# Maths NCERT solutions class 9 Chapter 2 Polynomials

## 2. POLYNOMIALS

Maths NCERT solutions class 9 Chapter 2 Polynomials

• 2.1 Introduction
• 2.2 Polynomials in One Variable
• 2.3 Zeroes of a Polynomial
• 2.4 Remainder Theorem
• 2.5 Factorisation of Polynomials
• 2.6 Algebraic Identities
• 2.7 Summary

Important formula

### Exercise 2.1

Q.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

Both are polynomial in one variable

Q.(iii) 3√t + t√2
(iv) y + 2/y
(v) x10 + y3 + t50

(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.

Q.2. Write the coefficients of x2 in each of the following:
(i) 2 + x2 + x
(ii) 2 – x2 + x3

(iv) √2x – 1

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

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

5x35+9 and 5x100

Q.4. Write the degree of each of the following polynomials:
(i) 5x3 + 4x2 + 7x
(ii) 4 – y2
(iii) 5t – √7
(iv) 3

(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.

Q.5. Classify the following as linear, quadratic and cubic polynomial:
(i) x2 + x

Q.(ii) x – x3
► Cubic Polynomial

Q.(iii) y + y2 +4

Q.(iv) 1 + x
► Linear Polynomial

Q.(v) 3t
►Linear Polynomial

Q.(vi) r2

Q.(vii) 7x3
► Cubic Polynomial
2.2

### Exercise 2.2

Q.1. Find the value of the polynomial at 5x + 4x2 + 3 at
(i) x = 0 (ii) x = – 1 (iii) x = 2

(i) p(x) = 5x + 4x2 + 3
p(0) = 5(0) + 4(0)2 + 3
= 3
(ii) p(x) = 5x + 4x2 + 3
p(-1) = 5(-1) + 4(-1)2 + 3
= -5 + 4(1) + 3 = 2
(iii) p(x) = 5x + 4x2 + 3
p(2) = 5(2) + 4(2)2 + 3
= 10 + 16 + 3 = 29

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

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

(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 = 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

Q.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

Q.(iii) p(x) = x2 – 1, x = 1, -1
(iv) p(x) = (x + 1) (x – 2), x = -1, 2
(v) p(x) = x2 , x = 0

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

(viii) If x = 1/2 is a zero of polynomial p(x) = 2x + 1 then p(1/2) should be 0.
At, p(1/2) = 2(1/2) + 1 = 1 + 1 = 2
Therefore, x = 1/2 is not a zero of given polynomial p(x) = 2x + 1.

Q.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, are real numbers.

(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.
s 2.3

### Exercise 2.3

Q.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

(i) x + 1
By long division,

Therefore, the remainder is 0.
(ii) x – 1/2
By long division,

Therefore, the remainder is 27/8.
(iii) x

Therefore, the remainder is 1.
(iv) x + π

Therefore, the remainder is [1 – 3π + 3π2 – π3].
(v) 5 + 2x

Therefore, the remainder is -27/8.

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

By Long Division,

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

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

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

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
(iv) x3 – x2 – (2 + √2)x + √2

(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 = 1As, 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.

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

(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.

Q.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

(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.

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

(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)

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

(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

Q.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) (y+ 3/2) (y– 3/2) (v) (3 – 2x) (3 + 2x)

(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) (y+ 3/2) (y– 3/2)
Using identity, (x + y) (x -y) = x2 – y2
Here, x = y2 and y = 3/2
(y+ 3/2) (y– 3/2) = (y2)– (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

Q.2. Evaluate the following products without multiplying directly:
(i) 103 × 107 (ii) 95 × 96 (iii) 104 × 96

(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)= 10000 – 16 = 9984

Q.3. Factorise the following using appropriate identities:
(i) 9x2 + 6xy + y(ii) 4y2 – 4y + 1 (iii) x– y2/100

(i) 9x2 + 6xy + y= (3x) 2 + (2×3x×y) + y2
Using identity, (a + b)2 = a2 + 2ab + b2
Here, a = 3x and b = y
9x2 + 6xy + y= (3x) 2 + (2×3x×y) + y= (3x + y)=(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) + 1= (2y – 1)=(2y – 1) (2y – 1)
(iii) x– y2/100 = x– (y/10)2
Using identity, a2 – b2 = (a + b) (a – b)
Here, a = x and b = (y/10)
x– y2/100 = x– (y/10)= (x- y/10) (x+ y/10)

Q.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)(v) (–2x + 5y – 3z)(vi) [1/4 a – 1/2 b + 1]2

(i) (x + 2y + 4z)2
Using identity, (a + b + c)= a2 + b2 + c2 + 2ab + 2bc + 2ca
Here, a = x, b = 2y and c = 4z
(x + 2y + 4z)= 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)= a2 + b2 + c2 + 2ab + 2bc + 2ca
Here, a = 2x, b = -y and c = z
(2x – y + z)= (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)= a2 + b2 + c2 + 2ab + 2bc + 2ca
Here, a = -2x, b = 3y and c = 2z
(–2x + 3y + 2z)= (-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)= 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)= 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)= 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

Q.5. Factorise:
(i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz
(ii) 2x2 + y2 + 8z2 – 2√2 xy + 4√2 yz – 8xz

(i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz
Using identity, (a + b + c)= 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)= 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)

Q.6. Write the following cubes in expanded form:
(i) (2x + 1)3 (ii) (2a – 3b)(iii) [3/2 x + 1]3 (iv) [x – 2/3 y]3

(i) (2x + 1)3
Using identity, (a + b)= a3 + b3 + 3ab(a + b)
(2x + 1)= (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)= a3 – b3 – 3ab(a – b)
(2a – 3b)= (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)= a3 + b3 + 3ab(a + b)
[3/2 x + 1]= (3/2 x)3 + 13 + (3×3/2 x×1)(3/2 x + 1)
= 27/8 x+ 1 + 9/2 x(3/2 x + 1)
= 27/8 x+ 1 + 27/4 x2 + 9/2 x
= 27/8 x+ 27/4 x2 + 9/2 x + 1
(iv) [x – 2/3 y]3
Using identity, (a – b)= 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

Q.
7. Evaluate the following using suitable identities:
(i) (99)3 (ii) (102)3 (iii) (998)3

(i) (99)3 = (100 – 1)3
Using identity, (a – b)= a3 – b3 – 3ab(a – b)
(100 – 1)= (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)= a3 + b3 + 3ab(a + b)
(100 + 2)= (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)= a3 – b3 – 3ab(a – b)
(1000 – 2)= (1000)3 – 23 – (3×1000×2)(1000 – 2)
= 100000000 – 8 – 6000(1000 – 2)
= 100000000 – 8- 600000 + 12000
= 994011992

Q.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

(i) 8a3 + b3 + 12a2b + 6ab2
Using identity, (a + b)= 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)= 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)= 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)= 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)= 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)

Q.9. Verify : (i) x3 + y3 = (x + y) (x2 – xy + y2) (ii) x3 – y3 = (x – y) (x2 + xy + y2)

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

Q.10. Factorise each of the following:
(i) 27y3 + 125z3 (ii) 64m3 – 343n3

(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

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

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)

Q.12. Verify that: x3 + y3 + z3 – 3xyz = 1/2(x + y + z) [(x – y)+ (y – z)+(z – x)2]

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) + (y+ z2 – 2yz) + (x2 + z– 2xz)]
= 1/2(x + y + z) [(x – y)+ (y – z)+(z – x)2]

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

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

Q.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

(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

Q.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 : 35 y2 + 13y – 12

(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)

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

Q.(i) Volume : 3x2 – 12x
(ii) Volume : 12ky2 + 8ky – 20

(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)

Maths NCERT solutions class 9 Chapter 2 Polynomials