x³ + x² + x + 1
p(–1) = (–1)³ + (–1)² + (–1) + 1
= –1 + 1 – 1 + 1
= 0
Since p(–1) = 0, (x + 1) is a factor of x³ + x² + x + 1.

(ii) Let p(x) = x⁴ + x³ + x² + x + 1
p(–1) = (–1)⁴ + (–1)³ + (–1)² + (–1) + 1
= 1 – 1 + 1 – 1 + 1
= 1, which is not 0
So (x + 1) is not a factor of x⁴ + x³ + x² + x + 1.

(iii) Let p(x) = x⁴ + 3x³ + 3x² + x + 1
p(–1) = (–1)⁴ + 3(–1)³ + 3(–1)² + (–1) + 1
= 1 – 3 + 3 – 1 + 1
= 1, which is not 0
So (x + 1) is not a factor of x⁴ + 3x³ + 3x² + x + 1.

(iv) Let p(x) = x³ – x² – (2 + √2)x + √2
p(–1) = (–1)³ – (–1)² – (2 + √2)(–1) + √2
= –1 – 1 + 2 + √2 + √2
= 2√2, which is not 0
So (x + 1) is not a factor of x³ – x² – (2 + √2)x + √2.

Q2. 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
(ii) p(x) = x³ + 3x² + 3x + 1, g(x) = x + 2
(iii) p(x) = x³ – 4x² + x + 6, g(x) = x – 3

Answer:
In each case, first find the zero of g(x), then put that value into p(x).

(i) g(x) = x + 1 = 0 gives x = –1. So the zero of g(x) is –1.
p(–1) = 2(–1)³ + (–1)² – 2(–1) – 1
= –2 + 1 + 2 – 1
= 0
Since p(–1) = 0, g(x) is a factor of p(x).

(ii) g(x) = x + 2 = 0 gives x = –2. So the zero of g(x) is –2.
p(–2) = (–2)³ + 3(–2)² + 3(–2) + 1
= –8 + 12 – 6 + 1
= –1, which is not 0
So g(x) is not a factor of p(x).

(iii) g(x) = x – 3 = 0 gives x = 3. So the zero of g(x) is 3.
p(3) = (3)³ – 4(3)² + (3) + 6
= 27 – 36 + 3 + 6
= 0
Since p(3) = 0, g(x) is a factor of p(x).

Q3. 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
(ii) p(x) = 2x² + kx + √2
(iii) p(x) = kx² – √2x + 1
(iv) p(x) = kx² – 3x + k

Answer:
The zero of (x – 1) is 1. If (x – 1) is a factor of p(x), then p(1) = 0.

(i) p(1) = (1)² + (1) + k = 0
1 + 1 + k = 0
2 + k = 0
k = –2

(ii) p(1) = 2(1)² + k(1) + √2 = 0
2 + k + √2 = 0
k = –(2 + √2)

(iii) p(1) = k(1)² – √2(1) + 1 = 0
k – √2 + 1 = 0
k = √2 – 1

(iv) p(1) = k(1)² – 3(1) + k = 0
k – 3 + k = 0
2k – 3 = 0
k = 3/2

Q4. Factorise:
(i) 12x² – 7x + 1
(ii) 2x² + 7x + 3
(iii) 6x² + 5x – 6
(iv) 3x² – x – 4

Answer:
All four parts are done by splitting the middle term. Find two numbers whose sum is the middle coefficient and whose product is (first coefficient × last term).

(i) 12x² – 7x + 1
Sum needed = –7, product needed = 12 × 1 = 12
The two numbers are –3 and –4, because –3 + (–4) = –7 and (–3)(–4) = 12.
12x² – 7x + 1 = 12x² – 4x – 3x + 1
= 4x(3x – 1) – 1(3x – 1)
= (4x – 1)(3x – 1)

(ii) 2x² + 7x + 3
Sum needed = 7, product needed = 2 × 3 = 6
The two numbers are 6 and 1.
2x² + 7x + 3 = 2x² + 6x + x + 3
= 2x(x + 3) + 1(x + 3)
= (2x + 1)(x + 3)

(iii) 6x² + 5x – 6
Sum needed = 5, product needed = 6 × (–6) = –36
The two numbers are 9 and –4.
6x² + 5x – 6 = 6x² + 9x – 4x – 6
= 3x(2x + 3) – 2(2x + 3)
= (2x + 3)(3x – 2)

(iv) 3x² – x – 4
Sum needed = –1, product needed = 3 × (–4) = –12
The two numbers are –4 and 3.
3x² – x – 4 = 3x² – 4x + 3x – 4
= x(3x – 4) + 1(3x – 4)
= (3x – 4)(x + 1)

Q5. Factorise:
(i) x³ – 2x² – x + 2
(ii) x³ – 3x² – 9x – 5
(iii) x³ + 13x² + 32x + 20
(iv) 2y³ + y² – 2y – 1

Answer:
Each part is a cubic. First find one zero by trial from the factors of the constant term, then divide to get a quadratic, and factorise that quadratic.

(i) Let p(x) = x³ – 2x² – x + 2
The constant term is 2, so try ±1 and ±2.
p(–1) = (–1)³ – 2(–1)² – (–1) + 2
= –1 – 2 + 1 + 2
= 0
So (x + 1) is a factor.
Divide p(x) by (x + 1). The quotient is x² – 3x + 2.
So p(x) = (x + 1)(x² – 3x + 2)
Now factorise the quadratic:
x² – 3x + 2 = x² – x – 2x + 2 = x(x – 1) – 2(x – 1) = (x – 1)(x – 2)
Therefore x³ – 2x² – x + 2 = (x + 1)(x – 1)(x – 2)

(ii) Let p(x) = x³ – 3x² – 9x – 5
The constant term is –5, so try ±1 and ±5.
p(5) = (5)³ – 3(5)² – 9(5) – 5
= 125 – 75 – 45 – 5
= 0
So (x – 5) is a factor.
Divide p(x) by (x – 5). The quotient is x² + 2x + 1.
So p(x) = (x – 5)(x² + 2x + 1)
Now factorise the quadratic:
x² + 2x + 1 = x² + x + x + 1 = x(x + 1) + 1(x + 1) = (x + 1)(x + 1)
Therefore x³ – 3x² – 9x – 5 = (x – 5)(x + 1)(x + 1)

(iii) Let p(x) = x³ + 13x² + 32x + 20
The constant term is 20, so try ±1, ±2, ±4, ±5, ±10 and ±20.
p(–1) = (–1)³ + 13(–1)² + 32(–1) + 20
= –1 + 13 – 32 + 20
= 0
So (x + 1) is a factor.
Divide p(x) by (x + 1). The quotient is x² + 12x + 20.
So p(x) = (x + 1)(x² + 12x + 20)
Now factorise the quadratic:
x² + 12x + 20 = x² + 2x + 10x + 20 = x(x + 2) + 10(x + 2) = (x + 2)(x + 10)
Therefore x³ + 13x² + 32x + 20 = (x + 1)(x + 2)(x + 10)

(iv) Let p(y) = 2y³ + y² – 2y – 1
Try the factors ±1 and ±2.
p(1) = 2(1)³ + (1)² – 2(1) – 1
= 2 + 1 – 2 – 1
= 0
So (y – 1) is a factor.
Divide p(y) by (y – 1). The quotient is 2y² + 3y + 1.
So p(y) = (y – 1)(2y² + 3y + 1)
Now factorise the quadratic:
2y² + 3y + 1 = 2y² + 2y + y + 1 = 2y(y + 1) + 1(y + 1) = (2y + 1)(y + 1)
Therefore 2y³ + y² – 2y – 1 = (y – 1)(2y + 1)(y + 1)

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 2