NCERT Solutions Class 9 Mathematics Chapter 2 Polynomials

Exercise 2.1

Question. 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
Answer : (i) 4x² - 3x + 7
There is only one variable x with whole number power so this polynomial in one variable.
(ii) y² + √2
There is only one variable y with whole number power so this polynomial in one variable.
(iii) 3√2 + t√2 
There is only one variable t but in 3√t power of t is 1/2 which is not a whole number so 3√t + t√2 is not a polynomial.
(iv) y + 2/y
There is only one variable y but 2/y = 2y-1 so the power is not a whole number so y + 2/y is not a polynomial.
(v) x¹⁰ + y³ + t⁵⁰
There are three variable x, y and t and there powers are whole number so this polynomial in three variable.

Question. 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
Answer : (i) coefficients of x² = 1
(ii) coefficients of x² = -1
(iii) coefficients of x² = π/2
(iv) coefficients of x² = 0

Question. Give one example each of a binomial of degree 35, and of a monomial of degree 100.
Answer : x³⁵ +,5 is a binomial of degree 35.
2y¹⁰⁰ is a monomial of degree 100.

Question. Write the degree of each of the following polynomials:
(i) 5x³ + 4x² + 7x 
(ii) 4 – y² 
(iii) 5t – √7
(iv) 3
Answer : (i) 5x³ has highest power in the given polynomial which power is 3. Therefore, degree of polynomial is 3.
(ii) – y²  has highest power in the given polynomial which power is 2. Therefore, degree of polynomial is 2.
(iii) 5t has highest power in the given polynomial which power is 1. Therefore, degree of polynomial is 1.
(iv) There is no variable in the given polynomial. Therefore, degree of polynomial is 0.

Question. Classify the following as linear, quadratic and cubic polynomial:
(i) x² + x
(ii) x - x³
(iii) y + y² +4
(iv) 1 + x
(v) 3t
(vi) r²
(vii) 7x³
Answer : (i) x² + x
Quadratic Polynomial
(ii) x - x³
Cubic Polynomial
(iii) y + y² +4
Quadratic Polynomial
(iv) 1 + x
Linear Polynomial
(v) 3t
Linear Polynomial
(vi) r²
Quadratic Polynomial
(vii) 7x³
Cubic Polynomial

Exercise 2.2

Question. Find the value of the polynomial 5x – 4x² + 3 at
(i) x = 0
(ii) x = – 1
(iii) x = 2
Answer : Let p(x) = 5x – 4x² + 3
(i) p(0) = 5(0) – 4(0)² + 3 = 0 – 0 + 3 = 3
Thus, the value of 5x – 4x² + 3 at x = 0 is 3.

(ii) p(-1) = 5(-1) – 4(-1)² + 3
= – 5x – 4x² + 3 = -9 + 3 = -6
Thus, the value of 5x – 4x² + 3 at x = -1 is -6.

(iii) p(2) = 5(2) – 4(2)² + 3 = 10 – 4(4) + 3
= 10 – 16 + 3 = -3
Thus, the value of 5x – 4x² + 3 at x = 2 is – 3.

Question. 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)
Answer : (i) Given that p(y) = y² – y + 1.
∴ P(0) = (0)² – 0 + 1 = 0 – 0 + 1 = 1
p(1) = (1)² – 1 + 1 = 1 – 1 + 1 = 1
p(2) = (2)² – 2 + 1 = 4 – 2 + 1 = 3

(ii) Given that p(t) = 2 + t + 2t² – t3
∴p(0) = 2 + 0 + 2(0)² – (0)³
= 2 + 0 + 0 – 0=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) Given that p(x) = x³
∴ p(0) = (0)3 = 0, p(1) = (1)3 = 1
p(2) = (2)3 = 8

(iv) Given that 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

Question. 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) = 1x + m, x = – m/1
(vii) P (x) = 3x² – 1, x = – 1/√3, 2/√3
(viii) p (x) = 2x + 1, x = 1/2 
Answer : (i) If x = -1/3 is a zero of polynomial p(x) = 3x + 1 then p(-1/3) should be 0.
At, p(-1/3) = 3(-1/3) + 1 = -1 + 1 = 0
Therefore, x = -1/3 is a zero of polynomial p(x) = 3x + 1.

(ii) If x = 4/5 is a zero of polynomial p(x) = 5x - π then p(4/5) should be 0.
At, p(4/5) = 5(4/5) - π = 4 - π
Therefore, x = 4/5 is not a zero of given polynomial p(x) = 5x - π.

(iii) If x = 1 and x = -1 are zeroes of polynomial p(x) = x² - 1, then p(1) and p(-1) should be 0.
At, p(1) = (1)² - 1 = 0 and
At, p(-1) = (-1)² - 1 = 0
Hence, x = 1 and -1 are zeroes of the polynomial  p(x) = x² - 1.

(iv) If x = -1 and x = 2 are zeroes of polynomial p(x) = (x +1) (x - 2), then p( - 1) and (2)should be 0.
At, p(-1) = (-1 + 1) (-1 - 2) = 0 (-3) = 0, and
At, p(2) = (2 + 1) (2 - 2) = 3 (0) = 0
Therefore, x = -1 and x = 2 are zeroes of the polynomial p(x) = (x +1) (x - 2).

(v) If x = 0 is a zero of polynomial p(x) = x², then p(0) should be zero.
Here, p(0) = (0)² = 0
Hence, x = 0 is a zero of the polynomial p(x) = x².

Question. 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, are real numbers.
Answer : (i) p(x) = x + 5 
p(x) = 0
x + 5 = 0
x = -5
Therefore, x = -5 is a zero of polynomial p(x) = x + 5 .

(ii) p(x) = x - 5
p(x) = 0
x - 5 = 0
x = 5

Therefore, x = 5 is a zero of polynomial p(x) = x - 5.
(iii) p(x) = 2x + 5
p(x) = 0
2x + 5 = 0
2x = -5
x = -5/2
Therefore, x = -5/2 is a zero of polynomial p(x) = 2x + 5.

(iv) p(x) = 3x - 2
p(x) = 0
3x - 2 = 0
x = 2/3
Therefore, x = 2/3 is a zero of polynomial p(x) = 3x - 2.

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

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

(vii) p(x) = cx + d
p(x) = 0
cx + d = 0
x = -d/c
Therefore, x = -d/c is a zero of polynomial p(x) = cx + d.

Exercises 2.3

Question. 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
Answer : (i) x + 1
By long division,

Question. Find the remainder when x³ - ax² + 6x - a is divided by x - a.
Answer : By Long Division,

Question. Check whether 7 + 3x is a factor of 3x³ + 7x.
Answer : We have to divide 3x³ + 7x by 7 + 3x. If remainder comes out to be 0 then 7 + 3x will be a factor of 3x³ + 7x.
By Long Division,

As remainder is not zero so 7 + 3x is not a factor of 3x³ + 7x.

Exercise 2.4

Question. 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) If (x + 1) is a factor of p(x) = x³ + x² + x + 1, p(-1) must be zero. 
Here, p(x) = x³ + x² + x + 1 
p(-1) = (-1)³ + (-1)² + (-1) + 1 
= -1 + 1 - 1 + 1 = 0
Therefore, x + 1 is a factor of this polynomial

(ii) If (x + 1) is a factor of p(x) = x⁴ + x³ + x² + x + 1, p(-1) must be zero. 
Here, 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(-1) ≠ 0
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, p(- 1) must be 0. 
p(-1) = (-1)⁴ + 3(-1)³ + 3(-1)² + (-1) + 1
= 1 - 3 + 3 - 1 + 1 = 1
As, p(-1) ≠ 0
Therefore, x + 1 is not a factor of this polynomial.

(iv) If (x + 1) is a factor of polynomial
p(x) = x³ - x² - (2 + √2)x + √2, p(- 1) must be 0.
p(-1) =  (-1)³ -  (-1)² -  (2 + √2) (-1) + √2
= -1 - 1 + 2 + √2 + √2
=2√2
As, p(-1) ≠ 0
Therefore,, x + 1 is not a factor of this polynomial.

Question. 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
Answer : (i) If g(x) = x + 1 is a factor of given polynomial p(x), 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
Hence, g(x) = x + 1 is a factor of given polynomial.

(ii) If g(x) = x + 2 is a factor of given polynomial p(x), 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(-2) ≠ 0
Hence g(x) = x + 2 is not a factor of given polynomial.

(iii) If g(x) = x - 3 is a factor of given polynomial p(x), p(3) must be 0.
p(x) = x³ - 4x² + x + 6
p(3) = (3)³ - 4(3)² + 3 + 6
= 27 - 36 + 9 = 0
Therefore,, g(x) = x - 3 is a factor of given polynomial.

Question. 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 : (i) If x - 1 is a factor of polynomial p(x) = x² + x + k, then
p(1) = 0
⇒ (1)² + 1 + k = 0
⇒ 2 + k = 0
⇒ k = - 2
Therefore, value of k is -2.

(ii) If x - 1 is a factor of polynomial p(x) = 2x² + kx +  √2, then
p(1) = 0
⇒ 2(1)² + k(1) + √2 = 0
⇒ 2 + k + √2 = 0
⇒ k = -2 - √2 = -(2 + √2)
Therefore, value of k is -(2 + √2).

(iii) If x - 1 is a factor of polynomial p(x) = kx² - √2x + 1, then
p(1) = 0
⇒ k(1)² - √2(1) + 1 = 0
⇒ k - √2 + 1 = 0
⇒ k = √2 - 1
Therefore, value of k is √2 - 1.

(iv) If x - 1 is a factor of polynomial p(x) = kx² - 3x + k, then
p(1) = 0
⇒ k(1)² + 3(1) + k = 0
⇒ k - 3 + k = 0
⇒ 2k - 3 = 0
⇒ k = 3/2
Therefore, value of k is 3/2.

Question. Factorise:
(i) 12x² + 7x + 1
(ii) 2x² + 7x + 3
(iii) 6x² + 5x - 6
(iv) 3x² - x - 4 
Answer : (i) 12x² + 7x + 1
= 12x² - 4x - 3x+ 1                   
= 4x (3x - 1) - 1 (3x - 1)
= (3x - 1) (4x - 1)

(ii) 2x² + 7x + 3
= 2x² + 6x + x + 3
= 2x (x + 3) + 1 (x + 3)
=  (x + 3) (2x + 1) 

(iii) 6x² + 5x - 6
= 6x² + 9x - 4x - 6
 = 3x (2x + 3) - 2 (2x + 3)
= (2x + 3) (3x - 2)

(iv) 3x² - x - 4
= 3x² - 4x + 3x - 4 
= x (3x - 4) + 1 (3x - 4)
= (3x - 4) (x + 1)

Question. Factorise:
(i) x³ - 2x² - x + 2
(ii) x³ - 3x² - 9x - 5
(iii) x³ + 13x² + 32x + 20 
(iv) 2y³ + y² - 2y - 1
Answer : (i) Let p(x) = x³ - 2x² - x + 2
Factors of 2 are ±1 and ± 2
By trial method, we find that
p(1) = 0
So, (x+1) is factor of p(x)
Now,
p(x) = x³ - 2x² - 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) (x² - 3x + 2)
= (x+1) (x² - x - 2x + 2)
= (x+1) {x(x-1) -2(x-1)}
= (x+1) (x-1) (x+2)

(ii) Let p(x) = x³ - 3x² - 9x - 5
Factors of 5 are ±1 and ±5
By trial method, we find that
p(5) = 0
So, (x-5) is factor of p(x)
Now,
p(x) = x³ - 2x² - x + 2
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) (x² + 2x + 1)
= (x-5) (x² + x + x + 1)
= (x-5) {x(x+1) +1(x+1)}
= (x-5) (x+1) (x+1)

(iii) Let p(x) = x³ + 13x² + 32x + 20
Factors of 20 are ±1, ±2, ±4, ±5, ±10 and ±20
By trial method, we find that
p(-1) = 0
So, (x+1) is factor of p(x)
Now,
p(x) =  x³ + 13x² + 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) (x² + 12x + 20)
= (x+1) (x² + 2x + 10x + 20)
= (x-5) {x(x+2) +10(x+2)}
= (x-5) (x+2) (x+10)

(iv) Let p(y) = 2y³ + y² - 2y - 1
Factors of ab = 2× (-1) = -2 are ±1 and ±2
By trial method, we find that
p(1) = 0
So, (y-1) is factor of p(y)
Now,
p(y) =  2y³ + y² - 2y - 1
p(1) = 2(1)³ + (1)² - 2(1) - 1 = 2 +1 - 2 - 1 = 0
Therefore, (y-1) is the factor of  p(y)

Now, Dividend = Divisor × Quotient + Remainder
(y-1) (2y² + 3y + 1)
= (y-1) (2y² + 2y + y + 1)
= (y-1) {2y(y+1) +1(y+1)}
= (y-1) (2y+1) (y+1)

Exercise 2.5

Question. 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)
Answer : (i) Using identity, (x + a) (x + b) = x² + (a + b) x + ab 
In (x + 4) (x + 10), a = 4 and b = 10
Now,
(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
Here, a = 8 and b = –10
(x + 8) (x – 10) = x² + {8 +(– 10)}x + {8×(– 10)}
                         = x² + (8 – 10)x – 80
                         = x² – 2x – 80

(iii) (3x + 4) (3x – 5)
Using identity, (x + a) (x + b) = x² + (a + b) x + ab
Here, x = 3x , a = 4 and b = -5
(3x + 4) (3x – 5) = (3x) ² + {4 + (-5)}3x + {4×(-5)}
                           = 9x² + 3x(4 - 5) - 20
                           = 9x² - 3x - 20

(iv) (y² + 3/2) (y² - 3/2)
Using identity, (x + y) (x -y) = x² - y²
Here, x = y² and y = 3/2
(y² + 3/2) (y² - 3/2) = (y²)² - (3/2)²
                                         = y⁴ - 9/4

(v) (3 - 2x) (3 + 2x)
Using identity, (x + y) (x -y) = x² - y²
Here, x = 3 and y = 2x
(3 - 2x) (3 + 2x) = 3² - (2x)²
                                   =  9 - 4x²

Question. Evaluate the following products without multiplying directly:
(i) 103 × 107              
(ii) 95 × 96              
(iii) 104 × 96
Answer : (i) 103 × 107 = (100 + 3) (100 + 7)
Using identity, (x + a) (x + b) = x² + (a + b) x + ab
Here, x = 100, a = 3 and b = 7
103 × 107 = (100 + 3) (100 + 7) = (100)² + (3 + 7)10 + (3 × 7)
                 = 10000 + 100 + 21 
                 = 10121

(ii) 95 × 96 = (90 + 5) (90 + 4)
Using identity, (x + a) (x + b) = x² + (a + b) x + ab 
Here, x = 90, a = 5 and b = 4
95 × 96 = (90 + 5) (90 + 4) = 90² + 90(5 + 6) + (5 × 6) 
             = 8100 + (11 × 90) + 30
             = 8100 + 990 + 30 = 9120

(iii) 104 × 96 = (100 + 4) (100 - 4)
Using identity, (x + y) (x -y) = x² - y²
Here, x = 100 and y = 4
104 × 96 = (100 + 4) (100 - 4) = (100)² - (4)² = 10000 - 16 = 9984 

Question. Factorise the following using appropriate identities:
(i) 9x² + 6xy + y²                 
(ii) 4y² - 4y + 1             
(iii) x² - y²/100
Answer : (i) 9x² + 6xy + y²  = (3x) ² + (2×3x×y) + y²
Using identity, (a + b)² = a² + 2ab + b²
Here, a = 3x and b = y
9x² + 6xy + y²  = (3x) ² + (2×3x×y) + y² = (3x + y)² = (3x + y) (3x + y)

(ii) 4y² - 4y + 1 = (2y)² - (2×2y×1) + 1²
Using identity, (a - b)² = a² - 2ab + b²
Here, a = 2y and b = 1
4y² - 4y + 1 = (2y)² - (2×2y×1) + 1² = (2y - 1)² = (2y - 1) (2y - 1)

(iii) x² - y²/100 = x² - (y/10)²
Using identity, a² - b² = (a + b) (a - b)
Here, a = x and b = (y/10)
x² - y²/100 = x² - (y/10)² = (x - y/10) (x + y/10)

Question. 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  
Answer : (i) (x + 2y + 4z)²
Using identity, (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca  
Here, a = x, b = 2y and c = 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
Using identity, (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca  
Here, a = 2x, b = -y and c = z
(2x – y + z)² = (2x)² + (-y)² + z² + (2×2x×-y) + (2×-y×z) + (2×z×2x) 
                     = 4x² + y² + z² - 4xy - 2yz + 4xz

(iii) (–2x + 3y + 2z)² 
Using identity, (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca  
Here, a = -2x, b = 3y and c = 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
Using identity, (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca  
Here, a = 3a, b = -7b and c = -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 - 6ac

(v) (–2x + 5y – 3z)²  
Using identity, (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca  
Here, a = -2x, b = 5y and c = -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 + 12xz

(vi) [1/4 a - 1/2 b + 1]² 
Using identity, (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca  
Here, a = 1/4 a, b = -1/2 b and c = 1
[1/4 a - 1/2 b + 1]² = (1/4 a)² + (-1/2 b)² + 1² + (2×1/4 a×-1/2 b) + (2×-1/2 b×1) + (2×1×1/4 a) 
                                = 1/16 a² + 1/4 b² + 1 - 1/4 ab - b + 1/2 a

Question. Factorise: (i) 4x² + 9y² + 16z² + 12xy - 24yz - 16xz
(ii) 2x² + y² + 8z² - 2√2 xy + 4√2 yz - 8xz
Answer : (i) 4x² + 9y² + 16z² + 12xy - 24yz - 16xz
Using identity, (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca 
4x² + 9y² + 16z² + 12xy - 24yz - 16xz
= (2x)² + (3y)² + (-4z)² + (2×2x×3y) + (2×3y×-4z) + (2×-4z×2x)
= (2x + 3y - 4z)²
=  (2x + 3y - 4z) (2x + 3y - 4z)

(ii) 2x² + y² + 8z² - 2√2 xy + 4√2 yz - 8xz
Using identity, (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
2x² + y² + 8z² - 2√2 xy + 4√2 yz - 8xz 
= (-√2x)² + (y)² + (2√2z)² + (2×-√2x×y) + (2×y×2√2z) + (2×2√2z×-√2x)
= (-√2x + y + 2√2z)²
=  (-√2x + y + 2√2z) (-√2x + y + 2√2z)

Question. 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 
Answer : (i) (2x + 1)³
Using identity, (a + b)³ = a³ + b³ + 3ab(a + b)
(2x + 1)³ = (2x)³ + 1³ + (3×2x×1)(2x + 1)
= 8x³ + 1 + 6x(2x + 1)
= 8x³ + 12x² + 6x + 1

(ii) (2a – 3b)³
Using identity, (a - b)³ = a³ - b³ - 3ab(a - b) 
(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 
Using identity, (a + b)³ = a³ + b³ + 3ab(a + b)
[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
Using identity, (a - b)³ = a³ - b³ - 3ab(a - b)
[x - 2/3 y]³ = (x)³ - (2/3 y)³ - (3×x×2/3 y)(x - 2/3 y)
= x³ - 8/27y³ - 2xy(x - 2/3 y)
= x³ - 8/27y³ - 2x²y + 4/3xy²

Question. Evaluate the following using suitable identities: 
(i) (99)³           
(ii) (102)³            
(iii) (998)³  
Answer : (i) (99)³ = (100 - 1)³
Using identity, (a - b)³ = a³ - b³ - 3ab(a - b) 
(100 - 1)³ = (100)³ - 1³ - (3×100×1)(100 - 1)
= 1000000 - 1 - 300(100 - 1)
= 1000000 - 1 - 30000 + 300
= 970299

(ii) (102)³ = (100 + 2)³
Using identity, (a + b)³ = a³ + b³ + 3ab(a + b)
(100 + 2)³ = (100)³ + 2³ + (3×100×2)(100 + 2)
= 1000000 + 8 + 600(100 + 2)
= 1000000 + 8 + 60000 + 1200
= 1061208

(iii) (998)³ 
Using identity, (a - b)³ = a³ - b³ - 3ab(a - b) 
(1000 - 2)³ = (1000)³ - 2³ - (3×1000×2)(1000 - 2)
= 100000000 - 8 - 6000(1000 - 2)
= 100000000 - 8- 600000 + 12000
= 994011992

Question. 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
Answer : (i) 8a³ + b³ + 12a²b + 6ab²
Using identity, (a + b)³ = a³ + b³ + 3a²b + 3ab²
8a³ + b³ + 12a²b + 6ab² 
= (2a)³ + b³ + 3(2a)²b + 3(2a)(b)²
= (2a + b)³
= (2a + b)(2a + b)(2a + b)

(ii) 8a³ - b³ - 12a²b + 6ab²
Using identity, (a - b)³ = a³ - b³ - 3a²b + 3ab²
8a³ - b³ - 12a²b + 6ab²= (2a)³ - b³ - 3(2a)²b + 3(2a)(b)²
= (2a - b)³
= (2a - b)(2a - b)(2a - b)

(iii) 27 - 125a³ - 135a + 225a²
Using identity, (a - b)³ = a³ - b³ - 3a²b + 3ab²
27 - 125a³ - 135a + 225a²= 3³ - (5a)³ - 3(3)²(5a) + 3(3)(5a)²
= (3 - 5a)³
= (3 - 5a)(3 - 5a)(3 - 5a)

(iv) 64a³ - 27b³ - 144a²b + 108ab²
Using identity, (a - b)³ = a³ - b³ - 3a²b + 3ab²
64a³ - 27b³ - 144a²b + 108ab²= (4a)³ - (3b)³ - 3(4a)²(3b) + 3(4a)(3b)²
= (4a - 3b)³
= (4a - 3b)(4a - 3b)(4a - 3b)

(v) 27p³ - 1/216 - 9/2 p² + 1/4 p 
 Using identity, (a - b)³ = a³ - b³ - 3a²b + 3ab²
 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
= (3p - 1/6)(3p - 1/6)(3p - 1/6)

Question. Verify : (i) x³ + y³ = (x + y) (x² - xy + y²)             (ii) x³ - y³ = (x - y) (x² + xy + y²)
Answer : (i) x³ + y³ = (x + y) (x² - xy + y²)
We know that, 
(x + y)³ = x³ + y³ + 3xy(x + y) 
⇒ x³ + y³ = (x + y)³ - 3xy(x + y)
⇒ x³ + y³ = (x + y)[(x + y)² - 3xy]                  {Taking (x+y) common}
⇒ x³ + y³ = (x + y)[(x² + y² + 2xy) - 3xy] 
⇒ x³ + y³ = (x + y)(x² + y² - xy) 

(ii) x³ - y³ = (x - y) (x² + xy + y² )
We know that, 
(x - y)³ = x³ - y³ - 3xy(x - y) 
⇒ x³ - y³ = (x - y)³ + 3xy(x - y)
⇒ x³ + y³ = (x - y)[(x - y)² + 3xy]                     {Taking (x-y) common}
⇒ x³ + y³ = (x - y)[(x² + y² - 2xy) + 3xy] 
⇒ x³ + y³ = (x + y)(x² + y² + xy)

Question. Factorise each of the following:
(i) 27y³ + 125z³                    
(ii) 64m³ - 343n³
Answer : (i) 27y³ + 125z³
Using identity, x³ + y³ = (x + y) (x² - xy + y²)
27y³ + 125z³ = (3y)³ + (5z)³
= (3y + 5z) {(3y)² - (3y)(5z) + (5z)²}
= (3y + 5z) (9y² - 15yz + 25z)² 

(ii) 64m³ - 343n³ 
Using identity, x³ - y³ = (x - y) (x² + xy + y² ) 
64m³ - 343n³ = (4m)³ - (7n)³
= (4m + 7n) {(4m)² + (4m)(7n) + (7n)²}
= (4m + 7n) (16m² + 28mn + 49n)² 

Question. Factorise : 27x³ + y³ + z³ - 9xyz
Answer : 27x³ + y³ + z³ - 9xyz = (3x)³ + y³ + z³ - 3×3xyz
Using identity, x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - xz)
27x³ + y³ + z³ - 9xyz
= (3x + y + z) {(3x)² + y² + z² - 3xy - yz - 3xz}
= (3x + y + z) (9x² + y² + z² - 3xy - yz - 3xz)

Question. Verify that: x³ + y³ + z³ - 3xyz = 1/2(x + y + z) [(x - y)² + (y - z)² + (z - x)²]
Answer : We know that,
x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - xz)
⇒ x³ + y³ + z³ - 3xyz = 1/2×(x + y + z) 2(x² + y² + z² - xy - yz - xz)
= 1/2(x + y + z) (2x² + 2y² + 2z² - 2xy - 2yz - 2xz) 
= 1/2(x + y + z) [(x² + y² -2xy) + (y² + z² - 2yz) + (x² + z² - 2xz)] 
= 1/2(x + y + z) [(x - y)² + (y - z)² + (z - x)²]

Question. If x + y + z = 0, show that x³ + y³ + z³ = 3xyz.
Answer : We know that,
x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - xz)
Now put (x + y + z) = 0,  
x³ + y³ + z³ - 3xyz = (0)(x² + y² + z² - xy - yz - xz) 
⇒ x³ + y³ + z³ - 3xyz = 0

Question. 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
 Let x = -12, y = 7 and z = 5
We observed that, x + y + z = -12 + 7 + 5 = 0
We know that if,
x + y + z = 0, then x³ + y³ + z³ = 3xyz
(-12)³ + (7)³ + (5)³ = 3(-12)(7)(5) = -1260

(ii) (28)³ + (–15)³ + (-13)³
 Let x = 28, y = -15 and z = -13
We observed that, x + y + z = 28 - 15 - 13 = 0
We know that if,
x + y + z = 0, then x³ + y³ + z³ = 3xyz
(28)³ + (–15)³ + (-13)³ = 3(28)(-15)(-13) = 16380

Question. 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 : 35 y² + 13y - 12
Answer : (i) Area : 25a² - 35a + 12
Since, area is product of length and breadth therefore by factorizing the given area, we can know the length and breadth of rectangle.
25a² - 35a + 12
= 25a² - 15a -20a + 12
= 5a(5a - 3) - 4(5a - 3)
= (5a - 4)(5a - 3)
Possible expression for length = 5a - 4
Possible expression for breadth = 5a - 3

(ii) Area : 35 y² + 13y - 12
35 y² + 13y - 12
= 35y² - 15y + 28y - 12
= 5y(7y - 3) + 4(7y - 3)
= (5y + 4)(7y - 3)
Possible expression for length = (5y + 4)
Possible expression for breadth = (7y - 3)

Question. 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 : 35 y² + 13y - 12
Answer : (i) Area : 25a² - 35a + 12
Since, area is product of length and breadth therefore by factorizing the given area, we can know the length and breadth of rectangle.
25a² - 35a + 12
= 25a² - 15a -20a + 12
= 5a(5a - 3) - 4(5a - 3)
= (5a - 4)(5a - 3)
Possible expression for length = 5a - 4
Possible expression for breadth = 5a - 3
(ii) Area : 35 y² + 13y - 12
35 y² + 13y - 12
= 35y² - 15y + 28y - 12
= 5y(7y - 3) + 4(7y - 3)
= (5y + 4)(7y - 3)
Possible expression for length = (5y + 4)
Possible expression for breadth = (7y - 3)

Question. 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
Answer : (i) Volume : 3x² - 12x 
Since, volume is product of length, breadth and height therefore by factorizing the given volume, we can know the length, breadth and height of the cuboid. 
3x² - 12x
= 3x(x - 4)
Possible expression for length = 3
Possible expression for breadth = x
Possible expression for height = (x - 4)

(ii) Volume : 12ky² + 8ky - 20k 
Since, volume is product of length, breadth and height therefore by factorizing the given volume, we can know the length, breadth and height of the cuboid. 
12ky² + 8ky - 20k
= 4k(3y² + 2y - 5)
= 4k(3y² +5y - 3y - 5)
= 4k[y(3y +5) - 1(3y + 5)]
= 4k (3y +5) (y - 1)
Possible expression for length = 4k
Possible expression for breadth = (3y +5)
Possible expression for height = (y - 1)