Solutions to NCERT Class 9 Mathematics Chapter 2 Polynomials:

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Solutions to Exercise 2.1 (Page 29) of NCERT Class 9 Mathematics Chapter 2 Polynomials:

1. Which of the following expressions are polynomials in one variable, and which are not? State reasons for your answer.

(i) 4x² – 3x + 7
(ii) y² + √2
(iii) 3√t + t√2
(iv) y + 2/y
(v) x¹⁰ + y³ + t⁵⁰

Answers:

(i) 4x² – 3x + 7
The polynomial contains only one variable x.
Therefore, 4x² – 3x + 7 is a polynomial in one variable.

(ii) y² + √2
The polynomial contains only one variable y.
Therefore, y² + √2 is a polynomial in one variable.

(iii) 3√t + t√2
The power of the variable t in the first term is 1/2 which is not a whole number. Hence, 3√t + t√2 is not a polynomial.

(iv) y + 2/y
y + 2/y = y + 2y^-1. The power of the variable y in the second term in -1, which is not a whole number. Hence, y + 2/y is not a polynomial.

(v) x¹⁰ + y³ + t⁵⁰
In the above polynomial, there are three variables x, y, t. Hence, it is not a polynomial in one variable.

Summary: (i) 4x² – 3x + 7 and (ii) y² + √2 are polynomials in one variable, while (iii) 3√t + t√2, (iv) y + 2/y, (v) x¹⁰ + y³ + t⁵⁰ are not polynomials in one variable.

2. Write the coefficients of x² in each of the following:

(i) 2 + x² + x
(ii) 2 – x² + x³
(iii) (π/2)x² + x
(iv) √2x – 1

Answers:

(i) 2 + x² + x

2 + x2 + x = 2 + (1)x2 + x
Therefore, the coefficient of x2 is 1.

(ii) 2 – x² + x³

2 – x² + x³ = 2 + (-1)x² + x³
Therefore, the coefficient of x² is (-1).

(iii) (π/2)x² + x

The coefficient of x² is π/2.

(iv) √2x – 1

√2x – 1 can be written as (0)x² + √2x – 1.
Therefore, the coefficient of x² is 0.

Summary:
The coefficients of x² in the expressions (i) 2 + x²+ x, (ii) 2 – x² + x³, (iii) (π/2) x² + x, and (iv) √2x – 1 are 1, -1, π/2, and 0 respectively.

3. Give one example each of a binomial of degree 35, and of a monomial of degree 100.

Answer:

An example of a binomial of degree 35 is x³⁵ + 5 and an example of a monomial of degree 100 is 5x¹⁰⁰.

4. Write the degree of each of the following polynomials:

(i) 5x³ + 4x² + 7x        
(ii) 4–y²      
(iii) 5t – √7                 
(iv) 3

Answer:

(i) 5x³ + 4x² + 7x  

Degree of a polynomial refers to the highest power of the variable in the polynomial. Therefore, the degree of 5x³ + 4x² + 7x is 3.

(ii) 4 – y²      

Degree of a polynomial refers to the highest power of the variable in the polynomial. Therefore, the degree of 4 – y² is 2.

(iii) 5t – √7   

Degree of a polynomial refers to the highest power of the variable in the polynomial. Therefore, the degree of 5t – √7 is 1.

(iv) 3

Degree of a polynomial refers to the highest power of the variable in the polynomial. We can write 3 = 3x⁰. Therefore, the degree of 3 = 3x⁰ = 0.

Summary: The degree of the polynomials (i) 5x³ + 4x² + 7x, (ii) 4 − y², (iii) 5t − √7, and (iv) 3 are 3, 2, 1, and 0 respectively

5. Classify the following as linear, quadratic and cubic polynomials:

(i) x² + x              
(ii) x – x³           
(iii) y + y² + 4           
(iv) 1 + x
(v) 3t                  
(vi) r²                
(vii) 7x³

Answers:

(i) x² + x

The degree of x² + x is 2. Hence, it is quadratic polynomial.

(ii) x – x³           

The degree of x – x³ is 3. Hence, it is a cubic polynomial.

(iii) y + y² + 4

The degree of y + y² + 4 is 2. Hence, it is a quadratic polynomial.

(iv) 1 + x

The degree of 1 + x is 1. Hence, it is a linear polynomial.

(v) 3t

The degree of 3t is 1. Hence, it is a linear polynomial.

(vi) r²

The degree of r² is 2. Hence, it is quadratic polynomial.

(vii) 7x³

The degree of 7x³ is 3. Hence, it is a cubic polynomial.

Summary: (iv) 1 + x and (v) 3t are linear polynomials, (i) x² + x, (iii) y + y² + 4, (vi) r² are quadratic polynomials and (ii) x – x³and (vii) 7x³ are cubic polynomials.

Solutions to Exercise 2.2 (Page 31) of NCERT Class 9 Mathematics Chapter 2 Polynomials:

1. Find the value of the polynomial 5x − 4x² + 3 at

(i) x = 0              
(ii) x = – 1           
(iii) x = 2

Answers:

(i) x = 0

Let p(x) = 5x − 4x² + 3

or, p(0) = 5(0) – 4(0)² + 3 = 0 – 0 + 3 = 3

(ii) x = – 1

Let p(x) = 5x − 4x² + 3

or, p(-1) = 5(-1) – 4(-1)² + 3 = – 5 – 4 + 3 = – 6

(iii) x = 2

Let p(x) = 5x − 4x² + 3

or, p(2) = 5(2) – 4(2)² + 3 = 10 – 16 + 3 = – 3

Summary: The values of the polynomial 5x − 4x² + 3 at (i) x = 0, (ii) x = – 1 and (iii) x = 2 are 3, – 6, -3 respectively.

2. Find p(0), p(1) and p(2) for each of the following polynomials:

(i) p(y) = y² – y + 1           
(ii) p(t) = 2 + t + 2t² – t³
(iii) p(x) = x³                     
(iv) p(x) = (x – 1)(x + 1)

Answers:

(i) p(y) = y² – y + 1           

p(0) = 0² – 0 + 1 = 1

p(1) = 1² – 1 + 1 = 1

p(2) = 2² – 2 + 1 = 3

(ii) p(t) = 2 + t + 2t² – t³

p(0) = 2 + 0 + 2(0)² – (0)³ = 2

p(1) = 2 + 1 + 2(1)² – (1)³ = 2 + 1 + 2 – 1 = 4

p(2) = 2 + 2 + 2(2)² – (2)³ = 2 + 2 + 8 – 8 = 4

(iii) p(x) = x³

p(0) = (0)³ = 0

p(1) = 1³ = 1

p(2) = 2³ = 8

(iv) p(x) = (x – 1)(x + 1)

p(0) = (0 – 1)(0 + 1) = (-1)(1) = – 1

p(1) = (1 – 1)(1 + 1) = (0)(2) = 0

p(2) = (2 – 1)(2 + 1) = (1)(3) = 3

Summary:

The values of p(0), p(1) and p(2) for (i) p(y) = y² – y + 1, (ii)p(t) = 2 + t + 2t² – t³, (iii) p(x) = x³ and (iv) p(x) = (x – 1)(x + 1) are (1, 1, 3), (2, 4, 4), (0, 1, 8) and (-1, 0, 3).

3. Verify whether the following are zeroes of the polynomial, indicated against them.

(i) p(x) = 3x + 1, x = −1/3
(ii) p(x) = 5x – π, x = 4/5
(iii) p(x) = x²−1, x = 1, −1
(iv) p(x) = (x + 1)(x – 2), x =−1, 2
(v) p(x) = x², x = 0
(vi) p(x) = lx + m, x = −m/l
(vii) p(x) = 3x² − 1, x = – 1/√3 , 2/√3
(viii) p(x) = 2x + 1, x = 1/2

Answers:

(i) p(x) = 3x + 1, x = −1/3

p(−1/3) = 3(−1/3) + 1 = – 1 + 1 = 0
Therefore, x = −1/3 is a zero of the polynomial p(x) = 3x + 1.

(ii) p(x) = 5x – π, x = 4/5

p(4/5) = 5(4/5) – π = 4 – π ≠ 0
Therefore, x = 4/5 is not a zero of the polynomial p(x) = 5x – π.

(iii) p(x) = x²−1, x = 1, −1

p(1) = 12 – 1 = 0
p(-1) = (-1)2 – 1 = 1 – 1 = 0
Therefore, x = 1, – 1 are both zeroes of the polynomial p(x) = x2−1.

(iv) p(x) = (x + 1)(x – 2), x =−1, 2

p(-1) = (-1 + 1)(-1 – 2) = (0)(-3) = 0
p(2) = (2 + 1)(2 – 2) = 3(0) = 0
Therefore, x =−1, 2 are both zeroes of the polynomial p(x) = (x + 1)(x – 2).

(v) p(x) = x², x = 0

p(0) = (0)² = 0
Therefore, x = 0 is a zero of the polynomial p(x) = x2.

(vi) p(x) = lx + m, x = −m/l

p(−m/l) = l(−m/l) + m = -m + m = 0
Therefore, x = −m/l is a zero of the polynomial p(x) = lx + m.

(vii) p(x) = 3x² − 1, x = – 1/√3 , 2/√3

p(-1/√3) = 3(-1/√3)² – 1 = 3(1/3) – 1 = 1 – 1 = 0
p(2/√3) = 3(2/√3)² – 1 = 3(4/3) – 1 = 4 – 1 = 3
Therefore, x = – 1/√3 is a zero of the polynomial p(x) = 3x² – 1, but x = 2/√3 is not a zero of p(x) = 3x² – 1.

(viii) p(x) = 2x + 1, x = 1/2

p(1/2) = 2(1/2) + 1 = 1 + 1 = 2
Therefore, x = 1/2 is not a zero of the polynomial p(x) = 2x + 1.

4. Find the zero of the polynomial in each of the following cases:

(i) p(x) = x + 5               
(ii) p(x) = x – 5               
(iii) p(x) = 2x + 5
(iv) p(x) = 3x – 2         
(v) p(x) = 3x                     
(vi) p(x) = ax, a≠0
(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.

Answers:

(i) p(x) = x + 5

p(x) = x + 5 = 0

or, x = – 5

Therefore, x = – 5 is a zero of the polynomial p(x) = x + 5.

(ii) p(x) = x – 5

p(x) = x – 5 = 0

or, x = 5

Therefore, x = 5 is a zero of the polynomial p(x) = x – 5.

(iii) p(x) = 2x + 5

p(x) = 2x + 5 = 0

or, 2x = – 5

or, x = -5/2

Therefore, x = -5/2 is a zero of the polynomial p(x) = 2x + 5.

(iv) p(x) = 3x – 2

p(x) = 3x – 2 = 0

or, 3x = 2

or, x = 2/3

Therefore, x = 2/3 is a zero of the polynomial p(x) = 3x – 2.

(v) p(x) = 3x

p(x) = 3x = 0

or, x = 0

Therefore, x = 0 is a zero of the polynomial p(x) = 3x.

(vi) p(x) = ax, a≠0

p(x) = ax = 0

or x = 0

Therefore, x = 0 is a zero of the polynomial p(x) = ax, a≠0.

(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.

p(x) = cx + d = 0

or, cx = -d

or, x = -d/c 

Therefore, x = -d/c is a zero of the polynomial p(x) = cx + d.

Summary:

The zeros of the polynomials in each of the following cases (i) p(x) = x + 5, (ii) p(x) = x – 5, (iii) p(x) = 2x + 5, (iv) p(x) = 3x – 2, (v) p(x) = 3x, (vi) p(x) = ax, (a ≠ 0), and (vii) p(x) = cx + d, (c ≠ 0 and c, d are real numbers) are -5, 5, -5/2, 2/3, 0, 0, and -d/c respectively.

Solutions to Exercise 2.3 (Page 35) of NCERT Class 9 Mathematics Chapter 2 Polynomials:

1. 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:

(i) x³ + x² + x + 1

x + 1 = 0

or, 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) x⁴ + x³ + x² + x + 1

x + 1 = 0

or, x = – 1

p(-1) = (-1)⁴ + (-1)³ + (-1)² – 1 + 1 = 1 – 1 + 1 – 1 + 1 = 1 ≠ 0.

Since p(-1) = 1 ≠ 0, x + 1 is a not factor of x⁴ + x³ + x² + x + 1.

(iii) x⁴ + 3x³ + 3x² + x + 1

x + 1 = 0

or, x = – 1

p(-1) = (-1)⁴ + 3(-1)³ + 3(-1)² – 1 + 1 = 1 – 3 + 3 – 1 + 1 = 1 ≠ 0.

Since p(-1) = 1 ≠ 0, x + 1 is a not factor of x⁴ + 3x³ + 3x² + x + 1.

(iv) x³ – x²– (2 + √2)x + √2

x + 1 = 0

or, x = – 1

p(-1) = (-1)³ – (-1)² – (2 + √2)(-1) + √2

= -1 – 1 + 2 + √2 + √2

= 2√2 ≠ 0 Since p(-1) = 2√2 ≠ 0, x + 1 is a not factor of x³ – x²– (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) = 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

Answers:

(i) p(x) = 2x³ + x² – 2x – 1, g(x) = x + 1

By Factor Theorem, x – a is a factor of p(x), if p(a) = 0.

g(x) = x + 1 = 0

or, x = – 1

p(−1) = 2(−1)³ + (−1)² – 2(−1) – 1

= −2 + 1 + 2 − 1

= 0

Since p(−1) = 0, g(x) = x + 1 is a factor of p(x) = 2x³ + x² – 2x – 1.

(ii) p(x) = x³ + 3x² + 3x + 1, g(x) = x + 2

g(x) = x + 2 = 0

or, x = – 2

p(-2) = (-2)³ + 3(-2)² + 3(-2) + 1

= – 8 + 12 – 6 + 1

= – 1 ≠ 0

Since, p(-2) ≠ 0, by Factor Theorem we can say that g(x) = x + 2 is not a factor of p(x) = x³ + 3x² + 3x + 1.

(iii) p(x) = x³ – 4x² + x + 6, g(x) = x – 3

g(x) = x – 3 = 0

or, x = 3

p(3) = (3)³ – 4(3)² + 3 + 6

= 27 – 36 + 3 + 6

= 0

Since p(3) = 0, by Factor Theorem we can say that g(x) = x – 3 is factor of p(x) = x³ – 4x² + x + 6.

Summary: Using factor theorem we determine that (i) g(x) = x + 1 is a factor of p(x) = 2x³ + x² – 2x – 1, (ii) g(x) = x + 2 is not a factor of p(x) = x³ + 3x² + 3x + 1, and (iii) g(x) = x – 3 is a factor of p(x) = x³ – 4x² + x + 6.

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

Answers:

(i) p(x) = x² + x + k

By Factor Theorem, if x – 1 is a factor of p(x) = x² + x + k, then p(1) = 0.

p(1) = 1² + 1 + k = 0

or, k = -2

(ii) p(x) = 2x² + kx + √2

By Factor Theorem, if x – 1 is a factor of p(x) = 2x² + kx + √2, then p(1) = 0.

p(1) = 2(1)² + k(1) + √2 = 0

or, 2 + k + √2 = 0

or, k = – 2 – √2

(iii) p(x) = kx² – √2x + 1

By Factor Theorem, if x – 1 is a factor of p(x) = kx² – √2x + 1, then p(1) = 0.

p(1) = k(1)² – √2(1) + 1 = 0

or, k – √2 + 1 = 0

or, k = √2 – 1

(iv) p(x) = kx² – 3x + k

By Factor Theorem, if x – 1 is a factor of p(x) = kx² – 3x + k, then p(1) = 0.

p(1) = k(1)² – 3(1) + k = 0

or, k – 3 + k = 0

or, 2k = 3

or, k = 3/2

Summary: The values 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 are -2, – 2 – √2, √2 – 1, and 3/2 respectively.

4. Factorise:

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

Answers:

(i) 12x² – 7x + 1

p + q = -7 and pq = 12 × 1 = 12

We get -3 and -4 as the numbers because – 3 – 4 = -7 and (-3)(-4) = 12.

12x² – 7x + 1

= 12x² + (-3 – 4)x + 1

= 12x² – 3x – 4x + 1

= 3x(4x – 1) – 1(4x – 1)

= (4x – 1)(3x – 1)

(ii) 2x² + 7x + 3

p + q = 7 and pq = 2 × 3 = 6

We get 6 and 1 as the numbers because 6 + 1 = 7 and (6)(1) = 6.

2x² + 7x + 3

= 2x² + (6 + 1)x + 3

= 2x² + 6x + x + 3

= 2x(x + 3) + 1(x + 3)

= (x + 3)(2x + 1)

(iii) 6x² + 5x – 6

p + q = 5 and pq = 6 × (-6) = -36

We get 9 and -4 as the numbers because 9 – 4 = 5 and (9)(-4) = -36.

6x² + 5x – 6

= 6x² + (9 – 4)x – 6

= 6x² + 9x – 4x – 6

= 3x(2x + 3) – 2(2x + 3)

= (2x + 3)(3x – 2)

(iv) 3x² – x – 4

p + q = -1 and pq = 3 × (-4) = -12

We get -4 and 3 as the numbers because -4 + 3 = -1 and (-4)(3) = -12.

3x² – x – 4

= 3x² + (-4 + 3)x – 4

= 3x² – 4x + 3x – 4

= x(3x – 4) + 1(3x – 4)

= (3x – 4)(x + 1)

Summary: The factorized form of (i) 12x² – 7x + 1, (ii) 2x² + 7x + 3, (iii) 6x² + 5x – 6 and (iv) 3x² – x – 4 are (4x – 1)(3x – 1), (x + 3)(2x + 1), (2x + 3)(3x – 2), (3x – 4)(x + 1) respectively.

5. Factorise:

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

Answers:

(i) x³ – 2x² – x + 2   

Let p(x) = x³ – 2x² – x + 2

Factors of 2 are ±1 and ± 2.

p(1) = (1)³ – 2(1)² – (1) + 2

= 1 – 2 – 1 + 2

= 0

Therefore, by factor theorem (x – 1) is a factor of p(x).

Now we divide p(x) by (x – 1):

x³ – 2x² – x + 2 = (x – 1)(x² – x – 2)

Now, x² – x – 2 can be factorised by splitting the middle term.

p + q = -1 and pq = (1)(-2) = -2

We get -2 and 1 as the numbers because -2 + 1 = -1 and (-2)(1) = -2.

x² – x – 2

= x² + (-2 + 1)x – 2

= x² – 2x + x – 2

= x(x – 2) + 1(x – 2)

= (x – 2)(x + 1)

x³ – 2x² – x + 2 = (x – 1)(x – 2)(x + 1)

(ii) x³ – 3x² – 9x – 5

p(x) = x³ – 3x² – 9x – 5

We observe,

p(-1) = (-1)³ – 3(-1)² – 9(-1) – 5 = – 1 – 3 + 9 – 5 = 0

By factor theorem, x + 1 is a factor of p(x).

Now we see that,

x³ – 3x² – 9x – 5

= x³ + x² – 4x² – 4x – 5x – 5

= x²(x + 1) – 4x(x + 1) – 5(x + 1)

= (x + 1)(x² – 4x – 5)

Now, x² – 4x – 5 can be factorised by splitting the middle term.

p + q = -4 and pq = (1)(-5) = -5

We get -5 and 1 as the numbers because -5 + 1 = -4 and (-5)(1) = -5.

x² – 4x – 5

= x² + (-5 + 1)x – 5

= x² – 5x + x – 5

= x(x – 5) + 1(x – 5)

= (x – 5)(x + 1)

x³ – 3x² – 9x – 5 = (x + 1)(x – 5)(x + 1)

(iii) x³ + 13x² + 32x + 20

p(x) = x³ + 13x² + 32x + 20

We observe,

p(-1) = (-1)³ + 13(-1)² + 32(-1) + 20 = – 1 + 13 – 32 + 20 = 0

By factor theorem, x + 1 is a factor of p(x).

Now we see that,

x³ + 13x² + 32x + 20

= x³ + x² + 12x² + 12x + 20x + 20

= x²(x + 1) + 12x(x + 1) + 20(x + 1)

= (x + 1)(x² + 12x + 20)

Now, x² + 12x + 20 can be factorised by splitting the middle term.

p + q = 12 and pq = (20)(1) = 20

We get 10 and 2 as the numbers because 10 + 2 = 12 and (10)(2) = 20.

x² + 12x + 20

= x² + (10 + 2)x + 20

= x² + 10x + 2x + 20

= x(x + 10) + 2(x + 10)

= (x + 10)(x + 2)

x³ + 13x² + 32x + 20 = (x + 1)(x + 10)(x + 2)

(iv) 2y³ + y² – 2y – 1

p(y) = 2y³ + y² – 2y – 1

We observe,

p(1) = 2(1)³ + (1)² – 2(1) – 1 = 2 + 1 –  2 – 1 = 0

By factor theorem, y – 1 is a factor of p(y).

Now we divide p(y) by (y – 1):

2y³ + y² – 2y – 1 = (y – 1)(2y² + 3y + 1)

Now, 2y² + 3y + 1 can be factorised by splitting the middle term.

p + q = 3 and pq = (2)(1) = 2

We get 2 and 1 as the numbers because 2 + 1 = 3 and (2)(1) = 2.

2y² + 3y + 1

= 2y² + (2 + 1)y + 1

= 2y² + 2y + y + 1

= 2y(y + 1) + 1(y + 1)

= (y + 1)(2y + 1)

Therefore,

2y³ + y² – 2y – 1 = (y – 1)(y + 1)(2y + 1)

Summary:

The factorized form of (i) x³ – 2x² – x + 2, (ii) x³ – 3x² – 9x – 5, (iii) x³ + 13x² + 32x + 20 and (iv) 2y³ + y² – 2y – 1 are (x – 1)(x – 2)(x + 1), (x + 1)(x – 5)(x + 1), (x + 1)(x + 10)(x + 2)and (y – 1)(y + 1)(2y + 1) respectively.

Solutions to Exercise 2.4 (Page 40) of NCERT Class 9 Mathematics Chapter 2 Polynomials:

1. Use suitable identities to find the following products:

(i) (x + 4)(x + 10)            
(ii) (x + 8)(x – 10)            
(iii) (3x + 4)(3x – 5)
(iv) (y² + 3/2)(y² – 3/2)          
(v) (3 – 2x)(3 + 2x)

Answers:

(i) (x + 4)(x + 10)            

Using the identity (x + a)(x + b) = x² + (a + b)x + ab we get:

(x + 4)(x + 10) = x² + (4 + 10)x + 4 × 10 = x² + 14x + 40

(ii) (x + 8)(x – 10)

Using the identity (x + a)(x + b) = x² + (a + b)x + ab we get:

(x + 8)(x – 10) = x² + (8 – 10)x + 8 × (-10) = x² – 2x – 80

(iii) (3x + 4)(3x – 5)

Using the identity (x + a)(x + b) = x² + (a + b)x + ab we get:

(3x + 4)(3x – 5) = (3x)² + [4 + (-5)]3x + (4)(-5) [Taking 3x in place of x]

= 9x² – 3x – 20

(iv) (y² + 3/2)(y² – 3/2)

Using identity (a + b)(a – b) = a² – b² we get:

(y² + 3/2)(y² – 3/2) = (y²)² – (3/2)² [Taking a = y² and b = ]

= y⁴ – 9/4

(v) (3 – 2x)(3 + 2x)

Using identity (a – b)(a + b) = a² – b² we get:

(3 – 2x)(3 + 2x) = 3² – (2x)² [Taking a = 3 and b = 2x]

= 9 – 4x²

Summary: Using suitable identities we find the products of (i) (x + 4)(x + 10), (ii) (x + 8)(x – 10), (iii) (3x + 4)(3x – 5), (iv) (y² + 3/2)(y² – 3/2), and (v) (3 – 2x)(3 + 2x) as x² + 14x + 40, x² – 2x – 80, 9x² – 3x – 20, y⁴ – 9/4, and 9 – 4x² respectively.

2. Evaluate the following products without multiplying directly:

(i) 103 × 107               
(ii) 95 × 96               
(iii) 104 × 96

Answers:

(i) 103 × 107

103 × 107 = (100 + 3) × (100 + 7)

We use the identity [(x + a)(x + b) = x² + (a + b)x + ab].

Here, x = 100, a = 3, b = 7

Now, 103 × 107 = (100 + 3) × (100 + 7)

= (100)² + (3 + 7)100 + (3 × 7)

= 10000 + 1000 + 21

= 11021

(ii) 95 × 96

95 × 96 = (100 – 5)(100 – 4)

We use the identity [(x + a)(x + b) = x² + (a + b)x + ab].

Here, x = 100, a = -5, b = -4

Now, 95 × 96 = (100 – 5) × (100 – 4)

(100)² + (-5 + (-4))100 + (-5) × (-4)

= 10000 – 900 + 20

= 9120

(iii) 104 × 96

104 × 96 = (100 + 4) × (100 – 4)

We use the identity [(a + b)(a – b)= a² – b²]

Here, a = 100, b = 4

Now, 104 × 96 = (100 + 4) × (100 – 4)

= (100)² – (4)²

= 10000 – 16

= 9984

Summary: The products without multiplying directly of (i)103 × 107, (ii)95 × 96 and (iii) 104 × 96 are 11021, 9120 and 9984 respectively.

3. Factorise the following using appropriate identities:

(i) 9x² + 6xy + y²                             
(ii) 4y² − 4y + 1                 
(iii)  x² – y²/100

Answers:

(i) 9x² + 6xy + y²

We use identity x² + 2xy + y² = (x + y)².

Here, x = 3x and y = y.

Therefore,

9x² + 6xy + y²

= (3x)² + 2(3x)(y) + y²

= (3x + y)²

(ii) 4y² − 4y + 1

We use identity x² – 2xy + y² = (x – y)².

Here, x = 2y and y = 1.

Therefore,

4y² − 4y + 1

= (2y)² – 2(2y)(1) + 1²

= (2y – 1)²

(iii)  x² – y²/100       

We use the identity [(a + b)(a – b)= a² – b²]

Here, a = x, b = y/10

Therefore,

x² – y²/100      

= (x)² – (y/10)²

= (x + y/10)(x – y/10)

Summary: The factorized form using appropriate identities of (i) 9x² + 6xy + y², (ii) 4y² – 4y + 1, and (iii) x² – y²/100 are (3x + y)², (2y – 1)², and (x + y/10) (x – y/10) respectively.

4. Expand each of the following using suitable identities:

(i) (x + 2y + 4z)²                
(ii) (2x – y + z)²                 
(iii) (−2x + 3y + 2z)²
(iv) (3a – 7b – c)²            
(v) (–2x + 5y – 3z)²          
(vi) [(1/4)a – (1/2)b + 1]²

Answers:

(i) (x + 2y + 4z)²

We use the identity (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx.

Here, x = x

y = 2y

z = 4z

(x + 2y + 4z)² = x² + (2y)² + (4z)² +( 2 × x × 2y) + (2 × 2y × 4z) + (2 × 4z × x)

= x² + 4y² + 16z² + 4xy + 16yz + 8xz

(ii) (2x – y + z)²

We use the identity (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx.

Here, x = 2x

y = -y

z = z

(2x – y + z)² = (2x)² + (-y)² + z² + 2 × 2x × (-y) + 2 × (-y) × z + 2 × z × 2x

= 4x² + y² +z ² – 4xy – 2yz + 4xz

(iii) (−2x + 3y + 2z)²

We use the identity (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx.

Here, x = -2x

y = 3y

z = 2z

(−2x + 3y + 2z)² = (-2x)² + (3y)² + (2z)² + 2 × (-2x) × (3y) + 2 × (3y) × (2z) + 2 × (2z) × (-2x)

= 4x² + 9y² + 4z² – 12xy + 12yz – 8xz

(iv) (3a – 7b – c)²

We use the identity (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx.

Here, x = 3a

y = -7b

z = -c

(3a – 7b – c)² = (3a)² + (-7b)² + (-c)² + 2 × (3a) × (-7b) + 2 × (-7b) × (-c) + 2 × (-c) × (3a)

= 9a² + 49b² + c²– 42ab + 14bc – 6ca

(v) (–2x + 5y – 3z)²

We use the identity (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx.

Here, x = -2x

y = 5y

z = -3z

(–2x + 5y – 3z)² = (–2x)² + (5y)² + (–3z)² + 2 × (-2x) × (5y) + 2 × (5y) × (-3z) + 2 × (-3z) × (-2x)

= 4x² + 25y² + 9z² – 20xy – 30yz + 12zx

(vi) [(1/4)a – (1/2)b + 1]²

We use the identity (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx.

Here, x = a/4

y = – b/2

z = 1

[(1/4)a – (1/2)b + 1]² = (a/4)² + (– b/2)² + 1² + 2 × (a/4) × (– b/2) + 2 × (– b/2) × 1 + 2 × (1) × (a/4)

= (1/16)a² + (1/4)b² + 1 – ab/4 – b + a/2

Summary: The expanded form of each of the following (i) (x + 2y + 4z)², (ii) (2x – y + z)², (iii) (−2x + 3y + 2z)², (iv) (3a – 7b – c)², (v) (–2x + 5y – 3z)² and (vi) [a – b + 1]² are x² + 4y² + 16z² + 4xy + 16yz + 8xz, 4x² + y² +z ² – 4xy – 2yz + 4xz, 4x² + 9y² + 4z² – 12xy + 12yz – 8xz, 9a² + 49b² + c²– 42ab + 14bc – 6ca, 4x² + 25y² + 9z² – 20xy – 30yz + 12zx, (1/16)a² + (1/4)b² + 1 – ab/4 – b + a/2 respectively.

5. Factorise:

(i) 4x² + 9y² + 16z² + 12xy – 24yz – 16xz

(ii) 2x² + y² + 8z² –2√2xy + 4√2yz – 8xz

Answers:

(i) 4x² + 9y² + 16z² + 12xy – 24yz – 16xz

We use the identity x² + y² + z² + 2xy + 2yz + 2zx = (x + y + z)².

4x² + 9y² + 16z² + 12xy – 24yz – 16xz

= (2x)² + (3y)² + (-4z)² + 2(2x)(3y) + 2(3y)(-4z) + 2(2x)(-4z)

Here x = 2x, y = 3y and z = -4z.

= (2x + 3y – 4z)²

(ii) 2x² + y² + 8z² –2√2xy + 4√2yz – 8xz

We use the identity x² + y² + z² + 2xy + 2yz + 2zx = (x + y + z)².

2x² + y² + 8z² –2√2xy + 4√2yz – 8xz

= (-√2x)² + y² + (2√2z)² + 2(-√2x)(y) + 2(y)( 2√2z) + 2(2√2z)(-√2x)

Here, x = -√2x, y = y and z = 2√2z.

= (-√2x + y + 2√2z)²

Summary: The factorised forms of (i) 4x² + 9y² + 16z² + 12xy – 24yz – 16xz and (ii) 2x² + y² + 8z² –2√2xy + 4√2yz – 8xz are (2x + 3y – 4z)² and (-√2x + y + 2√2z)² respectively.

6. Write the following cubes in expanded form:

(i) (2x + 1)³            
(ii) (2a − 3b)³          
(iii) [(3/2)x + 1]³               
(iv) [x – (2/3)y]³

Answers:

(i) (2x + 1)³

We use the identity (x + y)³ = x³ + y³ + 3xy(x + y):

Here x = 2x and y = 1.

(2x + 1)³= (2x)³ + 1³ + 3 × 2x × 1(2x + 1)

= 8x³ + 1 + 6x(2x + 1)

= 8x³ + 12x² + 6x + 1

(ii) (2a − 3b)³

We use the identity (x – y)³ = x³ – y³ – 3xy(x – y):

Here x = 2a and y = 3b.

(2a − 3b)³ = (2a)³ – (3b)³ – 3(2a)(3b)(2a – 3b)

= 8a³ – 27b³ – 18ab(2a – 3b)

= 8a³ – 27b³ – 36a²b + 54ab²

(iii) [(3/2)x + 1]³

We use the identity (x + y)³ = x³ + y³ + 3xy(x + y):

Here x = (3/2)x and y = 1.

[(3/2)x + 1]³ = ((3/2)x)³ + 1³ + 3.(3/2).x.1.((3/2)x + 1)

= (27/8)x³ + 1 + (9/2)x ((3/2)x + 1)

= (27/8)x³ + 1 + (27/4)x² + (9/2)x

= (27/8)x³ + (27/4)x² + (9/2)x + 1

(iv) [x – (2/3)y]³

We use the identity (x – y)³ = x³ – y³ – 3xy(x – y):

Here x = x and y = (2/3)y.

[x – (2/3)y]³ = x³ – ((2/3)y)³ – 3x((2/3)y)(x – (2/3)y)

= x³ – (8/27)y³ – 2xy(x – (2/3)y)

= x³ – (8/27)y³ – 2x²y + (4/3)xy²

Summary: The expanded forms of the following cubes (i) (2x + 1)³, (ii) (2a − 3b)³, (iii) [(3/2)x + 1]³ , (iv) [x – (2/3)y]³ are 8x³ + 12x² + 6x + 1, 8a³ – 27b³ – 36a²b + 54ab², (27/8)x³ + (27/4)x² + (9/2)x + 1, x³ – (8/27)y³ – 2x²y + (4/3)xy² .

7. Evaluate the following using suitable identities: 

(i) (99)³                      
(ii) (102)³                   
(iii) (998)³

Answers:

(i) (99)³

99 = 100–1

We use identity (x – y)³ = x³ – y³ – 3xy(x – y):

(99)³ = (100 – 1)³

= (100)³ – 1³ – (3 × 100 × 1)(100 – 1)

= 1000000 – 1 – 300(100 – 1)

= 1000000 – 1 – 30000 + 300

= 970299

(ii) (102)³

102 = 100 + 2

We use identity (x + y)³ = x³ + y³ + 3xy(x + y):

(100 + 2)³ = (100)³ + 2³ + (3 × 100 × 2)(100 + 2)

= 1000000 + 8 + 600(100 + 2)

= 1000000 + 8 + 60000 + 1200

= 1061208

(iii) (998)³

998 = 1000 – 2

We use identity (x – y)³ = x³ – y³ – 3xy(x – y):

(998)³ = (1000 – 2)³

= (1000)³ – (2)³ – 3(1000) (2) (1000 – 2)

= 1000000000 – 8 – 6000(1000 – 2)

= 1000000000 – 8 – 6000000 + 12000

= 994011992

Summary: The evaluated values using suitable identities of the following (i) (99)³, (ii) (102)³, and (iii) (998)³ are 970299, 1061208, and 994011992 respectively.

8. Factorise each of the following:

(i) 8a³ + b³ + 12a²b + 6ab²                                 (ii) 8a³ – b³ – 12a²b + 6ab²

(iii) 27 – 125a³ – 135a + 225a²                        (iv) 64a³ – 27b³ – 144a²b + 108ab²

(v) 27p³ – 1/216 – (9/2)p² + (1/4)p

Answers:

(i) 8a³ + b³ + 12a²b + 6ab²

8a³ + b³ + 12a²b + 6ab² = (2a)³ + b³ + 3(2a)²b + 3(2a)(b)²

This is of the form x³ + y³ + 3x²y + 3xy² = (x + y)³, where x = 2a, y = b.

= (2a + b)³

(ii) 8a³ – b³ – 12a²b + 6ab²

8a³ – b³ – 12a²b + 6ab² = (2a)³ – b³ – 3(2a)²b + 3(2a)(b)²

This is of the form x³ – y³ – 3x²y + 3xy² = (x – y)³, where x = 2a, y = b.

= (2a – b)³

(iii) 27 – 125a³ – 135a + 225a² 

27 – 125a³ – 135a + 225a² = 3³ – (5a)³ – 3(3)³5a + 3(3)(5a)²

This is of the form x³ – y³ – 3x²y + 3xy² = (x – y)³, where x = 3, y = 5a.

= (3 – 5a)³

(iv) 64a³ – 27b³ – 144a²b + 108ab²

64a³ – 27b³ – 144a²b + 108ab² = (4a)³ – (3b)³ – 3(4a)²(3b) + 3(4a)(3b)²

This is of the form x³ – y³ – 3x²y + 3xy² = (x – y)³, where x = 4a, y = 3b.

= (4a – 3b)³

(v) 27p³ – 1/216 – (9/2)p² + (1/4)p

27p³ – 1/216 – (9/2)p² + (1/4)p= (3p)³ – (1/6)³ – 3(3p)² (1/6) +3(3p)(1/6)²

= (3p – 1/6)³

Summary: The factorised form of the following (i) 8a³ + b³ + 12a²b + 6ab², (ii) 8a³ – b³ – 12a²b + 6ab², (iii) 27 – 125a³ – 135a + 225a², (iv) 64a³ – 27b³ – 144a²b + 108ab², 27p³ – 1/216 – (9/2)p² + (1/4)p are (2a + b)³, (2a – b)³, (3 – 5a)³, (4a – 3b)³ and (3p – 1/6)³ respectively.

9. Verify:

(i) x³ + y³ = (x + y)(x² – xy + y²)                  (ii) x³ – y³ = (x – y)(x² + xy + y²)

Answers:

(i) x³ + y³ = (x + y)(x² – xy + y²)

We use the identity (x + y)³ = x³ + y³ + 3xy(x + y):

x³ + y³ = (x + y)³ – 3xy(x + y)

or, x³ + y³ = (x + y)[(x + y)² – 3xy]

or, x³ + y³ = (x + y)[x² + y² + 2xy – 3xy]

or, x³ + y³ = (x + y)(x² + y² – xy)

(ii) x³ – y³ = (x – y)(x² + xy + y²)

We use the identity (x – y)³ = x³ – y³ – 3xy(x – y):

x³ – y³ = (x – y)³ + 3xy(x – y)

or, x³ – y³ = (x – y)[(x – y)² + 3xy]

or, x³ – y³ = (x – y)(x² + y² – 2xy + 3xy)

or, x³ – y³ = (x – y)(x² + xy + y²)

Summary: Hence it is verified that (i) x³ + y³ = (x + y)(x² – xy + y²) and (ii) x³ – y³ = (x – y)(x² + xy + y²).

10. Factorise each of the following:

(i) 27y³ + 125z³                   (ii) 64m³ – 343n³

[Hint: See Question 9.]

Answers:

(i) 27y³ + 125z³

= (3y)³ + (5z)³

Using identity x³ + y³ = (x + y)(x² – xy + y²) we get:

(3y)³ + (5z)³ = (3y + 5z)[(3y)² – (3y)(5z) + (5z)²]

= (3y + 5z)(9y² – 15yz + 25z²)

(ii) 64m³ – 343n³

= (4m)³ – (7n)³

Using identity x³ – y³ = (x – y)(x² + xy + y²) we get:

(4m)³ – (7n)³

= (4m – 7n)[(4m)² + (4m)(7n) + (7n)²]

= (4m – 7n)(16m² + 28mn + 49n²)

Summary:

The factorised form of (i) 27y³ + 125z³ and(ii) 64m³ – 343n³ are (3y + 5z)(9y² – 15yz + 25z²) and (4m – 7n)(16m² + 28mn + 49n²) respectively.

11. Factorise: 27x³ + y³ + z³ – 9xyz. 

Answer:

27x³ + y³ + z³ – 9xyz = (3x)³ + y³ + z³ +3(3x)(y)(z)

We use the identity x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx):

(3x)³ + y³ + z³ +3(3x)(y)(z)

= (3x + y + z)[(3x)² + y² + z² – (3x)(y) – (y)(z) – (z)(3x)]

= (3x + y + z)(9x² + y² + z² – 3xy – yz – 3xz)

Summary: The factorised form of 27x³ + y³ + z³ – 9xyz is (3x + y + z)(9x² + y² + z² – 3xy – yz – 3xz).

12. Verify that x³ + y³ + z³ – 3xyz = (x + y + z)[(x – y)² + (y – z)² + (z – x)²]

Answer:

We use the identity x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx):

or, x³ + y³ + z³ – 3xyz =  (x + y + z)[2(x² + y² + z² – xy – yz – zx)]

or, x³ + y³ + z³ – 3xyz =  (x + y + z)[2x² + 2y² + 2z² – 2xy – 2yz – 2zx]

or, x³ + y³ + z³ – 3xyz =  (x + y + z)[(x² + y² − 2xy) + (y² + z² – 2yz) + (x² + z² –2xz)]

or, x³ + y³ + z³ – 3xyz = (x + y + z)[(x – y)² + (y – z)² + (z – x)²] (Proved)

Summary: Hence, it is verified that x³ + y³ + z³ – 3xyz = (x + y + z)[(x – y)² + (y – z)² + (z – x)²].

13. If x + y + z = 0, show that x³ + y³ + z³ = 3xyz.

Answer:

Answer:

We know x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx)

or, x³ + y³ + z³ – 3xyz = (0) (x² + y² + z² – xy – yz – zx)

or, x³ + y³ + z³ – 3xyz = 0

or, x³ + y³ + z³ = 3xyz (Proved)

Summary: Hence it is proved that if x + y + z = 0, x³ + y³ + z³ = 3xyz.

14. Without actually calculating the cubes, find the value of each of the following:

(i) (−12)³ + (7)³ + (5)³

(ii) (28)³ + (−15)³ + (−13)³

Answer:

(i) (−12)³ + (7)³ + (5)³

Here, x = -12, y = 7 and z = 5 and x + y + z = -12 + 7 + 5 = 0.

We know that, if x + y + z = 0, x³ + y³ + z³ = 3xyz.

Therefore,

(−12)³ + (7)³ + (5)³ = 3(-12)(7)(5) = -1260

(ii) (28)³ + (−15)³ + (−13)³

Here, x = 28, y = -15 and z = -13 and x + y + z = 28 – 15 – 13 = 0.

We know that, if x + y + z = 0, x³ + y³ + z³ = 3xyz.

Therefore,

(28)³ + (-15)³ + (-13)³ = 3(28)(-15)(-13) = 16380

Summary: Without actually calculating the cubes the values of (i) (−12)³ + (7)³ + (5)³ and (ii) (28)³ + (−15)³ + (−13)³ are -1260 and 16380.

15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given: 

(i) Area: 25a² – 35a + 12

(ii) Area: 35y² + 13y – 12

Answers:

(i) Area: 25a² – 35a + 12

Area = Length × Breadth

Therefore, we have to factorise the expression for area.

p + q = -35 and pq = 25 × 12 = 300

By trial and error, we get p = -20 and q = -15.

25a² – 35a + 12

= 25a² – 20a – 15a + 12

= 5a(5a – 4) – 3(5a – 4)

= (5a – 3)(5a – 4)

Therefore, length = 5a – 3 and breath = 5a – 4.

(ii) Area: 35y² + 13y – 12

Area = Length × Breadth

Therefore, we have to factorise the expression for area.

p + q = 13 and pq = 35 × (-12) = -420

By trial and error we get p = -15 and q = 28.

35y² + 13y – 12

= 35y² – 15y + 28y – 12

= 5y(7y – 3) + 4(7y – 3)

= (5y + 4)(7y – 3)

Therefore, length = 5y + 4 and breadth = 7y – 3.

Summary: The possible expressions for the length and breadth of each of the following rectangles of (i) Area: 25a² – 35a + 12 and (ii) Area: 35y² + 13y – 12 are (Length = 5a – 3, Breadth = 5a – 4) and (Length = 5y + 4, Breadth = 7y – 3) respectively.

16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below? 

(i) Volume: 3x² – 12x

(ii) Volume: 12ky² + 8ky –20k

Answers:

(i) Volume: 3x² – 12x

3x² – 12x = 3x(x – 4)

Volume of cuboid = Length × Breadth × Height

Possible expressions are:

Length = 3, breadth = x and height = x – 4.

(ii) Volume: 12ky² + 8ky –20k

12ky² + 8ky –20k = 4k(3y² + 2y – 5)

Now let us factorise 3y² + 2y – 5.

p + q = 2 and pq = 3(-5) = -15.

By trial and error, we get p = 5 and q = -3.

3y² + 2y – 5

= 3y² + (5 – 3)y – 5

= 3y² + 5y – 3y – 5

= y(3y + 5) – 1(3y + 5)

= (y – 1)(3y + 5)

Therefore,

12ky² + 8ky –20k = 4k(y – 1)(3y + 5)

Possible expressions are:

Length = 4k, breadth = y – 1 and height = 3y + 5.

Summary:

The possible expressions for the dimensions of the cuboids whose volume are (i) 3x² – 12x and (ii) 12ky² + 8ky – 20k are 3, x, (x – 4) and 4k, (y – 1), (3y + 5) respectively.

Important Problems from Old NCERT Textbook:

Exercise 2.3 Page 35 (Old Textbook):

1. Find the remainder when x³ + 3x² + 3x + 1 is divided by

(i) x + 1              
(ii) x – 1/2           
(iii) x          
(iv) x + π           
(v) 5 + 2x

Answers:

(i) x + 1             

By remainder theorem, if p(x) is divided by linear polynomial x – a, the remainder is p(a).

x + 1 = 0

or, x = – 1

p(-1) = (-1)³ + 3(-1)² + 3(-1) + 1 = -1 + 3 – 3 + 1 = 0

Therefore, when x³ + 3x² + 3x + 1 is divided by x + 1the remainder = 0.

(ii) x – 1/2       

By remainder theorem, if p(x) is divided by linear polynomial x – a, the remainder is p(a).

x – 1/2 = 0

or, x = 1/2

p(1/2) = (1/2)³ + 3(1/2)² + 3(1/2) + 1 = 1/8 + 3/4 + 3/2 + 1 = (1 + 6 + 12 + 8)/8 = 27/8

Therefore, when x³ + 3x² + 3x + 1 is divided by x – 1/2, the remainder is 27/8.

(iii) x

By remainder theorem, if p(x) is divided by linear polynomial x – a, the remainder is p(a).

The root of x = 0 is 0.

p(0) = (0)³ + 3(0)² + 3(0) +1

= 0 + 0 + 0 + 1

= 1

Therefore, when x³ + 3x² + 3x + 1 is divided by x, the remainder is 1.

(iv) x + π

By remainder theorem, if p(x) is divided by linear polynomial x – a, the remainder is p(a).

x + π = 0

or, x = – π.

p(-π) = (-π)³ + 3(-π)² + 3(-π) + 1

= -π³ + 3π² – 3π + 1

Therefore, when x³ + 3x² + 3x + 1 is divided by x + π, the remainder is -π³ + 3π² – 3π + 1.

(v) 5 + 2x

By remainder theorem, if p(x) is divided by linear polynomial x – a, the remainder is p(a).

5 + 2x = 0 or x = -5/2

p(-5/2) = [((-5)/2)³ + 3((-5)/2)² + 3((-5)/2) + 1]

= [((-125)/8) + (75/4) + ((-15)/2) + 1]

= (-125 + 150 – 60 + 8)/8

= (-27)/8

Therefore, when x³ + 3x² + 3x + 1 is divided by 5 + 2x, the remainder is (-27)/8.

Summary: The remainder when x³ + 3x² + 3x + 1 is divided by (i) x + 1, (ii) x – 1/2, (iii) x, (iv) x + π, and (v) 5 + 2x respectively are 0, 27/8, 1, −π³ + 3π² – 3π + 1 and -27/8.

2. Find the remainder when x³ − ax² + 6x − a is divided by x – a.

Answer:

By remainder theorem, if p(x) is divided by linear polynomial x – a, the remainder is p(a).

x – a = 0

 or, x = a

p(a) = a³ – a(a)² + 6(a) – a

= a³ – a³ + 5a

= 5a Therefore, when x³ − ax² + 6x − a is divided by x – a, the remainder is 5a.

3. Check whether 7 + 3x is a factor of 3x³ + 7x.

Answer:

By remainder theorem, if p(x) is divided by linear polynomial x – a, the remainder is p(a).

7 + 3x = 0

or, x = -7/3

p(-7/3) = 3(-7/3)³ + 7(-7/3)

= (-343)/9 – 49/3

= (-343-49×3)/9

= – 490/9 ≠ 0

Therefore, 7 + 3x is not a factor of 3x³ + 7x since the remainder = – 490/9 ≠ 0.

Extra Questions to Complement Solutions to NCERT Class 9 Mathematics Chapter 2 Polynomials:

(A) Multiple Choice Questions (MCQ):

1. A rectangle has length x units and breadth 4 units. The area of the rectangle is a polynomial of degree:

(A) 1
(B) 2
(C) 0
(D) Not a polynomial

Answer: (A) 1

Area of the rectangle = Length × Breadth = 4x, which is a polynomial of degree 1 (highest power of x is 1).

2. The degree of the polynomial 0 is:

(A) 1
(B) 0
(C) Undefined
(D) Infinity

Answer: (C) Undefined

3. Degree of the polynomial 4x⁴ + 0x³ + 0x⁵ + 5x + 7 is (NCERT Exemplar)

(A) 4
(B) 5
(C) 3
(D) 7

Answer: (A) 4

4x⁴ + 0x³ + 0x⁵ + 5x + 7 = 4x⁴ + 5x + 7 The highest power of x is 4. So, the degree is 4.

4. Which of the following is a polynomial?

(A) 1/x³
(B) √2x³
(C) x – 1/x
(D) √x
Answer: (B) √2x³

The exponents of the terms have to be whole numbers. Therefore, √2×3 is the only polynomial.

5. P(x) = x² + x + 1. Find the value of [P(0)]².

(A) 4
(B) 1
(C) 0
(D) 2

Answer: (B) 1

P(0) = 0² + 0 + 1 = 1 [P(0)]² = P(0) × P(0) = 1 × 1 = 1

6. If x – 1 perfectly divides x² + kx + x + 1, what is the value of k?

(A) -2
(B) -3
(C) 1
(D) -1

Answer: (B) -3

By Remainder Theorem, if x – 1 exactly divides p(x), then p(1) = 0.

p(x) = x² + kx + x + 1

or, p(1) = 1² + k.1 + 1 + 1 = 0 or, k = -3

7. What is the remainder when 2x² – 1 is divided by x?

(A) 1
(B) 2x² – 1
(C) -1
(D) 0

Answer: (C) -1

x = x – 0

Therefore, by Remainder Theorem p(0) = r.

p(x) = 2x² – 1

or, p(0) = – 1 = r

8. The value of 300² – 299² is:

(A) 499
(B) 301
(C) 599
(D) 1

Answer: (C) 599

Using the identity a² – b² = (a + b)(a – b) we get:

300² – 299² = (300 + 299)(300 – 299) = 599.1 = 599

9. Which of the following is a factor of (2x + 1)² – (2x – 1)²?

(A) 9x
(B) 4x
(C) 3x
(D) 10x

Answer: (B) 4x

Using the identity a² – b² = (a + b)(a – b) we get:

(2x + 1)² – (2x – 1)²

= (2x + 1 + 2x – 1)(2x + 1 – 2x + 1)

= 4x × 2

= 8x

4x is a factor of 8x.

10. Which of the following can be expressed as the sum of two cubes:

(A) (3x + 1)(4x² – 2x + 1)
(B) (2x + 1)(4x² – 2x + 1)
(C) (2x + 1)(4x² – 4x + 1)
(D) (x + 1)(4x² – 2x + 1)

Answer: (B) (2x + 1)(4x² – 2x + 1)

We know,

(x + y)(x² – xy + y²) = x³ + y³

Here x = 2x and y = 1

(2x + 1)(4x² – 2x + 1) = 8x³ + 1³ (Sum of two cubes)

11. If x = -y, which of the following is true:

(A) x³ – y³ – 3xy(x – y) = 0
(B) x³ – y³ + 3xy(x – y) = 0
(C) x³ + y³ + 3xy(x + y) = 0
(D) x³ + y³ – 3xy(x – y) = 0

Answer: (C) x³ + y³ + 3xy(x + y) = 0

x = -y

or, x + y = 0

or, (x + y)³ = 0

or, x³ + y³ + 3xy(x + y) = 0

12. The coefficient of x² in the expansion of (2x + 1)³ is:

(A) 12
(B) 6
(C) 9
(D) 3

Answer: (A) 12

(2x + 1)³

= (2x)³ + 3(2x)²(1) + 3(2x)(1)² + 1³

= 8x³ + 12x² + 6x + 1

13. Factorise x² – x – 12:

(A) (x + 1)(x – 12)
(B) (x + 3)(x – 4)
(C) (x – 3)(x – 4)
(D) (x – 3)(x + 4)

Answer: (B) (x + 3)(x – 4)

x² – x – 12

p + q = – 1 and pq = -12

By trial and error we get,

p = – 4 and q = 3

x² – x – 12

= x² +(-4 + 3)x – 12

= x² – 4x + 3x – 12

= x(x – 4) + 3(x – 4) = (x + 3)(x – 4)

14. Find the value of (3x+3y)³/(x+y)³

(A) 9
(B) 8
(C) 27
(D) Undeterminable

Answer: (C) 27

(3x+3y)³/(x+y)³ =[3(x+y)]³/(x+y)³ = 3³ (x+y)³/(x+y)³ = 27

15. Find the value of 1/102³ without actually multiplying:

(A) 1/1000000
(B) 1/1001208
(C) 1/1061208
(D) 1/1041208

102³ = (100 + 2)³ = (100)³ + 3(100)²(2) + 3(100)(2)² + 2³

= 1000000 + 60000 + 1200 + 8

= 1061208

(B) Short Answer Type Questions:

Is x² – 3x + 1 a multiple of x – 1/2?

Answer:

p(x) = x² – 3x + 1

p(1/2) = (1/2)² – 3(1/2) + 1

= 1/4 – 3/2 + 1

= (1-6+4)/4

= (-1)/4

Since p(1/2) ≠ 0, there is a remainder when x² – 3x + 1 is divided by x – 1/2.

We can say that x² – 3x + 1 is not a multiple of x – 1/2.

2. Factorise: p³ + 6 + 12/p³ + 8/p⁶

Answer:

p³ + 6 + 12/p³ + 8/p⁶

= p³ + 3p²(2/p²) + 3p(2/p²)² + (2/p²)³

= (p3 + 2/p²)² [Factorised]

3. Express as the sum of two cubes and factorise: 3√3a³ + 3√3b³

Answer:

3√3a³ + 3√3b³

= (√3a)³ + (√3b)³ (Expressed as the sum of two cubes)

= (√3a + √3b)[(√3a)² – (√3a)(√3b) + (√3b)²]

= (√3a + √3b)(3a² – 3ab + 3b²)

= 3√3 (a + b)(a² – ab + b²) [Factorised]

4. Without finding the cubes, factorise:

(x – y)³ + (y – z)³ + (z – x)³

Answer:

We use identity x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx)

Here, x = x – y, y = y – z, z = z – x

Here, (x – y) + (y – z) + (z – x) = 0

Therefore, (x – y)³ + (y – z)³ + (z – x)³ – 3(x – y)(y – z)(z – x) = 0 or, (x – y)³ + (y – z)³ + (z – x)³ = 3(x – y)(y – z)(z – x) (Factorised)

5. Express the following as the sum of two squares:

(p + 1)³ – (p – 1)³

= p³ + 3p²(1) + 3p(1)² + 1³ – (p³ – 3p²(1) + 3p(1)² – 1³)

= p³ + 3p² + 3p + 1 – p³ + 3p² – 3p + 1

= 6p² + 2

= (√6p)² + (√2)² [Sum of two squares]

6. Express the following as a cube:

(1 – x)³ + (1 – y)³ – 3(x + y – 2)(1 – x)(1 – y)

Answer:

(1 – x)³ + (1 – y)³ – 3(x + y – 2)(1 – x)(1 – y)

= (1 – x)³ + (1 – y)³ + 3(1 – x)(1 – y)(2 – x – y)

= (1 – x)³ + (1 – y)³ + 3(1 – x)(1 – y)[(1 – x) + (1 – y)]

We use the identity x³ + y³ + 3xy(x + y) = (x + y)³.

Here, x = 1 – x, y = 1 – y.

(1 – x)³ + (1 – y)³ + 3(1 – x)(1 – y)[(1 – x) + (1 – y)]

= (1 – x + 1 – y)³ = (2 – x – y)³ [Expressed as a cube)

7. Simplify the expression (3x³-24)/(x-2) where x ≠ 2.

Answer:

(3x³-24)/(x-2)

= (3(x³-8))/(x-2)

= (3(x-2)(x²+x.2+ 2²))/(x-2)

= 3(x² + 2x + 4)

8. Simplify the expression: (x²+x-6)/(x²+5x+6) where x≠-2, -3.

Answer:

(x²+x-6)/(x²+5x+6)

For x² + x – 6 we get:

p + q = 1 and pq = -1 × 6

By trial and error we get p = 3, q = -2.

For x² + 5x + 6 we get:

p + q = 5, pq = 1 × 6

By trial and error we get p = 3, q = 2.

(x²+x-6)/(x²+5x+6)

= (x²+(3-2)x-6)/(x²+(3+2)x+6)

= (x²+3x-2x-6)/(x²+3x+2x+6)

= (x(x+3)-2(x+3))/(x(x+3)+2(x+3))

= (x+3)(x-2)/(x+3)(x+2)

= (x-2)/(x+2) (Answer)

9. Find the value of x³ + y³ – 27xy + 8, when x + y = – 9.

Answer:

x³ + y³ – 27xy + 8

= x³ + y³ + 3xy(-9) + 8

= x³ + y³ + 3xy(x + y) + 8 [Since x + y = – 9]

= (x + y)³ + 8

= (-9)³ + 8

= -729 + 8 = -721 (Answer)

10. If y + 1 = x, find 2y³ + 6y² + 6y + 54 in terms of x.

Answer:

2y³ + 6y² + 6y + 54

= 2(y³ + 3y² + 3y + 27)

= 2(y³ + 3y².1 + 3y.1² + 1³ + 26)

= 2(y + 1)³ + 2 × 26

= 2(y + 1)³ + 52

= 2x³ + 52 (Answer)

(C) Long Answer Type Questions:

1. If a + b = 4 and ab = 2, find the value of (1/a + 1/b)³ – 1/a³ – 1/b³ .

Answer:

(1/a + 1/b)³ – 1/a³ – 1/b³ .

= 1/a³ + 1/b³ + 3(1/a)( 1/b)(1/a + 1/b) – 1/a³ – 1/b³

= (3/ab)(1/a + 1/b)

= 3/ab((a + b)/ab)

= (3(a + b))/(a² b² )

Putting a + b = 4 and ab = 2 we get:

(3(a + b))/(a² b²) = (3 × 4 )/4 = 3

Therefore,

(1/a + 1/b)³ – 1/a³ – 1/b³ = 3

2. If x + 1/x = 3, find the value of (x⁶+1+3x⁴+3x²)/x³ .

Answer:

(x⁶+1+3x⁴+3x²)/x³

= x⁶/x³ + 1/x³ +3x⁴/x³ + 3x²/x³

= x³ + 1/x³ + 3x + 3/x

= x³ + 1/x³ + 3×1/x (x + 1/x)

= (x + 1/x)³

= 3³

= 27

3. If x – 1 is a factor of x² – px – p and x – 2 is a factor of x² + qx + 2, find the value of p + q.

Answer:

p(x) = x² – px – p

By remainder theorem,

p(1) = 1² – p(1) – p = 0

or, 2p = 1

or, p = 1/2

q(x) = x² + qx + 2

By remainder theorem,

q(2) = 2² + q(2) + 2 = 0

or, 2q = – 6

or, q = -3

Therefore, p + q = 1/2 – 3 = -5/2

4. Express the following as the sum of squares:

2(x – 1)² + 2(y – 1)² + 2(z – 1)² 2(x – 1)(y – 1) + 2(y – 1)(z – 1) + 2 (z – 1)(x – 1)

Answer:

2(x – 1)² + 2(y – 1)² + 2(z – 1)² + 2(x – 1)(y – 1) + 2(y – 1)(z – 1) + 2 (z – 1)(x – 1)

= (x – 1)² + 2(x – 1)(y – 1) + (y – 1)² + (y – 1)² + 2(y – 1)(z – 1) + (z – 1)² + (z – 1)² + 2 (z – 1)(x – 1) + (x – 1)²

Using identity a² + 2ab + b² = (a + b)² we get:

= ( x – 1 + y – 1)² + (y – 1 + z – 1)² + (z – 1 + x – 1)² = (x + y – 2)² + (y + z – 2)² + (z + x – 2)² [Expressed as the sum of three squares]

(D) Fill in the Blanks:

(a) A polynomial of degree __________ is called a linear polynomial.

(b) A polynomial with three terms is called a _________.

(c) In the polynomial a₀ + a₁x + a₂x², if a₀ = a₁ = a₂ = 0 then the polynomial is a __________ polynomial.

(d) The polynomial x² + xy + z² has __________ variables.

(e) The expression (x – 2)(x² + 5x + 6) has __________ factors.

Answers:

(a) A polynomial of degree 1 is called a linear polynomial.

(b) A polynomial with three terms is called a trinomial.

(c) In the polynomial a₀ + a₁x + a₂x², if a₀ = a₁ = a₂ = 0 then the polynomial is a zero polynomial.

(d) The polynomial x² + xy + z² has three variables.

(e) The expression (x – 2)(x² + 5x + 6) has three factors.

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Frequently Asked Questions (FAQs) on NCERT Solutions to Class 9 Mathematics Chapter 2 Polynomials: