# Important Questions Class 9 Maths Chapter 2

## Important Questions Class 9 Mathematics Chapter 2- Polynomials.

Mathematics Chapter 2 of Class 9  is about Polynomials. Polynomial consists of two terms, namely Poly (meaning “many”) and Nominal (meaning “terms.”). A polynomial is explained as an expression which is composed of variables, constants and exponents that are combined using mathematical operations like addition, subtraction, multiplication and division (No division operation by a variable). Based on the number of terms present in the expression, it is classified as monomial, binomial, and trinomial.

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## Important Questions Class 9 Mathematics Chapter 2 – With Solutions

Our in-house Mathematics faculty experts  have collated a complete list of Important Questions Class 9 Mathematics Chapter 2 by referring to various sources. The subject experts have meticulously prepared  illustration for individual questions that will enable students to comprehend the notions used in each question. Furthermore, the questions are selected in a way that would cover all the topics. So by practising from our question bank, students will be able to revise the chapter and understand their strong and weak topics. And enhance their preparation by further concentrating on weaker sections of the chapter.

Given below are a few of the questions and answers from our question bank of Important Questions Class 9 Mathematics Chapter 2:

Question 1: Calculate the value of 9x² + 4y² if xy = 6 and 3x + 2y = 12.

Answer 1: Consider the equation 3x + 2y = 12

Now, square both sides:

(3x + 2y)² = 12²

=> 9x² + 12xy + 4y² = 144

=>9x² + 4y² = 144 – 12xy

From the questions, xy = 6

So,

9x² + 4y² = 144 – 72

Thus, the value of 9x² + 4y² = 72

Question 2:Evaluate the following using suitable identity

(102) ³

Answer 2: We can write 102 as 100+2

Using identity,(x+y) ³ = x ³ +y ³ +3xy(x+y)

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

= 1000000 + 8 + 600(100 + 2)

= 1000000 + 8 + 60000 + 1200

= 1061208

Question 3:Without any actual division, prove that the following 2x⁴

– 5x³ + 2x² – x + 2 is divisible by x² – 3x + 2.

[Hint: Factorise x² – 3x + 2]

x²-2x-1x+2

x(x-2)-1(x-2)

(x-2)(x-1)

Therefore,(x-2)(x-1)are the factors.

Considering (x-2),

x-2=0

x=2

Then, p(x) becomes,

p(x)=2

p(x)=2x⁴-5x³+2x²-x+2

p(2)=2(2)⁴-5(2)³+2(2)²-2+2

=32-40+8

= -40+40=0

Therefore, (x-2) is a factor.

Considering (x-1),

x-1=0

x=1

Then, p(x) becomes,

p(x)=1

p(x)=2x⁴-5x³+2x²-x+2

p(1)=2(1)⁴-5(1)³+2(1)²-1+2

=2-5+2-1+2

=6-6

=0

Therefore, (x-1) is a factor.

Question 4: Using the Factor Theorem to determine whether g(x) is a factor of p(x) in the following case

(i) p(x) = 2x³+x²–2x–1, g(x) = x+1

Answer 4:p(x) = 2x³+x²–2x–1, g(x) = x+1

g(x) = 0

⇒ x+1 = 0

⇒ x = −1

∴Zero of g(x) is -1.

Now,

p(−1) = 2(−1)³+(−1)²–2(−1)–1

= −2+1+2−1

= 0

∴By the given factor theorem, g(x) is a factor of p(x).

Question 5: Obtain an example of a monomial and a binomial having degrees of 82 and 99, respectively.

Answer 5: An example of a monomial having a degree of 82 = x⁸²

An example of a binomial having a required degree of 99 = x⁹⁹ + 7

Question 6: If the two  x – 2 and x – ½ are the given factors of px²

+ 5x + r, show that p = r.

Answer 6: Given, f(x) = px²+5x+r and factors are x-2, x – ½

g1(x) = 0,

x – 2 = 0

x = 2

Substituting x = 2 in place of the equation, we get

f(x) = px²+5x+r

f(2) = p(2)²+5(2)+r=0

= 4p + 10 + r = 0 … eq.(i)

x – ½ = 0

x = ½

Substituting x = ½ in place of the equation, we get,

f(x) = px²+5x+r

f( ½ ) = p( ½ )² + 5( ½ ) + r =0

= p/4 + 5/2 + r = 0

= p + 10 + 4r = 0 … eq(ii)

On solving eq(i) and eq(ii),

We get,

4p + r = – 10 and p + 4r = – 10

the RHS of both equations are the same,

We get,

4p + r = p + 4r

3p=3r

p = r.

Hence Proved.

Question 7: Identify constant, linear, quadratic, cubic and quartic polynomials from the following.

(i) – 7 + x

(ii) 6y

(iii) – ? ³

(iv) 1 – y – ? ³

(v) x – ? ³ + ?⁴

(vi) 1 + x + ?²

(vii) -6?²

(viii) -13

(ix) –p

Answer 7: (i) – 7 + x

The degree of – 7 + x is 1.

Hence, it is a linear polynomial.

(ii) 6y

The degree of 6y is 1.

Therefore, it is a linear polynomial.

(iii) – ? ³

We know that the degree of – ? ³ is 3.

Therefore, it is a cubic polynomial.

(iv) 1 – y – ? ³

We know that the degree of 1 – y – ? ³ is 3.

Therefore, it is a cubic polynomial.

(v) x – ? ³ + ?⁴

We know that the degree of x – ? ³ + ?⁴ is 4.

Therefore, it is a quartic polynomial.

(vi) 1 + x + ?²

We know that the degree of 1 + x + ?² is 2.

Therefore, it is a quadratic polynomial.

(vii) -6?²

We know that the degree of -6?² is 2.

Therefore, it is a quadratic polynomial.

(viii) -13

We know that -13 is a constant.

Therefore, it is a constant polynomial.

(ix) – p

We know that the degree of –p is 1.

Therefore, it is a linear polynomial.

Question 8: Observe the value of the polynomial 5x – 4x² + 3 at x = 2 and x = –1.

Answer 8: Let the polynomial be f(x) = 5x – 4x² + 3

Now, for x = 2,

f(2) = 5(2) – 4(2)² + 3

=> f(2) = 10 – 16 + 3 = –3

Or, the value of the polynomial 5x – 4x² + 3 at x = 2 is -3.

Similarly, for x = –1,

f(–1) = 5(–1) – 4(–1)² + 3

=> f(–1) = –5 –4 + 3 = -6

The value of the polynomial 5x – 4x² + 3 at x = -1 is -6.

Question 9:Expanding each of the following, using all the 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

Using identity, (x+y+z)² = x²+²+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

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

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)

= 9a² + 49b² + c²– 42ab+14bc–6ca

(v) (–2x+5y–3z)²

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

Question 10: If the polynomials az³ + 4z² + 3z – 4 and z³ – 4z + leave the same remainder when divided by z – 3, find the value of a.

Answer 10: Zero of the polynomial,

g1(z) = 0

z-3 = 0

z = 3

Hence, zero of g(z) = – 2a

Let p(z) = az³+4z²+3z-4

Now, substituting the given value of z = 3 in p(z), we get,

p(3) = a (3)³ + 4 (3)² + 3 (3) – 4

⇒p(3) = 27a+36+9-4

⇒p(3) = 27a+41

Let h(z) = z³-4z+a

Now, by substituting the value of z = 3 in h(z), we get,

h(3) = (3)³-4(3)+a

⇒h(3) = 27-12+a

⇒h(3) = 15+a

As per the question,

The two polynomials, p(z) and h(z), leave the same remainder when divided by z-3

So, h(3)=p(3)

⇒15+a = 27a+41

⇒15-41 = 27a – a

⇒-26 = 26a

⇒a = -1

Question 11: Compute the perimeter of a rectangle whose area is 25x² – 35x + 12.

Answer 11: Area of rectangle = 25x² – 35x + 12

We know the area of a rectangle = length × breadth

So, by factoring 25x² – 35x + 12, the length and breadth can be obtained.

25x² – 35x + 12 = 25x² – 15x – 20x + 12

=> 25x² – 35x + 12 = 5x(5x – 3) – 4(5x – 3)

=> 25x² – 35x + 12 = (5x – 3)(5x – 4)

Thus, the length and breadth of a rectangle are (5x – 3)(5x – 4).

So, the perimeter = 2(length + breadth)

Therefore, the perimeter of the given rectangle = 2[(5x – 3)+(5x – 4)]

= 2(5x – 3 + 5x – 4)

= 2(10x – 7)

= 20x – 14

Hence, the perimeter of the rectangle = 20x – 14

Question 12: 2x²+y²+²–2√2xy+4√2yz–8xz

Answer 12: Using identity, (x +y+z)² = x²+y²+z²+2xy+2yz+2zx

We can say that, x²+²+²+2xy+2yz+2zx = (x+y+z)²

2x²+y²+8z²–2√2xy+4√2yz–8xz

= (-√2x)²+(y)²+(2√2z)²+(2×-√2x×y)+(2×y×2√2z)+(2×2√2×−√2x)

= (−√2x+y+2√2z)²

= (−√2x+y+2√2z)(−√2x+y+2√2z)

Question 13: If ? + 2? is a factor of ? ⁵ – 4?²?³ + 2? + 2? + 3, find a.

Answer 13: According to the question,

Let p(x) = x ⁵ – 4a²x³ + 2x + 2a + 3 and g(x) = x + 2a

g(x) = 0

⟹ x + 2a = 0

⟹ x = – 2a

Hence, zero of g(x) = – 2a

As per the factor theorem,

If g(x) is a factor of p(x), then p( – 2a) = 0

So, substituting the value of x in p(x), we get,

p ( – 2a) = ( – 2a) ⁵ – 4a²( – 2a)³ + 2( – 2a) + 2a + 3 = 0

⟹ – 32a ⁵ + 32a ⁵ – 2a + 3 = 0

⟹ – 2a = – 3

⟹ a = 3/2

Question 14: Find the value of x³+ y ³ + z ³ – 3xyz if x² + y² + z² = 83 and x + y + z = 1

Answer 14: Consider the equation x + y + z = 15

From algebraic identities, we know that (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

So,

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

From the question, x² + y² + z²= 83 and x + y + z = 15

So,

152 = 83 + 2(xy + yz + xz)

=> 225 – 83 = 2(xy + yz + xz)

Or, xy + yz + xz = 142/2 = 71

Using algebraic identity a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca),

x ³ + y ³ + z ³ – 3xyz = (x + y + z)(x² + y² + z² – (xy + yz + xz))

Now,

x + y + z = 15, x² + y² + z² = 83 and xy + yz + xz = 71

So, x ³ + y ³ + z ³ – 3xyz = 15(83 – 71)

=> x ³ + y ³ + z ³ – 3xyz = 15 × 12

Or, x ³ + y ³ + z ³ – 3xyz = 180

Question 15:Verify that:

(i) x³+y³ = (x+y)(x²–xy+y²)

(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)

(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 16: For what value of m is ?³ – 2??² + 16 divisible by x + 2?

Answer 16: According to the question,

Let p(x) = x³ – 2mx² + 16, and g(x) = x + 2

g(x) = 0

⟹ x + 2 = 0

⟹ x = – 2

Hence, zero of g(x) = – 2

As per the factor theorem,

if p(x) is divisible by g(x), then the remainder of p(−2) should be zero.

Thus, substituting the value of x in p(x), we obtain,

p( – 2) = 0

⟹ ( – 2)³ – 2m( – 2)² + 16 = 0

⟹ 0 – 8 – 8m + 16 = 0

⟹ 8m = 8

⟹ m = 1

Question 17:If a + b + c = 15 and a² + b² + c² = 83, find the value of a³ + b³ + c³ – 3abc.

a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca) ….(i)

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca ….(ii)

Given, a + b + c = 15 and a² + b² + c² = 83

From (ii), we have

152 = 83 + 2(ab + bc + ca)

⇒ 225 – 83 = 2(ab + bc + ca)

⇒ 142/2 = ab + bc + ca

⇒ ab + bc + ca = 71

Now, (i) can be written as

a³ + b³ + c³ – 3abc = (a + b + c)[(a² + b² + c² ) – (ab + bc + ca)]

a³ + b³+ c³ – 3abc = 15 × [83 – 71] = 15 × 12 = 180.

Question 18: Factorise: 27x³+y³+z³–9xyz

Answer 18: The expression27x³+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+³–3xyz = (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+²–3xy–yz–3xz)

Question 19: If (x – 1/x) = 4, then evaluate (x² + 1/x²) and (x⁴ + 1/x⁴).

Answer 19: Given, (x – 1/x) = 4

Squaring both sides, we get,

(x – 1/x)² = 16

⇒ x² – 2.x.1/x + 1/x² = 16

⇒ x² – 2 + 1/x² = 16

⇒ x² + 1/x² = 16 + 2 = 18

∴ (x² + 1/x²) = 18 ….(i)

Again, squaring both sides of (i), we get

(x² + 1/x²)² = 324

⇒ x⁴ + 2.x².1/x² + 1/x⁴= 324

⇒ x⁴ + 2 + 1/x⁴ = 324

⇒ x⁴ + 1/x⁴ = 324 – 2 = 322

∴ (x⁴ + 1/x⁴) = 322.

Question 20: Factorise

64m³–343n³

Answer 20: The expression 64m³–343n³ can 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)

Question 21: Find out the values of a and b so that (2x³ + ax² + x + b) has (x + 2) and (2x – 1) as factors.

Answer 21: Let p(x) = 2x³ + ax² + x + b. Then, p( –2) = and p(½) = 0.

p(2) = 2(2)³ + a(2)² + 2 + b = 0

⇒ –16 + 4a – 2 + b = 0 ⇒ 4a + b = 18 ….(i)

p(½) = 2(½)³ + a(½)² + (½) + b = 0

⇒ a + 4b = –3 ….(ii)

On solving (i) and (ii), we get a = 5 and b = –2.

Hence, a = 5 and b = –2.

Question 22: Explain that p – 1 is a factor of p¹⁰ – 1 and p¹¹ – 1.

Answer 22: According to the question,

Let h(p) = ?¹⁰ − 1,and g(p) = ? – 1

zero of g(p) ⇒ g(p) = 0

p – 1 = 0

p = 1

Therefore, zero of g(x) = 1

We know that,

According to the factor theorem, if g(p) is a factor of h(p), then h(1) should be zero

So,

h(1) = (1) ¹⁰ − 1 = 1 − 1 = 0

⟹ g (p) is a factor of h(p).

Here, we have h(p) = ?¹¹ − 1, g (p) = ? – 1

Putting g (p) = 0 ⟹ ? − 1 = 0 ⟹ ? = 1

As per the factor theorem, if g (p) is a factor of h(p),

Then h(1) = 0

⟹ (1) ¹¹ – 1 = 0

Hence, g(p) = ? – 1 is the factor of h(p) = ? ¹⁰ – 1

Question 23: Examine whether (7 + 3x) is a factor of (3×3 + 7x).

Answer 23: Let p(x) = 3×3 + 7x and g(x) = 7 + 3x. Now g(x) = 0 ⇒ x = –7/3.

By the remainder theorem, p(x) is divided by g(x), and then the remainder is p(–7/3).

Now, p(–7/3) = 3(–7/3)3 + 7(–7/3) = –490/9 ≠ 0.

∴ g(x) is not a factor of p(x).

Question 24:Prove that:

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

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)(2×2+2y²+²–2xy–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 25: Find out which of the following polynomials has x – 2 a factor:

(i) 3?² + 6?−24.

(ii) 4?² + ?−2.

Answer 25: (i) According to the question,

Let p(x) =3?² + 6?−24 and g(x) = x – 2

g(x) = x – 2

zero of g(x) ⇒ g(x) = 0

x – 2 = 0

x = 2

Hence, zero of g(x) = 2

Thus, substituting the value of x in p(x), we get,

p(2) = 3(2)² + 6 (2) – 24

= 12 + 12 – 24

= 0

the remainder = zero,

We can derive that,

g(x) = x – 2 is factor of p(x) = 3?² + 6?−24

(ii) According to the question,

Let p(x) = 4?² + ?−2 and g(x) = x – 2

g(x) = x – 2

zero of g(x) ⇒ g(x) = 0

x – 2 = 0

x = 2

Hence, zero of g(x) = 2

Thus, substituting the value of x in p(x), we get,

p(2) = 4(2)² + 2−2

= 16 ≠ 0

Since the remainder ≠ zero,

We can say that,

g(x) = x – 2 is not a factor of p(x) = 4?² + ?−2

Question 26: Factorise x² + 1/x² + 2 – 2x – 2/x.

Answer 26: x² + 1/x² + 2 – 2x – 2/x = (x² + 1/x² + 2) – 2(x + 1/x)

= (x + 1/x)² – 2(x + 1/x)

= (x + 1/x)(x + 1/x – 2).

Question 27: Factorise

8a³+b³+12a²b+6ab²

Answer 27: 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.

Question 28: By Remainder Theorem, find out the remainder when p(x) is divided by g(x), where

(i) p(?) = ?³ – 2?² – 4? – 1, g(?) = ? + 1

(ii) p(?) = ?³ – 3?² + 4? + 50, g(?) = ? – 3

(iii) p(?) = 4?³ – 12?² + 14? – 3, g(?) = 2? – 1

(iv) p(?) = ?³ – 6?² + 2? – 4, g(?) = 1 – 3/2 ?

Answer 28: (i) Given p(x) = ?³ – 2?² – 4? – 1 and g(x) = x + 1

Here zero of g(x) = – 1

By applying the remainder theorem

P(x) divided by g(x) = p( – 1)

P ( – 1) = ( – 1)³ – 2 ( – 1)² – 4 ( – 1) – 1 = 0

Therefore, the remainder = 0

(ii) given p(?) = ?³ – 3?² + 4? + 50, g(?) = ? – 3

Here zero of g(x) = 3

By applying the remainder theorem p(x) divided by g(x) = p(3)

p(3) = 3³ – 3 × (3)² + 4 × 3 + 50 = 62

Therefore, the remainder = 62

(iii) p(x) = 4x³ – 12x² + 14x – 3, g(x) = 2x – 1

Here zero of g(x) = ½

By applying the remainder theorem p(x) divided by g(x) = p (½)

P( ½ ) = 4( ½ )³ – 12( ½ )² + 14 ( ½ ) – 3

= 4/8 – 12/4 + 14/2 – 3

= ½ + 1

= 3/2

Hence, the remainder = 3/2

(iv) p(?) = ?³ – 6?² + 2? – 4, g(?) = 1 – 3/2 ?

so, zero of g(x) = 2/3

By applying the remainder theorem p(x) divided by g(x) = p(2/3)

p(2/3) = (2/3)³ – 6(2/3)² + 2(2/3) – 4

= – 136/27

Therefore, the remainder = – 136/27

Question 29:Factorise x² – 1 – 2a – a².

Answer 29: x² – 1 – 2a – a² = x² – (1 + 2a + a²)

= x² – (1 + a)²

= [x – (1 – a)][x + 1 + a]

= (x – 1 – a)(x + 1 + a)

∴ x² – 1 – 2a – a² = (x – 1 – a)(x + 1 + a).

Question 30:Evaluate the following using suitable identity

(998)³

Answer 30: We can write 99 as 1000–2

Using identity,(x–y)³ = x³ –y³ –3xy(x–y)

(998)³ =(1000–2)³

=(1000)³ –2³ –(3×1000×2)(1000–2)

= 1000000000–8–6000(1000– 2)

= 1000000000–8- 6000000+12000

= 994011992

Question 31: Find the zeroes of the polynomial:

p(?)= (? –2)² −(? + 2)²

Answer 31: p(x) = (? –2)² −(? + 2)²

We know that,

Zero of the polynomial p(x) = 0

Hence, we get,

⇒ (x–2)² −(x + 2)² = 0

Expanding using the identity, a² – b² = (a – b) (a + b)

⇒ (x – 2 + x + 2) (x – 2 –x – 2) = 0

⇒ 2x ( – 4) = 0

⇒ – 8 x= 0

Therefore, the zero of the polynomial = 0

### Benefits Of Solving Important Questions Class 9 Mathematics Chapter 2

Consistently solving questions is a vital element of mastering   Mathematics. By solving Mathematics Class 9 Chapter 2 important questions, students can get a further understanding of the polynomials chapter.

A few other advantages of solving Important Questions Class 9 Mathematics Chapter 2 are:

• Class 9 Mathematics Chapter 2 important questions provide details about the types of questions that may be expected in the exams, which increases their confidence in achieving a high grade. .
• Studying something once may help you understand the concepts, but through revisions and guided practice make them aware of their mistakes and help to get the best results. Your chances of getting good grades on exams increase as you solve all the questions from our question bank of Important Questions Class 9 Mathematics Chapter 2.
• The questions and answers provided are based on the NCERT books and the latest CBSE syllabus and examination guidelines. So the students can have complete faith and trust in Extramarks because CBSE itself recommends NCERT books and students can rely on the authenticity of our solutions.
• By solving our Chapter 2 Class 9 Mathematics important questions, students will understand the question paper pattern.. Practising questions identical to the exam questions would help the students become confident in solving advanced level questions with ease and get 100% results in their exams.

Extramarks leaves no stone unturned to give the best learning material to students while combining fun and learning activities through its own study materials to enhance their learning experience.  It provides comprehensive learning solutions for students from Class 1 to Class 12. Our website has abundant resources, along with important questions and solutions. Students can click on the links given below to access some of these resources:

Q.1 By actual division, find the quotient and the remainder when x5 + 1 is divided by x 1

Marks:3
Ans

$\begin{array}{l}\text{}{\mathrm{x}}^{4}+{\mathrm{x}}^{3}+{\mathrm{x}}^{2}+\mathrm{x}+1\\ \mathrm{x}1\overline{){\mathrm{x}}^{5}+1\text{}}\\ \text{}{\mathrm{x}}^{5}{\mathrm{x}}^{4}\\ \text{}\text{+}\\ \text{}\overline{\begin{array}{l}{\text{x}}^{4}+1\\ {\text{x}}^{4}{\mathrm{x}}^{3}\\ \text{+}\\ \overline{\begin{array}{l}{\text{x}}^{3}\text{+ 1}\\ {\text{x}}^{3}{\mathrm{x}}^{2}\\ \text{+}\end{array}}\\ \text{}\overline{\begin{array}{l}{\text{x}}^{2}+1\\ {\text{x}}^{2}\mathrm{x}\\ \text{+}\\ \overline{\begin{array}{l}\text{x + 1}\\ \text{x 1}\\ \text{+}\\ \overline{\text{2}}\end{array}}\end{array}}\end{array}}\\ \mathrm{Quotient}:\text{}{\mathrm{x}}^{4}+{\mathrm{x}}^{3}+{\mathrm{x}}^{2}+\mathrm{x}+1\\ \mathrm{Remainder}:\text{2}\end{array}$

Q.2 Find the value of k if x 5 is a factor of kx2 + 3x + 7.

Marks:2
Ans

$\begin{array}{l}\mathrm{Zero}\text{of x 5 is 5}\mathrm{as}\text{}\mathrm{x}\text{}\text{5}=\text{}0\text{}\mathrm{gives}\text{}\mathrm{x}\text{}=\text{5}.\\ {\text{p(x) = kx}}^{2}+3\mathrm{x}+7\\ \mathrm{p}\left(5\right)=0\\ 25\mathrm{k}+15+7=0\\ 25\mathrm{k}+22=0\\ \mathrm{k}=\frac{22}{25}\end{array}$

Q.3 If x + y + z = 6 and xy + yz + zx = 11, then find the value of x2 + y2 + z2.

Marks:3
Ans

$\begin{array}{l}\mathrm{Given}:\text{}\mathrm{x}+\mathrm{y}+\mathrm{z}=6\text{and}\mathrm{xy}+\mathrm{yz}+\mathrm{zx}=11\\ \mathrm{Squaring}\text{}\mathrm{both}\text{}\mathrm{sides},\text{}\mathrm{we}\text{}\mathrm{get}\\ \text{}{\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)}^{2}={\left(6\right)}^{2}\\ \text{}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}+2\mathrm{xy}+2\mathrm{yz}+2\mathrm{zx}=36\\ \text{}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}+2\left(\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\right)\text{}=\text{}36\\ \text{}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}+2\left(11\right)\text{}=\text{}36\text{}\left[\text{Since}\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\text{}=\text{}11\right]\\ \text{}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}+22=36\\ \text{}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}\text{}=\text{}3622\text{}=\text{}14\end{array}$

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### 1. What are the four types of polynomials?

The 4 types of polynomials are zero polynomial, linear polynomial, quadratic polynomial, and cubic polynomial.

### 2. Where can I get important questions for Class 9 Mathematics Chapter 2 online?

On the Extramarks website, you can find all of the important questions for Class 9 Mathematics Chapter 2, along with their answers. On the website, you can also find important questions and NCERT solutions for all classes from 1 to 12.

### 3. What are the important chapters in Class 9 Mathematics?

The NCERT Mathematics book has 15 chapters. Each chapter is equally important when it comes to learning the fundamentals and taking the test. Additionally, because CBSE does not specify the distribution of marks for each chapter, students are advised to fully study all chapters. Each and every chapter must be completely understood to acquire a good grade in exams.

All the fifteen chapters of CBSE Class 9 Mathematics syllabus are given below:

1. Chapter – Number Systems
2. Chapter – Polynomials
3. Chapter – Coordinate Geometry
4. Chapter – Linear Equations In Two Variables
5. Chapter – Introduction To Euclid’s Geometry
6. Chapter – Lines And Angles
7. Chapter – Triangles