Polynomials

Polynomials are algebraic expressions containing one or more terms with non-negative integer exponents. They are built using variables, constants, and arithmetic operations (addition, subtraction, and multiplication).

Key Concepts

• Terms: The components of a polynomial separated by addition or subtraction.
• Coefficients: The numerical factors of the terms.
• Variables: The alphabetical symbols representing unknown values.
• Types of Polynomials: Based on the number of terms (monomial, binomial, trinomial) and degree (linear, quadratic, cubic).

Operations on Polynomials

• Addition and Subtraction: Combining like terms.
• Multiplication: Using the distributive property.
• Division: Long division or synthetic division (for dividing by linear expressions).

Important Theorems

• Remainder Theorem: 

When a polynomial p(x) is divided by (x – a), the remainder is p(a).

• Factor Theorem: If p(a) = 0, then (x – a) is a factor of p(x).

Applications

Polynomials are used in various fields, including algebra, calculus, and engineering. They are essential for modeling real-world phenomena and solving equations.

In essence, the chapter on polynomials lays the foundation for understanding algebraic expressions and their manipulation, which is crucial for further mathematical studies.

Exercise 2.1

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

(i) 4×2 – 3x + 7

(ii) y2 + √2

(iii) 3 √t + t√2

(iv) y+ 2/y

(v) x10+ y3+t50

Ans : 

i) 4x² – 3x + 7

• Contains only one variable (x)
• All exponents of x are non-negative integers (2, 1, and 0)
• Polynomial in one variable

ii) y² + √2

• Contains only one variable (y)
• The exponent of y is a non-negative integer (2)
• Polynomial in one variable

iii) 3√t + t√2

• Contains only one variable (t)
• However, the exponent of t in the first term is 1/2 (since √t = t^(1/2)) which is not a non-negative integer.
• Not a polynomial

iv) y + 2/y

• Contains only one variable (y)
• But, the second term can be written as 2y^(-1), where the exponent is -1, which is not a non-negative integer.
• Not a polynomial

v) x^10 + y^3 + t^50

• Contains three variables (x, y, and t)
• A polynomial must contain only one variable.
• Not a polynomial 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) √2 x – 1

Ans : 

(i) 2 + x² + x

• The coefficient of x² is 1.

(ii) 2 – x² + x³

• The coefficient of x² is -1.

(iii) (π/2)x² + x

• The coefficient of x² is π/2.

(iv) √2x – 1

• There is no x² term in this expression. So, the coefficient of x² is 0.

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

Ans : 

Binomial of degree 35:

• x^35 – 7

A binomial has two terms, and the highest power (degree) of the variable is 35 in this example.

Monomial of degree 100:

• 5y^100

A monomial has only one term, and the degree is determined by the highest power of the variable, which is 100 in this case.

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

(i) 5x3+4x2 + 7x

(ii) 4 – y2

(iii) 5t – √7

(iv) 3

Ans : 

(i) 5x³ + 4x² + 7x

• The highest power of x is 3.
• Degree: 3

(ii) 4 – y²

• The highest power of y is 2.
• Degree: 2

(iii) 5t – √7

• The highest power of t is 1 (as 5t can be written as 5t¹).
• Degree: 1

(iv) 3

• The constant term can be written as 3x⁰.
• The highest power of x is 0.
• Degree: 0

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

Ans : 

Classifying the Given Polynomials

(i) x² + x

• Degree: 2 (highest power of x)
• Quadratic polynomial

(ii) x – x³

• Degree: 3 (highest power of x)
• Cubic polynomial

(iii) y + y² + 4

• Degree: 2 (highest power of y)
• Quadratic polynomial

(iv) 1 + x

• Degree: 1 (highest power of x)
• Linear polynomial

(v) 3t

• Degree: 1 (highest power of t)
• Linear polynomial

(vi) r²

• Degree: 2 (highest power of r)
• Quadratic polynomial

(vii) 7x³

• Degree: 3 (highest power of x)
• Cubic polynomial

Exercise 2.2

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

(i) x = 0

(ii) x = – 1

(iii) x = 2

Ans : 

(i) x = 0

• 5(0) – 4(0)² + 3 = 0 – 0 + 3 = 3

(ii) x = -1

Substituting x = -1 in the polynomial, we get:

• 5(-1) – 4(-1)² + 3 = -5 – 4 + 3 = -6

(iii) x = 2

Substituting x = 2 in the polynomial, we get:

• 5(2) – 4(2)² + 3 = 10 – 16 + 3 = -3

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 +1 + 2t2 -t3

(iii) P (x) = x3

(iv) p (x) = (x-1) (x+1)

Ans : 

i) p(y) = y² – y + 1

• p(0) = (0)² – (0) + 1 = 1
• p(1) = (1)² – (1) + 1 = 1
• p(2) = (2)² – (2) + 1 = 4 – 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) = (1)³ = 1
• p(2) = (2)³ = 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

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

(i) p(x) = 3x + 1,x = – 13

(ii) p (x) = 5x – π, x = 45

(iii) p (x) = x2 – 1, x = x – 1

(iv) p (x) = (x + 1) (x – 2), x = – 1,2

(v) p (x) = x2, x = 0

(vi) p (x) = 1x + m, x = – m1

(vii) P (x) = 3×2 – 1, x = –13 , 23

(viii) p (x) = 2x + 1, x = 12

Ans : 

i) p(x) = 3x + 1, x = -1/3

p(-1/3) = 3(-1/3) + 1 = -1 + 1 = 0

Yes, x = -1/3 is a zero of p(x).

ii) p(x) = 5x – π, x = 4/5

p(4/5) = 5(4/5) – π 

= 4 – π ≠ 0

No, x = 4/5 is not a zero of p(x).

iii) 

p(1) = (1)² – 1 = 1 – 1 = 0

p(-1) = (-1)² – 1 = 1 – 1 = 0

Yes, both x = 1 and x = -1 are zeroes of p(x).

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

Yes, both x = -1 and

 x = 2 are zeroes of p(x).

v) p(x) = x², x = 0

p(0) = (0)² = 0

Yes, x = 0 is a zero of p(x).

vi) p(x) = lx + m, x = -m/l

p(-m/l) = l(-m/l) + m = -m + m = 0

Yes, x = -m/l is a zero of p(x).

vii) 

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 ≠ 0

Only x = -1/√3 is a zero of p(x).

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

p(1/2) = 2(1/2) + 1 = 1 + 1 = 2 ≠ 0

No, x = 1/2 is not a zero of p(x).

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 where c and d are real numbers.

Ans : 

i) p(x) = x + 5 To find the zero, we set p(x) = 0: x + 5 = 0 x = -5

ii) p(x) = x – 5 x – 5 = 0 x

 = 5

iii) p(x) = 2x + 5 2x + 5 = 0 2x = -5 x = -5/2

iv) p(x) = 3x – 2 3x – 2 = 0 3x = 2 x = 2/3

v) p(x) = 3x 3x = 0 x = 0

vi) p(x) = ax, a ≠ 0 ax = 0 Since a ≠ 0, x must be 0 for the product to be 0. x = 0

vii) p(x) = cx + d, c ≠ 0 cx + d = 0 cx = -d x = -d/c

Exercise 2.3

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

Ans : 

i) p(x) = x³ + x² + x + 1

• p(-1) = (-1)³ + (-1)² + (-1) + 1 = -1 + 1 – 1 + 1 = 0
• (x + 1) is a factor.

ii) p(x) = x⁴ + x³ + x² + x + 1

• p(-1) = (-1)⁴ + (-1)³ + (-1)² + (-1) + 1 = 1 – 1 + 1 – 1 + 1 = 1
• (x + 1) is not a factor.

iii) 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
• (x + 1) is not a factor.

iv) 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 ≠ 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

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

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

Ans : 

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

• For g(x) = x + 1, x = -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) p(x) = x³ + 3x² + 3x + 1, g(x) = x + 2

• For g(x) = x + 2, x = -2.
• p(-2) = (-2)³ + 3(-2)² + 3(-2) + 1 = -8 + 12 – 6 + 1 = -1
• Since p(-2) ≠ 0,
•  g(x) is not a factor of p(x).

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

• For g(x) = x – 3, x = 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).

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 – √2 x + 1

(iv) p (x) = kx2 – 3x + k

Ans : 

i) p(x) = x² + x + k If (x – 1) is a factor, then p(1) = 0. So, (1)² + (1) + k = 0 1 + 1 + k = 0 k = -2

ii) p(x) = 2x² + kx + √2 If (x – 1) is a factor, then p(1) = 0. So, 2(1)² + k(1) + √2 = 0 2 + k + √2 = 0 k = -2 – √2

iii) p(x) = kx² – √2x + 1 If (x – 1) is a factor, then p(1) = 0. So, k(1)² – √2(1) + 1 = 0 k – √2 + 1 = 0 k = √2 – 1

iv) p(x) = kx² – 3x + k If (x – 1) is a factor, then p(1) = 0. So, k(1)² – 3(1) + k = 0 k – 3 + k = 0 2k = 3 k = 3/2

4. Factorise

(i) 12x2 – 7x +1

(ii) 2x2 + 7x + 3

(iii) 6x2 + 5x – 6

(iv) 3x2 – x – 4

Ans : 

i) 12x² – 7x + 1

• We need to find two numbers whose product is 12 and whose sum is -7.
• These numbers are -3 and -4.
• 12x² – 7x + 1 = 12x² – 3x – 4x + 1 = 3x(4x – 1) – 1(4x – 1) = (4x – 1)(3x – 1)

ii) 2x² + 7x + 3

• These numbers are 6 and 1.
• 2x² + 7x + 3 = 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (x + 3)(2x + 1)

iii) 6x² + 5x – 6

• These 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

• We need to find two numbers whose product is -12 and whose sum is -1.
• These numbers are -4 and 3.
• 3x² – x – 4 = 3x² – 4x + 3x – 4 = x(3x – 4) + 1(3x – 4) = (3x – 4)(x + 1)

5. Factorise

(i) x3 – 2x2 – x + 2

(ii) x3 – 3x2 – 9x – 5

(iii) x3 + 13x2 + 32x + 20

(iv) 2y3 + y2 – 2y – 1

Ans : 

i) x³ – 2x² – x + 2

• Dividing x³ – 2x² – x + 2 by (x – 1), we get x² – x – 2.
• Factorizing x² – x – 2, we get (x – 2)(x + 1).
• Therefore, x³ – 2x² – x + 2 = (x – 1)(x – 2)(x + 1)

ii) 

• We can see that p(-1) = 0, so (x + 1) is a factor.
• Dividing x³ – 3x² – 9x – 5 by (x + 1), we get x² – 4x – 5.
• Factorizing x² – 4x – 5, we get (x – 5)(x + 1).
• Therefore, x³ – 3x² – 9x – 5 = (x + 1)(x – 5)(x + 1) = (x + 1)²(x – 5)

iii) x³ + 13x² + 32x + 20

• We can see that p(-1) = 0, so (x + 1) is a factor.
• Dividing x³ + 13x² + 32x + 20 by (x + 1), we get x² + 12x + 20.
• Factorizing x² + 12x + 20, we get (x + 10)(x + 2).
• Therefore, x³ + 13x² + 32x + 20 = (x + 1)(x + 10)(x + 2) = (x + 1)²(x + 10)

iv) 2y³ + y² – 2y – 1

• We can group the terms: (2y³ + y²) – (2y + 1)
• Factor out common terms: y²(2y + 1) – 1(2y + 1)
• Factor out (2y + 1): (2y + 1)(y² – 1)
• Factorize y² – 1 as a difference of squares: (2y + 1)(y + 1)(y – 1)
• Therefore, 2y³ + y² – 2y – 1 = (y – 1)(2y + 1)(y + 1)

Exercise 2.4

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)

Ans : 

i) (x + 4)(x + 10) Using the identity (a + b)(a + c), we get:

• x² + x(4 + 10) + 4 * 10
• x² + 14x + 40

ii) (x + 8)(x – 10) Using the identity (a + b)(a – b) = a² – b², we get:

• x² – 10²
• x² – 100

iii) (3x + 4)(3x – 5) Using the identity (a + b)(a – b) = a² – b², we get:

• (3x)² – (5)²
• 9x² – 25

iv) (y² + 3/2)(y² – 3/2) Using the identity (a + b)(a – b) = a² – b², we get:

• (y²)² – (3/2)²
• y⁴ – 9/4

v) (3 – 2x)(3 + 2x) Rearranging the terms to match the identity, we get:

• (3 + (-2x))(3 – (-2x)) Using the identity (a + b)(a – b) = a² – b², we get:
• 3² – (-2x)²
• 9 – 4x²

2. Evaluate the following products without multiplying directly

(i) 103 x 107

(ii) 95 x 96

(iii) 104 x 96

Ans : 

i) 103 x 107

• We can write 103 as (100 + 3) and 107 as (100 + 7)
• So, 103 x 107 = (100 + 3)(100 + 7)
• Applying the identity, we get:
  • 100² + (3 + 7) * 100 + (3 * 7)
  • 10000 + 1000 + 21
  • 11021

ii) 95 x 96

• We can write 95 as (100 – 5) and 96 as (100 – 4)
• So, 95 x 96 = (100 – 5)(100 – 4)
• Applying the identity, we get:
  • 100² + (-5 – 4) * 100 + (-5 * -4)
  • 10000 – 900 + 20
  • 9120

iii) 104 x 96

• We can write 104 as (100 + 4) and 96 as (100 – 4)
• This is in the form (a + b)(a – b), which equals a² – b²
• So, 104 x 96 = (100 + 4)(100 – 4) = 100² – 4² = 10000 – 16
•  = 9984

3. Factorise the following using appropriate identities

(i) 9x2 + 6xy + y2

(ii) 4y2-4y + 1

(iii) x2 –  – y2/100

Ans : 

i) 9x² + 6xy + y²

• This expression can be written as (3x)² + 2(3x)(y) + y²
• Using the identity (a + b)² = a² + 2ab + b², we get:
  • (3x + y)²

ii) 4y² – 4y + 1

• This expression can be written as (2y)² – 2(2y)(1) + 1²
• Using the identity (a – b)² = a² – 2ab + b², we get:
  • (2y – 1)²

iii) x² – y²/100

• This expression can be written as x² – (y/10)²
• Using the identity a² – b² = (a + b)(a – b), we get:
  • (x + y/10)(x – y/10)

4. Expand each of the following, using suitable identity

(i) (x+2y+ 4z)2

(ii) (2x – y + z)2

(iii) (- 2x + 3y + 2z)2

(iv) (3a -7b – c)z

(v) (- 2x + 5y – 3z)2

(vi)  [(1/4)a – (1/4)b + 1]²

Ans : 

i) (x + 2y + 4z)² Applying the identity, we get:

• x² + (2y)² + (4z)² + 2(x)(2y) + 2(2y)(4z) + 2(x)(4z)
• x² + 4y² + 16z² + 4xy + 16yz + 8xz

ii) (2x – y + z)² We can rewrite this as (2x + (-y) + z)² Applying the identity, we get:

• (2x)² + (-y)² + z² + 2(2x)(-y) + 2(-y)(z) + 2(2x)(z)
• 4x² + y² + z² – 4xy – 2yz + 4xz

iii) (-2x + 3y + 2z)² We can rewrite this as ((-2x) + 3y + 2z)² Applying the identity, we get:

• (-2x)² + (3y)² + (2z)² + 2(-2x)(3y) + 2(3y)(2z) + 2(-2x)(2z)
• 4x² + 9y² + 4z² – 12xy + 12yz – 8xz

iv) (3a – 7b – c)z This is a simple multiplication, not an expansion using identities:

• 3az – 7bz – cz

v) (-2x + 5y – 3z)² Applying the identity, we get:

• (-2x)² + (5y)² + (-3z)² + 2(-2x)(5y) + 2(5y)(-3z) + 2(-2x)(-3z)
• 4x² + 25y² + 9z² – 20xy – 30yz + 12xz

vi) [(1/4)a – (1/4)b + 1]² Applying the identity, we get:

• (1/4a)² + (-1/4b)² + 1² + 2(1/4a)(-1/4b) + 2(-1/4b)(1) + 2(1/4a)(1)
• a²/16 + b²/16 + 1 – ab/8 – b/2 + a/2

5. Factorise

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

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

Ans : 

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

• We can observe that the given expression is in the form of a² + b² + c² + 2ab + 2bc + 2ca
• Comparing with the identity, we can see that a = 2x, b = 3y, and c = -4z
• Therefore, the expression can be factored as (2x + 3y – 4z)²

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

• We can rewrite the expression as: (√2x)² + y² + (2√2z)² + 2(√2x)(y) + 2(y)(2√2z) + 2(√2x)(2√2z)
• This is in the form of a² + b² + c² + 2ab + 2bc + 2ca
• Comparing with the identity, we can see that a = √2x, b = y, and c = 2√2z
• Therefore, the expression can be factored as (√2x + y + 2√2z)²

6. Write the following cubes in expanded form

Ans : 

We’ll use the following identity:

• (a + b)³ = a³ + 3a²b + 3ab² + b³
• (a – b)³ = a³ – 3a²b + 3ab² – b³

Expanding the Expressions

i) (2x + 1)³ Applying the identity, we get:

• (2x)³ + 3(2x)²(1) + 3(2x)(1)² + 1³
• 8x³ + 12x² + 6x + 1

ii) (2a – 3b)³ Applying the identity for (a – b)³, we get:

• (2a)³ – 3(2a)²(3b) + 3(2a)(3b)² – (3b)³
• 8a³ – 36a²b + 54ab² – 27b³

iii) ((3/2)x + 1)³ Applying the identity for (a + b)³, we get:

• (3/2x)³ + 3(3/2x)²(1) + 3(3/2x)(1)² + 1³
• (27/8)x³ + (27/4)x² + (9/2)x + 1

7. Evaluate the following using suitable identities

(i) (99)3

(ii) (102)3

(iii) (998)3

Ans : 

We will use the identity:

• (a – b)³ = a³ – b³ – 3ab(a – b)
• (a + b)³ = a³ + b³ + 3ab(a + b)

Evaluating the Expressions

i) (99)³

• We can write 99 as (100 – 1)
• Using the identity (a – b)³, we get:
  • (100 – 1)³ = 100³ – 1³ – 3 * 100 * 1 (100 – 1) = 1000000 – 1 – 30000 + 300 = 970299

ii) (102)³

• We can write 102 as (100 + 2)
• Using the identity (a + b)³, we get:
  • (100 + 2)³ = 100³ + 2³ + 3 * 100 * 2 (100 + 2) = 1000000 + 8 + 60000 + 1200 

= 1061208

iii) (998)³

• We can write 998 as (1000 – 2)
• Using the identity (a – b)³, we get:
  • (1000 – 2)³ = 1000³ – 2³ – 3 * 1000 * 2 (1000 – 2) = 1000000000 – 8 – 6000000 + 12000 

= 994011992

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) 27p 3 – 1/ 216 – 9 / 2 p2+1/ 4 p 

Ans : 

We’ll primarily use the following identities:

• (a + b)³ = a³ + b³ + 3ab(a + b)
• (a – b)³ = a³ – b³ – 3ab(a – b)

Factorization

i) 8a³ + b³ + 12a²b + 6ab²

• Rearrange the terms: 8a³ + b³ + 6ab(2a + b)
• This matches the expansion of (a + b)³
• Therefore, the factorization is (2a + b)³

ii) 8a³ – b³ – 12a²b + 6ab²

• Rearrange the terms: 8a³ – b³ – 6ab(2a – b)
• This matches the expansion of (a – b)³
• Therefore, the factorization is (2a – b)³

iii) 27 – 125a³ – 135a + 225a²

• Rearrange the terms: 27 + 225a² – 125a³ – 135a
• This can be written as 3³ + (15a)² – 3(3)(15a)(3 – 5a)
• This matches the expansion of (a – b)³
• Therefore, the factorization is (3 – 5a)³

iv) 64a³ – 27b³ – 144a²b + 108ab²

• Rearrange the terms: 64a³ – 27b³ – 36ab(4a – 3b)
• This matches the expansion of (a – b)³
• Therefore, the factorization is (4a – 3b)³

v) 27p³ – 1/216 – 9/2 * p²

• Rewrite the expression as: (3p)³ – (1/6)³ – 3(3p)(1/6)
• This matches the expansion of (a – b)³
• Therefore, the factorization is (3p – 1/6)³

9. Verify

(i) x3 + y3 = (x + y)-(x2 – xy + y2)

(ii) x3 – y3 = (x – y) (x2 + xy + y2)

Ans : 

i) Let’s expand the right-hand side:

• (x + y)(x² – xy + y²) = x(x² – xy + y²) + y(x² – xy + y²)
•  = x³ – x²y + xy² + x²y – xy² + y³ 
• = x³ + y³

Hence, the equation is verified.

ii)  Let’s expand the right-hand side:

• (x – y)(x² + xy + y²) = x(x² + xy + y²) – y(x² + xy + y²) 
• = x³ + x²y + xy² – x²y – xy² – y³ 
• = x³ – y³

10. Factorise each of the following

(i) 27y3 + 125z3

(ii) 64m3 – 343n3

[Hint See question 9]

Ans : 

i) 27y³ + 125z³

• This can be written as (3y)³ + (5z)³
• Using the identity a³ + b³ = (a + b)(a² – ab + b²), we get:
  • (3y + 5z)(9y² – 15yz + 25z²)

ii) 64m³ – 343n³

• This can be written as (4m)³ – (7n)³
• Using the identity a³ – b³ = (a – b)(a² + ab + b²), we get:
  • (4m – 7n)(16m² + 28mn + 49n²)

11. Factorise 27x3 +y3 +z3 -9xyz.

Ans : 

The given expression is 27x³ + y³ + z³ – 9xyz.

We can rewrite it as (3x)³ + y³ + z³ – 3(3x)(y)(z).

Comparing this with the identity, we can see that a = 3x, b = y, and c = z.

Therefore, the factorization of 27x³ + y³ + z³ – 9xyz is:

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

12. Verify that

x3 +y3 +z3 – 3xyz = 1/2 (x + y+z)[(x-y)2 + (y – z)2 +(z – x)2]

Ans : 

Left-hand side (LHS): x³ + y³ + z³ – 3xyz

Right-hand side (RHS): (1/2)(x + y + z)[(x – y)² + (y – z)² + (z – x)²]

Let’s expand the RHS and simplify it.

Expanding the RHS:

• (1/2)(x + y + z)[(x² – 2xy + y²) + (y² – 2yz + z²) + (z² – 2zx + x²)]
• (1/2)(x + y + z)[3x² + 3y² + 3z² – 2xy – 2yz – 2zx]
• (1/2)[3x³ + 3y³ + 3z³ – 2x²y – 2xy² – 2x²z – 2y²z – 2yz² – 2z²x + 3x²y + 3y²z + 3z²x]
• (1/2)[3x³ + 3y³ + 3z³ + x²y + xy² + x²z + y²z + yz² + z²x]

Simplifying the RHS:

• (3/2)x³ + (3/2)y³ + (3/2)z³ + (1/2)x²y + (1/2)xy² + (1/2)x²z + (1/2)y²z + (1/2)yz² + (1/2)z²x

Now, let’s group the terms on the RHS to match the LHS:

• (3/2)x³ + (3/2)y³ + (3/2)z³ – (3/2)x²y – (3/2)xy² – (3/2)x²z – (3/2)y²z – (3/2)yz² – (3/2)z²x + 2x²y + 2xy² + 2x²z + 2y²z + 2yz² + 2z²x
• (3/2)x³ + (3/2)y³ + (3/2)z³ + (1/2)x²y + (1/2)xy² + (1/2)x²z + (1/2)y²z + (1/2)yz² + (1/2)z²x

Simplifying further:

• x³ + y³ + z³ – 3xyz

This is equal to the LHS.

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

Ans :

Given: x + y + z = 0

To prove: x³ + y³ + z³ = 3xyz

Proof:

Cube both sides of the given equation:

(x + y + z)³ = 0³

Using the identity (a + b + c)³ = a³ + b³ + c³ + 3(a + b)(b + c)(c + a), 

we get:

x³ + y³ + z³ + 3(x + y)(y + z)(z + x) = 0

Since x + y = -z from the given equation, substitute -z for x + y:

x³ + y³ + z³ + 3(-z)(y + z)(z – x) = 0

Simplify:

x³ + y³ + z³ – 3z(y² – xz + yz – zx) = 0

x³ + y³ + z³ – 3z(y² – x² + yz – zx) = 0

x³ + y³ + z³ – 3z(y – x)(y + x) = 0

Substitute -z for x + y again:

x³ + y³ + z³ – 3z(y – x)(-z) = 0

x³ + y³ + z³ + 3xyz = 0

Rearrange the terms:

x³ + y³ + z³ = 3xyz

Hence proved.

Therefore, if x + y + z = 0, then 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

Ans : 

i) (-12)³ + (7)³ + (5)³

Here, a = -12, 

b = 7, and

 c = 5.

We can see that a + b + c = -12 + 7 + 5 = 0.

Therefore, (-12)³ + (7)³ + (5)³ = 3(-12)(7)(5) = -1260.

ii) (28)³ + (-15)³ + (-13)³

Here, a = 28, 

b = -15, and 

c = -13.

We can see that a + b + c = 28 – 15 – 13 = 0.

Therefore, (28)³ + (-15)³ + (-13)³ = 3(28)(-15)(-13) = 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

Ans : 

i) Area: 25a² – 35a + 12

To find the length and breadth, we need to factorize the given expression.

• 25a² – 35a + 12 = 25a² – 20a – 15a + 12 = 5a(5a – 4) – 3(5a – 4) = (5a – 4)(5a – 3)

Therefore, possible expressions for length and breadth are:

• Length = 5a – 4
• Breadth = 5a – 3

ii) Area: 35y² + 13y – 12

• 35y² + 13y – 12 = 35y² + 28y – 15y – 12 = 7y(5y + 4) – 3(5y + 4) = (5y + 4)(7y – 3)

Therefore, possible expressions for length and breadth are:

• Length = 5y + 4
• Breadth = 7y – 3

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

Ans : 

i) Volume: 3x² – 12x

To find the dimensions, we need to factorize the given expression.

3x² – 12x = 3x(x – 4)

Possible dimensions of the cuboid are:

Length = 3x

Breadth = x

Height = x – 4

Note: There can be other possible combinations of dimensions, but the product of these dimensions should always equal the given volume.

ii) Volume: 12ky² + 8ky – 20k

To find the dimensions, we need to factorize the given expression.

12ky² + 8ky – 20k = 4k(3y² + 2y – 5) = 4k(3y – 2)(y + 1)

Possible dimensions of the cuboid are:

Length = 4k

Breadth = 3y – 2

Height = y + 1