NCERT Solutions Class 9 Chapter 2 Polynomials Ex 2.3

Polynomials Class 9 Exercise 2.3 focuses on algebraic operations like adding, subtracting, and multiplying polynomials. This will help students understand expressions including factorisation and algebraic identities. These NCERT solutions explain each method in a clear and student-friendly manner. 

Class 9 Chapter 2 Polynomials Exercise 2.3 Questions

Class 9 Chapter 2 Polynomials Exercise 2.3 Questions include operations such as adding polynomials, subtracting expressions, and multiplying binomials or monomials with polynomials. Below are the Polynomials Class 9 Exercise 2.3 NCERT Solutions:

1. Determine which of the following polynomials has (x + 1) a factor:

(i) x 3 +x 2 +x+1

Solution: Let p(x) = x 3 +x 2 +x+1

The zero of x+1 is -1.
[x+1 = 0 means x = -1] p(−1)
= (−1) 3 +(−1) 2 +(−1)+1
= −1+1−1+1 = 0
∴ By factor theorem, x+1 is a factor of x 3 +x 2 +x+1

(ii) x 4 +x 3 +x 2 +x+1

Solution: Let p(x)= x 4 +x 3 +x 2 +x+1
The zero of x+1 is -1. [x+1= 0 means x = -1] p(−1)
= (−1) 4 +(−1) 3 +(−1) 2 +(−1)+1
= 1−1+1−1+1 = 1 ≠ 0
∴ By factor theorem, x+1 is not a factor of x 4 + x 3 + x 2 + x + 1

(iii) x 4 +3x 3 +3x 2 +x+1

Solution: Let p(x)= x 4 +3x 3 +3x 2 +x+1
The zero of x+1 is -1. p(−1)=(−1) 4 +3(−1) 3 +3(−1) 2 +(−1)+1 =1−3+3−1+1 =1 ≠ 0
∴ By factor theorem, x+1 is not a factor of x 4 +3x 3 +3x 2 +x+1

(iv) x 3 – x 2 – (2+√2)x +√2

Solution: Let p(x) = x 3 –x 2 –(2+√2)x +√2
The zero of x+1 is -1. p(−1)
= (-1) 3 –(-1) 2 –(2+√2)(-1) + √2
= −1−1+2+√2+√2 = 2√2 ≠ 0
∴ By factor theorem, x+1 is not a factor of x 3 –x 2 –(2+√2)x +√2

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) = 2x 3 +x 2 –2x–1, g(x) = x+1

Solution: p(x) = 2x 3 +x 2 –2x–1, g(x)
= x+1 g(x) = 0 ⇒ x+1 = 0 ⇒ x = −1
∴ Zero of g(x) is -1. Now, p(−1)
= 2(−1) 3 +(−1) 2 –2(−1)–1 = −2+1+2−1 = 0
∴ By factor theorem, g(x) is a factor of p(x).

(ii) p(x)=x 3 +3x 2 +3x+1, g(x) = x+2

Solution: p(x) = x 3 +3x 2 +3x+1, g(x)
= x+2 g(x) = 0 ⇒ x+2 = 0
⇒ x = −2 ∴ Zero of g(x) is -2.
Now, p(−2) = (−2) 3 +3(−2) 2 +3(−2)+1 = −8+12−6+1 = −1 ≠ 0
∴ By factor theorem, g(x) is not a factor of p(x).

(iii) p(x)=x 3 –4x 2 +x+6, g(x) = x–3

Solution: p(x) = x 3
–4x 2 +x+6, g(x) = x -3 g(x) = 0
⇒ x−3 = 0 ⇒ x = 3
∴ Zero of g(x) is 3. Now, p(3) = (3) 3 −4(3) 2 +(3)+6 = 27−36+3+6 = 0
∴ By factor theorem, g(x) is a factor of p(x).

3. Find the value of k, if x–1 is a factor of p(x) in each of the following cases:

(i) p(x) = x 2 +x+k

Solution: If x-1 is a factor of p(x), then p(1) = 0
By Factor Theorem ⇒ (1) 2 +(1)+k = 0
⇒ 1+1+k = 0 ⇒ 2+k = 0
⇒ k = −2

(ii) p(x) = 2x 2 +kx+ √2

Solution: If x-1 is a factor of p(x), then p(1) = 0
⇒ 2(1) 2 +k(1)+√2 = 0
⇒ 2+k+√2 = 0
⇒ k = −(2+√2)

(iii) p(x) = kx 2 – √ 2x+1

Solution: If x-1 is a factor of p(x), then p(1)=0
By Factor Theorem ⇒ k(1) 2 -√2(1)+1=0
⇒ k = √2-1

(iv) p(x)=kx 2 –3x+k

Solution: If x-1 is a factor of p(x), then p(1) = 0
By Factor Theorem ⇒ k(1) 2 –3(1)+k = 0
⇒ k−3+k = 0
⇒ 2k−3 = 0
⇒ k= 3/2

4. Factorise:

(i) 12x 2 –7x+1

Solution: Using the splitting the middle term method, We have to find a number whose sum = -7 and product =1×12 = 12
We get -3 and -4 as the numbers
[-3+-4=-7 and -3×-4 = 12]
12x 2 –7x+1
= 12x 2 -4x-3x+1
= 4x(3x-1)-1(3x-1)
= (4x-1)(3x-1)

(ii) 2x 2 +7x+3

Solution: Using the splitting the middle term method, We have to find a number whose sum = 7 and product = 2×3 = 6
We get 6 and 1 as the numbers [6+1 = 7 and 6×1 = 6]
2x 2 +7x+3 = 2x 2 +6x+1x+3
= 2x (x+3)+1(x+3) = (2x+1)(x+3)

(iii) 6x 2 +5x-6

Solution: Using the splitting the middle term method, We have to find a number whose sum = 5 and product = 6×-6 = -36
We get -4 and 9 as the numbers [-4+9 = 5 and -4×9 = -36]
6x 2 +5x-6
= 6x 2 +9x–4x–6
= 3x(2x+3)–2(2x+3)
= (2x+3)(3x–2)

(iv) 3x 2 –x–4

Solution: Using the splitting the middle term method, We have to find a number whose sum = -1 and product = 3×-4 = -12
We get -4 and 3 as the numbers [-4+3 = -1 and -4×3 = -12]
3x 2 –x–4 = 3x 2 –4x+3x–4
= x(3x–4)+1(3x–4) = (3x–4)(x+1)

5. Factorise:

(i) x 3 –2x 2 –x+2

Solution: Let p(x) = x 3 –2x 2 –x+2
Factors of 2 are ±1 and ± 2 Now, p(x) = x 3 –2x 2 –x+2 p(−1) = (−1) 3 –2(−1) 2 –(−1)+2 = −1−2+1+2 = 0 Therefore, (x+1) is the factor of p(x)
Now, Dividend = Divisor × Quotient + Remainder (x+1)(x 2 –3x+2) = (x+1)(x 2 –x–2x+2) = (x+1)(x(x−1)−2(x−1)) = (x+1)(x−1)(x-2)

(ii) x 3 –3x 2 –9x–5

Solution: Let p(x) = x 3 –3x 2 –9x–5 Factors of 5 are ±1 and ±5 By the trial method, we find that p(5) = 0 So, (x-5) is factor of p(x)
Now, p(x) = x 3 –3x 2 –9x–5 p(5) = (5) 3 –3(5) 2 –9(5)–5 = 125−75−45−5 = 0 Therefore, (x-5) is the factor of  p(x)
Now, Dividend = Divisor × Quotient + Remainder (x−5)(x 2 +2x+1) = (x−5)(x 2 +x+x+1) = (x−5)(x(x+1)+1(x+1)) = (x−5)(x+1)(x+1)

(iii) x 3 +13x 2 +32x+20

Solution: Let p(x) = x 3 +13x 2 +32x+20
Factors of 20 are ±1, ±2, ±4, ±5, ±10 and ±20
By the trial method, we find that p(-1) = 0
So, (x+1) is factor of p(x)
Now, p(x)= x 3 +13x 2 +32x+20 p(-1) = (−1) 3 +13(−1) 2 +32(−1)+20 = −1+13−32+20 = 0
Therefore, (x+1) is the factor of p(x)

Now, Dividend = Divisor × Quotient +Remainder (x+1)(x 2 +12x+20) = (x+1)(x 2 +2x+10x+20) = (x+1)x(x+2)+10(x+2) = (x+1)(x+2)(x+10)

(iv) 2y 3 +y 2 –2y–1

Solution: Let p(y) = 2y 3 +y 2 –2y–1
Factors = 2×(−1)= -2 are ±1 and ±2
By the trial method, we find that p(1) = 0
So, (y-1) is factor of p(y) Now, p(y) = 2y 3 +y 2 –2y–1 p(1) = 2(1) 3 +(1) 2 –2(1)–1 = 2+1−2 = 0
Therefore, (y-1) is the factor of p(y)

Now, Dividend = Divisor × Quotient + Remainder (y−1)(2y 2 +3y+1) = (y−1)(2y 2 +2y+y+1) = (y−1)(2y(y+1)+1(y+1)) = (y−1)(2y+1)(y+1)

NCERT Maths Class 9 Polynomials Exercise 2.3 PDF

The NCERT Maths Class 9 Polynomials Exercise 2.3 PDF provides clear solutions to every question. It shows step-by-step methods for performing algebraic operations.

You can use it to revise how to combine like terms, apply distributive property, and simplify expressions accurately. The PDF also helps students build confidence in handling multi-step polynomial operations. 

NCERT Solutions Maths Class 9 PDF

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