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NCERT Solutions for Class 9 Maths Chapter 2
Polynomials

Exercise 2
...
Which of the following expressions are polynomials in one variable, and which are not? State reasons for
your answer
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e
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(ii) y2+√2
Solution:
The equation y2+√2 can be written as y2+√2y0
Since y is the only variable in the given equation and the powers of y (i
...
, 2 and 0) are whole numbers, we can
say that the expression y2+√2 is a polynomial in one variable
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e
...
Hence, we can
say that the expression 3√t+t√2 is not a polynomial in one variable
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e
...
Hence, we can
say that the expression y+2/y is not a polynomial in one variable
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Hence, it is not a polynomial in one variable
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Write the coefficients of x2 in each of the following:
(i) 2+x2+x
Solution:
The equation 2+x2+x can be written as 2+(1)x2+x

NCERT Solutions for Class 9 Maths Chapter 2
Polynomials
We know that the coefficient is the number which multiplies the variable
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(ii) 2–x2+x3
Solution:
The equation 2–x2+x3 can be written as 2+(–1)x2+x3
We know that the coefficient is the number (along with its sign, i
...
– or +) which multiplies the variable
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(iii) (π/2)x2+x
Solution:
The equation (π/2)x2 +x can be written as (π/2)x2 + x
We know that the coefficient is the number (along with its sign, i
...
– or +) which multiplies the variable
...

Hence, the coefficient of x2 in (π/2)x2 +x is π/2
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e
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Here, the number that multiplies the variable x2is 0
Hence, the coefficient of x2 in √2x-1 is 0
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Give one example each of a binomial of degree 35, and of a monomial of degree 100
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For example, 3x35+5
Monomial of degree 100: A polynomial having one term and the highest degree 100 is called a monomial of
degree 100
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Write the degree of each of the following polynomials:
(i) 5x3+4x2+7x
Solution:

NCERT Solutions for Class 9 Maths Chapter 2
Polynomials
The highest power of the variable in a polynomial is the degree of the polynomial
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(ii) 4–y2
Solution:
The highest power of the variable in a polynomial is the degree of the polynomial
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(iii) 5t–√7
Solution:
The highest power of the variable in a polynomial is the degree of the polynomial
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(iv) 3
Solution:
The highest power of the variable in a polynomial is the degree of the polynomial
...

5
...

Quadratic polynomial: A polynomial of degree two is called a quadratic polynomial
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(i) x2+x
Solution:
The highest power of x2+x is 2
The degree is 2

NCERT Solutions for Class 9 Maths Chapter 2
Polynomials
Hence, x2+x is a quadratic polynomial
(ii) x–x3
Solution:
The highest power of x–x3 is 3
The degree is 3
Hence, x–x3 is a cubic polynomial
(iii) y+y2+4
Solution:
The highest power of y+y2+4 is 2
The degree is 2
Hence, y+y2+4 is a quadratic polynomial
(iv) 1+x
Solution:
The highest power of 1+x is 1
The degree is 1
Hence, 1+x is a linear polynomial
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(vi) r2
Solution:
The highest power of r2 is 2
The degree is 2
Hence, r2is a quadratic polynomial
...


NCERT Solutions for Class 9 Maths Chapter 2
Polynomials

Exercise 2
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Find the value of the polynomial (x)=5x−4x2+3
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Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p(y)=y2−y+1
Solution:
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
Solution:
p(t) = 2+t+2t2−t3
∴ p(0) = 2+0+2(0)2–(0)3 = 2

Page: 34

NCERT Solutions for Class 9 Maths Chapter 2
Polynomials
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
Solution:
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)
Solution:
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
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(i) p(x)=3x+1, x = −1/3
Solution:
For, x = -1/3, p(x) = 3x+1
∴ p(−1/3) = 3(-1/3)+1 = −1+1 = 0
∴ -1/3 is a zero of p(x)
...

(iii) p(x) = x2−1, x = 1, −1
Solution:
For, x = 1, −1;
p(x) = x2−1
∴ p(1)=12−1=1−1 = 0
p(−1)=(-1)2−1 = 1−1 = 0
∴ 1, −1 are zeros of p(x)
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(v) p(x) = x2, x = 0
Solution:
For, x = 0 p(x) = x2
p(0) = 02 = 0
∴ 0 is a zero of p(x)
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(vii) p(x) = 3x2−1, x = -1/√3 , 2/√3
Solution:
For, x = -1/√3 , 2/√3 ; p(x) = 3x2−1
∴ 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
∴ -1/√3 is a zero of p(x), but 2/√3 is not a zero of p(x)
...

4
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(ii) p(x) = x–5
Solution:
p(x) = x−5
⇒ x−5 = 0
⇒x=5
∴ 5 is a zero polynomial of the polynomial p(x)
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(iv) p(x) = 3x–2
Solution:
p(x) = 3x–2
⇒ 3x−2 = 0
⇒ 3x = 2
⇒x = 2/3
∴ x = 2/3 is a zero polynomial of the polynomial p(x)
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(vi) p(x) = ax, a≠0

NCERT Solutions for Class 9 Maths Chapter 2
Polynomials
Solution:
p(x) = ax
⇒ ax = 0
⇒x=0
∴ x = 0 is a zero polynomial of the polynomial p(x)
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Solution:
p(x) = cx + d
⇒ cx+d =0
⇒ x = -d/c
∴ x = -d/c is a zero polynomial of the polynomial p(x)
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3
1
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Find the remainder when x3−ax2+6x−a is divided by x-a
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Check whether 7+3x is a factor of 3x3+7x
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4
1
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[x+1 = 0 means x = -1]
p(−1) = (−1)3+(−1)2+(−1)+1
= −1+1−1+1
=0
∴ By factor theorem, x+1 is a factor of x3+x2+x+1
(ii) x4+x3+x2+x+1
Solution:
Let p(x)= x4+x3+x2+x+1
The zero of x+1 is -1
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p(−1)=(−1)4+3(−1)3+3(−1)2+(−1)+1
=1−3+3−1+1
=1 ≠ 0
∴ By factor theorem, x+1 is not a factor of x4+3x3+3x2+x+1
(iv) x3 – x2– (2+√2)x +√2
Solution:
Let p(x) = x3–x2–(2+√2)x +√2
The zero of x+1 is -1
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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
Solution:
p(x) = 2x3+x2–2x–1, g(x) = x+1
g(x) = 0
⇒ x+1 = 0
⇒ x = −1
∴ Zero of g(x) is -1
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(ii) p(x)=x3+3x2+3x+1, g(x) = x+2
Solution:
p(x) = x3+3x2+3x+1, g(x) = x+2
g(x) = 0
⇒ x+2 = 0
⇒ x = −2
∴ Zero of g(x) is -2
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(iii) p(x)=x3–4x2+x+6, g(x) = x–3
Solution:
p(x) = x3–4x2+x+6, g(x) = x -3
g(x) = 0
⇒ x−3 = 0

NCERT Solutions for Class 9 Maths Chapter 2
Polynomials
⇒x=3
∴ Zero of g(x) is 3
...

3
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Factorise:
(i) 12x2–7x+1
Solution:
Using the splitting the middle term method,
We have to find a number whose sum = -7 and product =1×12 = 12
We get -3 and -4 as the numbers [-3+-4=-7 and -3×-4 = 12]
12x2–7x+1= 12x2-4x-3x+1
= 4x(3x-1)-1(3x-1)
= (4x-1)(3x-1)
(ii) 2x2+7x+3
Solution:
Using the splitting the middle term method,
We have to find a number whose sum = 7 and product = 2×3 = 6
We get 6 and 1 as the numbers [6+1 = 7 and 6×1 = 6]
2x2+7x+3 = 2x2+6x+1x+3
= 2x (x+3)+1(x+3)
= (2x+1)(x+3)
(iii) 6x2+5x-6
Solution:
Using the splitting the middle term method,
We have to find a number whose sum = 5 and product = 6×-6 = -36
We get -4 and 9 as the numbers [-4+9 = 5 and -4×9 = -36]
6x2+5x-6 = 6x2+9x–4x–6
= 3x(2x+3)–2(2x+3)
= (2x+3)(3x–2)
(iv) 3x2–x–4

NCERT Solutions for Class 9 Maths Chapter 2
Polynomials
Solution:
Using the splitting the middle term method,
We have to find a number whose sum = -1 and product = 3×-4 = -12
We get -4 and 3 as the numbers [-4+3 = -1 and -4×3 = -12]
3x2–x–4 = 3x2–4x+3x–4
= x(3x–4)+1(3x–4)
= (3x–4)(x+1)
5
...
5
1
...
Evaluate the following products without multiplying directly:

Page: 48

NCERT Solutions for Class 9 Maths Chapter 2
Polynomials
(i) 103×107
Solution:
103×107= (100+3)×(100+7)
Using identity, [(x+a)(x+b) = x2+(a+b)x+ab
Here, x = 100
a=3
b=7
We get, 103×107 = (100+3)×(100+7)
= (100)2+(3+7)100+(3×7)
= 10000+1000+21
= 11021
(ii) 95×96
Solution:
95×96 = (100-5)×(100-4)
Using identity, [(x-a)(x-b) = x2-(a+b)x+ab
Here, x = 100
a = -5
b = -4
We get, 95×96 = (100-5)×(100-4)
= (100)2+100(-5+(-4))+(-5×-4)
= 10000-900+20
= 9120
(iii) 104×96
Solution:
104×96 = (100+4)×(100–4)
Using identity, [(a+b)(a-b)= a2-b2]
Here, a = 100
b=4
We get, 104×96 = (100+4)×(100–4)
= (100)2–(4)2
= 10000–16
= 9984

NCERT Solutions for Class 9 Maths Chapter 2
Polynomials
3
...
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

NCERT Solutions for Class 9 Maths Chapter 2
Polynomials
(v) (–2x+5y–3z)2
(vi) ((1/4)a-(1/2)b +1)2
Solution:
(i) (x+2y+4z)2
Using 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
Using 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+z2–4xy–2yz+4xz
(iii) (−2x+3y+2z)2
Solution:
Using 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
Solution:
Using 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)

NCERT Solutions for Class 9 Maths Chapter 2
Polynomials
= 9a2 + 49b2 + c2– 42ab+14bc–6ca
(v) (–2x+5y–3z)2
Solution:
Using 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
Solution:
Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx
Here, x = (1/4)a
y = (-1/2)b
z=1

5
...
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
Solution:
(i) (2x+1)3
Using identity,(x+y)3 = x3+y3+3xy(x+y)
(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,(x–y)3 = x3–y3–3xy(x–y)
(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,(x+y)3 = x3+y3+3xy(x+y)
((3/2)x+1)3=((3/2)x)3+13+(3×(3/2)x×1)((3/2)x +1)

(iv) (x−(2/3)y)3
Using identity, (x –y)3 = x3–y3–3xy(x–y)

NCERT Solutions for Class 9 Maths Chapter 2
Polynomials

7
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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
Solutions:
(i) 8a3+b3+12a2b+6ab2
Solution:
The expression, 8a3+b3+12a2b+6ab2 can be written as (2a)3+b3+3(2a)2b+3(2a)(b)2
8a3+b3+12a2b+6ab2 = (2a)3+b3+3(2a)2b+3(2a)(b)2
= (2a+b)3
= (2a+b)(2a+b)(2a+b)
Here, the identity, (x +y)3 = x3+y3+3xy(x+y) is used
...


(iii) 27–125a3–135a+225a2
Solution:
The expression, 27–125a3–135a +225a2 can be written as 33–(5a)3–3(3)2(5a)+3(3)(5a)2
27–125a3–135a+225a2 =
33–(5a)3–3(3)2(5a)+3(3)(5a)2

NCERT Solutions for Class 9 Maths Chapter 2
Polynomials
= (3–5a)3
= (3–5a)(3–5a)(3–5a)
Here, the identity, (x–y)3 = x3–y3-3xy(x–y) is used
...

(v) 27p3– (1/216)−(9/2) p2+(1/4)p
Solution:
The expression, 27p3–(1/216)−(9/2) p2+(1/4)p can be written as
(3p)3–(1/6)3−(9/2) p2+(1/4)p = (3p)3–(1/6)3−3(3p)(1/6)(3p – 1/6)
Using (x – y)3 = x3 – y3 – 3xy (x – y)
27p3–(1/216)−(9/2) p2+(1/4)p = (3p)3–(1/6)3−3(3p)(1/6)(3p – 1/6)
Taking x = 3p and y = 1/6
= (3p–1/6)3
= (3p–1/6)(3p–1/6)(3p–1/6)
9
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Factorise each of the following:
(i) 27y3+125z3
(ii) 64m3–343n3
Solutions:
(i) 27y3+125z3
The expression, 27y3+125z3 can be written as (3y)3+(5z)3
27y3+125z3 = (3y)3+(5z)3
We know that, 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+25z2)
(ii) 64m3–343n3
The expression, 64m3–343n3can be written as (4m)3–(7n)3
64m3–343n3 =
(4m)3–(7n)3
We know that, 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+49n2)
11
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Solution:
The expression 27x3+y3+z3–9xyz can be written as (3x)3+y3+z3–3(3x)(y)(z)
27x3+y3+z3–9xyz = (3x)3+y3+z3–3(3x)(y)(z)
We know that, x3+y3+z3–3xyz = (x+y+z)(x2+y2+z2–xy –yz–zx)
27x3+y3+z3–9xyz = (3x)3+y3+z3–3(3x)(y)(z)
= (3x+y+z)[(3x)2+y2+z2–3xy–yz–3xz]
= (3x+y+z)(9x2+y2+z2–3xy–yz–3xz)
12
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If x+y+z = 0, show that x3+y3+z3 = 3xyz
...
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
Solution:
(i) (−12)3+(7)3+(5)3
Let a = −12
b=7
c=5
We know that if x+y+z = 0, then x3+y3+z3=3xyz
...

Here, x+y+z = 28–15–13 = 0
(28)3+(−15)3+(−13)3 = 3xyz
= 0+3(28)(−15)(−13)
= 16380
15
...
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
Solution:
(i) Volume: 3x2–12x
3x2–12x can be written as 3x(x–4) by taking 3x out of both the terms
...

12ky2+8ky–20k = 4k(3y2+2y–5)
[Here, 3y2+2y–5 can be written as 3y2+5y–3y–5 using splitting the middle term method

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