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Exercise 2.1 

 

Q1. 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

Solution:

(i) 4x² – 3x + 7 

It is polynomial in one variable. Because the power of variable is natural number. 

(ii) y² + √2 

It is polynomial in one variable. Because the power of variable is natural number. 

(iii)3√t + t√2 

It is not a polynomial in one variable. Because the power of variable is not a natural number, it is a fractional number. 

(iv) y +2/y

It is not a polynomial in one variable.

(v) x¹⁰ + y³ + t^​50

It is not a polynomial in one variable. while It is a polynomial in three variable. 

Q2. 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

Solution: 

(i) 2 + x² + x 

Coefficient of x² = 1 

(ii) 2 – x² + x3 

Coefficient of x² = –1 ​

(iii) π/2 x² + x

Coefficient of x² = π/2

(iv) √2x −1

Coefficients of x² = 0 [Because There is no x² So coefficient will be 0]

Q3. Give one example each of a binomial of degree 35, and of a monomial of degree 100.

Solution: 

A binomial of degree 35 

⇒ 2x³⁵ + 5y ​

Note: Binomial means the expression of two terms. Like x + 5, 3a - 2b, 3t + 7 etc. 

A monomial of degree 100

⇒ 3y¹⁰⁰

Note: Monomial means the expression of single term, like 3x, 5t, y, 3xy etc. 

Q4. Write the degree of each of the following polynomials:
(i) 5x³ + 4x² + 7x

(ii) 4 – y²

(iii) 5t – √7

(iv) 3

Solution: 

(i) 5x³ + 4x² + 7x 

Ans: The degree of polynomial = 3               [Highest power]

(ii) 4 – y^​2

Ans: The degree of polynomial = 2

(iii) 5t – √7 

Ans: The degree of polynomial = 1

(iv) 3

Ans: The degree of polynomial = 0

[Note: There is no variable therefore degree = 0] 

Q5. Classify the following as linear, quadratic and cubic polynomials:

(i) x^​2 + x 

(ii) x – x³ 

(iii) y + y² + 4

(iv) 1 + x

(v) 3t

(vi) r² 

(vii) 7x²

Solution:

(i) x^​2 + x 

Ans: Quadratic polynomial

(ii) x – x³ 

Ans: Cubic polynomial

(iii) y + y² + 4 

Ans: Quadratic polynomial

(iv) 1 + x

Ans : Linear polynomial

(v) 3t 

Ans : Linear polynomial

(vi) r² 

Ans: Quadratic polynomial

(vii) 7x²

Ans: Cubic polynomial

 

2. Polynomials

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Exercise 2.2

Q1. Find the value of the polynomial 5x – 4x² + 3 at

(i) x = 0 (ii) x = –1 (iii) x = 2

Solution:

(i) p(x) = 5x - 4x² + 3

The value of the polynomial p(x) at x = 0 is given by

P(0) = 5(0) - 4(0)² + 3

       = 0 - 0 + 3

       = 3

(ii) p(x) = 5x - 4x² + 3

The value of the polynomial p(x) at x = 1 is given by

P(1) = 5(1) - 4(1)² + 3

        = 5 - 4 + 3

        = 4

(iii) p(x) = 5x - 4x² + 3

The value of the polynomial p(x) at x = 2 is given by

P(2) = 5(2) - 4(2)² + 3

       = 10 -16 + 3

       = - 9

Q2.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 + 2t ² – t ³

(iii) p(x) = x³

(iv) p(x) = (x – 1) (x + 1)

Solution:

(i) p(y) = y² - y + 1

    P(0) = (0)²- 0 + 1

            =1

  P(1) = (1)²- 1 + 1

          = 1 - 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

            = 4

   P(2) = 2 + 2 + 2(2)²- (2)³

            = 4 + 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(1)  =0

       P(2)= (2-1) (2+1)=1(3)  =3 

Q3. Verify whether the following are zeroes of the polynomial, indicated against them.

Solution:  

Q4. Find the zero of the polynomial in each of the following cases: 

(i)  P(x) = x + 5                                

(ii) P(x) = x – 5                           

(iii) Px) = 2x + 5                              

(iv) P(x) = 3x – 2                          

(v) P(x) = 3x                                    

(vi) P(x) = ax, a ≠ 0

Solution (i) :

(i)   P(x) = x + 5

     ⇒ x + 5 = 0

     ⇒ x = - 5 

The zero of the polynomial is - 5. 

 Solution (ii) :

 (ii) P(x) = x – 5

      ⇒ x – 5  = 0

     ⇒ x = 5

 The zero of the polynomial is  5.

 

The zero of the polynomial is - 5/2.

(iv)  P(x) = 3x - 2

       3x - 2 = 0

              

The zero of the polynomial is 2/3.

  

Chapter 2. Polynomials 

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Exercise 2.3

​

Solution: 

(i)By remainder theorem, the required remainder is equal to p(x) = (-1)

              P (-1) = x³+ 3x²+ 3x+1

                        = (-1)³ + 3 (-1)² + 3(-1) + 1

                        = -1 + 3 – 3 + 1 = 0 

              Required remainder is p (-1) = 0

          

          Required remainder is p (1/2) = 0

(iv) By remainder theorem, the required remainder is equal to p(x) = - π

                   P(x) = x³+ 3x²+ 3x+1

                   P(π) = (- π)³ + 3(-π)² + 3(-π) + 1

                           = (-π)³ + 3π² + 3π + 1

    Required remainder is p (π) = 0 

 Required remainder is p (- 5/2) = 0 

Q.2. find the remainder  when x³- ax² + 6x - a is divided by x- a.

Solution: Let P(x) = x³- ax²+ 6x- a .   P(x) = a.

                   By remainder theorem 

              P (a) = (a)³- a(a)²+ 6(a) - a

                       =a³-a³ + 6a - a

     Remainder = 5a

Q3. Check whether 7 + 3x is a factor of 3x³ + 7x. 

Solution: 

g(x) = 7 + 3x = 0 

       ⇒ 3x = - 7

       

P(x) ≠ 0

Therefore,  7 + 3x is not a factor of P(x). 

 

Chapter 2. Polynomials

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Exercise 2.4

Q.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                             

Solution:

(i) If (x + 1) is a factor of p(x) = x³+ x²+ x + 1, then p (−1) must be zero, otherwise (x + 1) is not a factor of p(x).

P(x)  = x³ + x² + x + 1

P(−1)= (−1)³ + (−1)² + (−1) + 1

         = − 1 + 1 − 1 − 1

        = 0

∵ P(x) = 0 

Hence, x + 1 is a factor of this polynomial.

(ii) If (x + 1) is a factor of p(x) = x⁴+ x³+ x²+ x + 1, then p (−1) must be zero, Otherwise (x + 1) is not a factor of p(x).

P(x) = x⁴+ x³+ x²+ x + 1

P(−1) = (−1)⁴ + (−1)³ + (−1)² + (−1) + 1

          = 1 − 1 + 1 −1 + 1

          = 1

As P(x) ≠ 0, (− 1)

Therefore, x + 1 is not a factor of this polynomial.

(iii) If (x + 1) is a factor of polynomial p(x) = x⁴+ 3x³+ 3x²+ x + 1, then p (−1) must  be 0, otherwise (x + 1) is not a factor of this polynomial.

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

As P(x) ≠ 0, (−1) 

Therefore (x+1) is not a factor of this polynomial .

Q.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 − 4 x² +  x + 6, g(x) = x − 3 

Solution:

(i)   If g(x) = x + 1 is a factor of the given polynomial p(x), then p (−1) must be zero.

P (x) = 2x³ + x² − 2x − 1

P (−1) = 2(−1)³+ (−1)²− 2(−1) − 1

           = 2(−1) + 1 + 2 – 1   

           = 0

∵ P(x) = 0 

Hence, g(x) = x + 1 is a factor of the given polynomial.

(ii) If g(x) = x + 2 is a factor of the given polynomial p(x), then p (−2) must be 0.

P (x) = x³+3x²+ 3x + 1

P (−2) = (−2)³+ 3(−2)²+ 3(−2) + 1

           = − 8 + 12 − 6 + 1  

           = −1

As P(x) ≠ 0, 

Hence, g(x) = x + 2 is not a factor of the given polynomial.

(iii) If g(x) = x − 3 is a factor of the given polynomial p(x), then P(3) must be 0.

        P(x) = x³− 4 x²+ x + 6

        P(3) = (3)³ − 4(3)² + 3 + 6

                = 27 −36 + 9

                = 0

Hence, g(x) = x − 3 is a factor of the given polynomial. 

Solution:

: If x − 1 is a factor of polynomial p(x), then P(1) must be 0.

(i) P(x) = x²+ x + k

    P(1) = (1)²+ 1 + k

            = 1+1+ 

            = 2+k

        k  =−2 

(iv) P(x) = kx²-3x + k    

      P(1) = k(1)²-3(1) + k

             = k-3+k

        2k = 3

         K = 3/2

Question 4: Factorise: 

(i) 12x²− 7x + 1                           (ii) 2x²+ 7x + 3

(iii) 6x²+ 5x – 6                           (iv) 3x²− x − 4 

Solution:

(i) 12x²− 7x + 1 we can find two numbers,

Such that pq = 12 × 1 = 12 and p + q = −7.

They are p = −4 and q = −3

Here, 12x²− 7x + 1

      = 12x²− 4x − 3x + 1

      = 4x (3x − 1) − 1 (3x − 1)

      = (3x − 1) (4x − 1)

(ii) 2x²+ 7x + 3 we can find two numbers such that pq = 2 × 3= 6 and p + q = 7.

     They are p = 6 and q = 1.

     Here, 2x² + 7x + 3

    = 2x²+ 6x + x + 3

    = 2x (x + 3) + 1 (x + 3)

    = (x + 3) (2x+ 1) 

(iii) 6x²+ 5x − 6 we can find two numbers such that pq = −36 and p + q = 5.

     They are p = 9 and q = −4.

     Here, 6x²+ 5x – 6

    = 6x²+ 9x − 4x – 6

    = 3x (2x + 3) − 2 (2x + 3)

    = (2x + 3) (3x − 2) 

(iv)  3x²− x − 4 we can find two numbers,

such that pq = 3 × (−4) = −12 and p + q = −1.

They are p = −4 and q = 3

Here, 3x²− x − 4

       = 3x²− 4x + 3x – 4

       = x (3x − 4) + 1 (3x − 4)

       = (3x − 4) (x + 1)

Question 5. Factorize:

 (i) x³− 2x²− x + 2                 (ii) x³+ 3x²−9x − 5

(iii) x³+ 13x²+ 32x + 20         (iv) 2y³+ y²− 2y – 1 

Solution:

 (i) Let P(x) = x³− 2x²− x + 2 all the factor are there. These are ± 1, ± 2.

By trial method, P (1) = (1)³− 2(1)²− 1 + 2

         = 1 − 2 − 1+ 2

         = 0 Therefore, (x − 1) is factor of polynomial p(x)

         Let us find the quotient on dividing x³− 2x²− x + 2 by x − 1.

         By long division method

         

                           

   

Now,

Dividend = Divisor × Quotient + remainder  

x³− 2x²− x + 2 = (x – 1) ( X²– x – 2) + 0

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

                        = (x – 1) [x (x–2) + 1(x–2)]

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

(ii)   Let p(x) = x³ – 3x²−9x – 5 all the factor are there. These are ± 1, ± 2.

        By trial method, p (–1) = (–1)³– 3(1)²− 9(1) – 5

              = –1– 3–9–5     =0

        Therefore (x+1) is the factor of polynomial p(x).

       .   Let us find the quotient on dividing x³– 3x²−9x – 5 by x+1.

            By long division method

                                    

Now,

Dividend = Divisor × Quotient + remainder  

x³– 3x²−9x – 5 = (x +1) ( X²–4x – 5) + 0

                       =(x + 1) (x²–5x+x–5)

                        =(x + 1) [x (x–5) +1(x–5)]

                        =(x + 1) (x + 1) (x – 5) 

(iii)  Let p(x) = x³+ 13x²+ 32x + 20    all the factor are there.

These are ± 1, ± 2, ± 3, ± 4.

       By trial method, p (–1) = (–1)³+13(–1)²+ 32(–1) +20

                                           = –1+13–32+20   = 0    

       Therefore (x+1) is the factor of polynomial p(x).

        Let us find the quotient on dividing x³+ 13x²+ 32x + 20 by x+1

       By long division method

                                       

Now,

Dividend = Divisor × Quotient + remainder  

x³ +13x² + 32x + 20 = (x +1) ( x² + 12x + 20) + 0

                                 =(x + 1) (x²+10x+2x+20)

                                 =(x + 1) [x (x+10) +2(x+10)]

                                 =(x + 1) (x + 2) (x + 10) 

(iv) Let p(y) = 2y³+ y²− 2y – 1 all the factor are there. These are ± 1, ± 2.

       By trial method, p (1) =2(1)³ + (1)² – 2(1) – 1

                                        =2 + 1 – 2 – 1    =0

     Therefore (y–1) is the factor of polynomial p(y).

    Let us find the quotient on dividing 2y³+ y²− 2y – 1 by y–1.

    By long division method

                             

Now,

Dividend = Divisor × Quotient + remainder

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

                           = (y − 1) (2y²+2y

                          = (y − 1) [2y (y+1) + 1 (y + 1)]

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

 

Algebraic Identities: 

 ∵ ∴

(1) (x + y)² = x² + 2xy + y²

(2)  (x - y)² = x² - 2xy + y²

(3)  x² - y² = (x + y) (x - y) 

(4)  (x + a) (x + b) = x² + (a + b)x + ab 

(5)  (x + y)³ = x³ + 3x²y + 3xy² + y³

(6)  (x - y)³ = x³ - 3x²y + 3xy² - y³

(7)  x³ + y³ = (x + y) (x² - xy + y²)

(8)  x³ - y³ = (x - y) (x² + xy + y²)

(9)  (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

(10) x³ + y³ + z³ - 3xyz = ( x + y + z) (x² + y² + z² - xy - yz - zx)

 

Exercise 2.5 

Q1. Use suitable identities to find the following products:

(i) (x + 4) (x + 10)

(ii) (x + 8) (x – 10)

(iii) (3x + 4) (3x – 5)

(v) (3 – 2x) (3 + 2x)

Solution: 

(i) (x + 4) (x + 10) 

Using identity;  (x + a) (x + b) = x² + (a + b)x + ab 

(x + 4) (x + 10) = x² + (4 + 10)x + (4)(10) 

= x² + 14x + 40 

(ii) (x + 8) (x – 10) 

Using identity;  (x + a) (x + b) = x² + (a + b)x + ab  

(x + 8) (x – 10) = x² + [8 + (-10)]x + (8)(-10) 

= x² - 2x - 80  

(iii) (3x + 4) (3x – 5)

Using identity;  (x + a) (x + b) = x² + (a + b)x + ab  

(3x + 4) (3x – 5) = (3x)² + [4 + (-5)]3x + (4)(-5) 

= 9x² - 3x - 20  

Using identity;  (x + y) (x - y) = x² - y² 

(v) (3 – 2x) (3 + 2x)

Using identity; (x + y) (x - y) = x² - y² 

(3 – 2x) (3 + 2x) = (3)² - (2x)²

= 9 - 4x²

Q2. Evaluate the following products without multiplying directly:

(i) 103 × 107

(ii) 95 × 96

(iii) 104 × 96

Solution: 

(i) 103 × 107 = (100 + 3) (100 + 7) 

Using identity;  (x + a) (x + b) = x² + (a + b)x + ab  

(100 + 3) (100 + 7) = (100)²​ + (3 + 7)100 + 3×7 

=10000 + 1000 + 21     

= 11021

(ii) 95 × 96 = (90 + 5) (90 + 6) 

Using identity;  (x + a) (x + b) = x² + (a + b)x + ab  

(90 + 5) (90 + 6) = (90)²​ + (5 + 6)90 + 5×6 

=8100 + 990 + 30     

= 9120

(iii)  104 × 96 = (100 + 4) (100 - 4) 

Using identity; (x + y) (x - y) = x² - y²  

(100)² - (4)²

=10000 - 16      

= 9984

3. Factorise the following using appropriate identities:

(i) 9x² + 6xy + y² 

(ii) 4y² – 4y + 1

Solution:

(i) 9x² + 6xy + y² 

= (3x)² + 2.3x.y + (y)²     [ ∵ x² + 2xy + y² = (x + y)²]

∴ = (3x + y)² 

=  (3x + y)  (3x + y)

(ii) 4y² - 4y + 1 

= (2y)² - 2.2y.1 + (1)²     [ ∵ x² - 2xy + y² = (x - y)²]

∴ = (2y - 1)² 

=  (2y - 1)  (2y - 1)

  

 

[ ∵ x² - y² = (x + y) (x - y) ​]

Q4. 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

Solution:

(i) (x + 2y + 4z)²  

Here let as a = x, b = 2y, c = 4z and putting the values of a, b and c in the

Identity (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

∴ (x + 2y + 4z)² = (x)² + (2y)² + (4z)² + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)

   = x² + 4y² + 16z² + 4xy + 16yz + 8zx                   

(ii) (2x – y + z)² 

Here let as a = 2x, b = - y, c = z and putting the values of a, b and c in the

Identity (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

∴ (2x – y + z)² = (2x)² + (- y)² + (z)² + 2(2x)(- y) + 2(- y)(z) + 2(z)(2x)

   = 4x² + y² + z² - 4xy - 2yz + 4zx           

(iii) (–2x + 3y + 2z)²

Here let as a = - 2x, b = 3y, c = 2z and putting the values of a, b and c in the

Identity (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

∴ (–2x + 3y + 2z)² 

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

   = 4x² + 9y² + 4z² – 12xy  + 12yz – 8zx  

(iv) (3a – 7b – c)² 

Here let as x = 3a, y = – 7b, z = – c and putting the values of x, y and z in the

Identity (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

∴ (3a – 7b – c)²  

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

   = 9a² + 49b² + c² – 42ab  + 14bc – 6ac  

(v) (–2x + 5y – 3z)²

Here let as a = - 2x, b = 5y, c = –3z and putting the values of a, b and c in the

Identity (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

∴ (–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  

Q5. Factorise:
(i) 4x² + 9y² + 16z² + 12xy – 24yz – 16xz

Solution:

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

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

 [∵ a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)² ]

= (2x + 3y + 4z)² 

= (2x + 3y + 4z) (2x + 3y + 4z)

Q6. Write the following cubes in expanded form:
(i) (2x + 1)³ 

(ii) (2a – 3b)³

Solution:

(i) (2x + 1)³ 

[using identity (a + b)³ = a³ + 3a²b + 3ab² + b³]

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

               = 8x³ + 12x² + 6x + 1

(ii) (2a – 3b)³

[Using identity  (x - y)³ = x³ - 3x²y + 3xy² - y³]

(2a - 3b)³ = (2a)³ - 3(2a)²(3b) + 3(2a)(3b)² - (3b)³

               = 8a³ - 36a²b + 54ab² - 27b³

[using identity (a + b)³ = a³ + 3a²b + 3ab² + b³]

[using identity (a - b)³ = a³ - 3a²b + 3ab² - b³]

Q7. Evaluate the following using suitable identities:

(i) (99)³ 

(ii) (102)³ 

(iii) (998)³

Solution: 

(i) (99)³ 

= (100 - 1)³

[using identity (a - b)³ = a³ - 3a²b + 3ab² - b³]

(100 - 1)³ = (100)³ - 3(100)²(1) + 3(100)(1)² - (1)³

                = 1000000 - 30000 + 300 - 1

                = 1000300 - 30001

                = 970299

(ii) (102)³ 

= (100 + 2)³

[using identity (a + b)³ = a³ + 3a²b + 3ab² + b³]

(100 + 2)³ = (100)³ + 3(100)²(2)+ 3(100)(2)² + (2)³

                = 1000000 + 60000 + 1200 + 8

                = 1061208

(iii) (998)³

= (1000 - 2)³

[using identity (a - b)³ = a³ - 3a²b + 3ab² - b³]

(1000 - 2)³ = (1000)³ - 3(1000)²(2)+ 3(1000)(2)² - (2)³

                = 1000000000 - 6000000 + 12000 - 8

                = 1000012000 - 6000008

                = 994011992

Q8. Factorise each of the following:

(i) 8a³ + b3 + 12a²b + 6ab² 

(ii) 8a² – b² – 12a²b + 6ab²

(iii) 27 – 125a³ – 135a + 225a²

(iv) 64a³ – 27b³ – 144a²b + 108ab²  

Solution:

(i) 8a³ + b3 + 12a²b + 6ab² 

= (2a)³ +(b)³ + 3(2a)²(b) + 3(2a)(b)²

[Using identity  x³  + y³ + 3x²y + 3xy² = (x + y)³ ]

= (2a)³ +(b)³ + 3(2a)²(b) + 3(2a)(b)² = (2a + b)³

= (2a + b)(2a + b)(2a + b)

(ii) 8a² – b² – 12a²b + 6ab²

= (2a)³ - (b)³ - 3(2a)²(b) + 3(2a)(b)²

[Using identity  x³ - y³ - 3x²y + 3xy² = (x - y)³ ]

= (2a)³ - (b)³ - 3(2a)²(b) + 3(2a)(b)² = (2a - b)³​

= (2a - b)(2a - b)(2a - b)

(iii) 27 – 125a³ – 135a + 225a²

= (3)³ - (5a)³ - 3(3)²(5a) + 3(3)(5a)²

[Using identity  x³ - y³ - 3x²y + 3xy² = (x - y)³ ]

= (3)³ - (5a)³ - 3(3)²(5a) + 3(3)(5a)²= (3 - 5a)³​

= (3 - 5a)(3 - 5a)(3 - 5a)

(iv) 64a³– 27b³ – 144a²b + 108ab²  

= (4a)³ - (3b)³ - 3(4a)²(3b) + 3(4a)(3b)²

[Using identity  x³ - y³ - 3x²y + 3xy² = (x - y)³ ]

= (4a)³ - (3b)³ - 3(4a)²(3b) + 3(4a)(3b)² = (4a - 3b)³​

= (4a - 3b)(4a - 3b)(4a - 3b)

[Using identity  x³ - y³ - 3x²y + 3xy² = (x - y)³ ]

Q9. Verify:

(i) x³ + y³ = (x + y) (x² – xy + y²)

Solution:

RHS = (x + y) (x² – xy + y²)

= x(x² – xy + y²) + y (x² – xy + y²)

= x³ – x²y + xy² + x²y – xy² + y³ 

= x³ + y³

 ∵ LHS = RHS Verified 

(ii) x³ – y³ = (x – y) (x² + xy + y²)

Solution:

RHS = (x - y) (x² + xy + y²)

x(x² + xy + y²) - y(x² + xy + y²)

= x³ + x²y + xy² – x²y – xy² – y³ 

= x³ – y³

∵ LHS = RHS Verified 

Q10. Factorise each of the following:

(i) 27y³ + 125z³ 

(ii) 64m³ – 343n³

Solution: 

(i) 27y³ + 125z³ 

= (3y)³ + (5z)³

[Using identity x³ + y³ = (x + y) (x² – xy + y²) ]

(3y)³ + (5z)³​ = (3y + 5y) [(3y)² - (3y)(5z) + (5z)²]

= (3y + 5y) (9y² - 15yz + 25z²)

(ii) 64m³ – 343n³

Solution: 

(ii) 64m³ – 343n³

= (4m)³ – (7n)³

[Using identity x³ – y³ = (x – y) (x² + xy + y²) ]

(4m)³ – (7n)³​ = (4m – 7n) [(4m)² + (4m)(7n) + (7n)²]

= (4m – 7n) (16m² + 28mn + 49n^​2)

Q11. Factorise : 27x³ + y³ + z³ – 9xyz

Solution: 

= (3x)³ + (y)³ + (z)³ - 9xyz 

∵ x³ + y³ + z³ - 3xyz =  (x + y + z) (x² + y² + z^​2 - xy - yz - zx)

Using identity: 

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

= (3x + y + z) (9x² + y² + z² - 3xy - yz - 3zx)

Q12. Verify that:

x³ + y³ + z³ - 3xyz =½ (x + y + z) [(x -y)² + (y - z)² + (z - x)²]

LHS =  ½(x + y + z) [x² - 2xy + y² + y² - 2yz + z² + z² - 2xz + x²]

        = ½(x + y + z) (2x² + 2y² + 2z² - 2xy - 2yz - 2xz)

        = ½ × 2(x + y + z)(x² + y² + z² - xy - yz - xz)

        = (x + y + z)(x² + y² + z² - xy - yz - xz) 

        = x³ + y³ + z³ - 3xyz                 [Using Identity]

LHS = RHS 

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