**Important questions for Class 9 Maths Chapter 2 Polynomials** are provided here to help the CBSE students score well in their Class 9 Maths exam. The practice questions given here from polynomials chapter (NCERT) will help the students to create a better understanding of the concepts and, thus, develop their problem-solving skills.

Students can find the CBSE Class 9 Important questions from Chapter 2 Polynomials of the subject Maths here. These questions help students to be familiar with the question types and thus face the exam more confidently.

## Important Polynomials Questions For Class 9- Chapter 2 (With Solutions)

Some important questions from polynomials are given below with solutions. These questions will help the 9th class students to get acquainted with a wide variety of questions and develop the confidence to solve polynomial questions more efficiently.

**1. Give an example of a monomial and a binomial having degrees of 82 and 99, respectively.**

**Solution:**

An example of a monomial having a degree of 82 = x^{82}

An example of a binomial having a degree of 99 = x^{99 }+ x

**2. Compute the value of 9x ^{2} + 4y^{2} if xy = 6 and 3x + 2y = 12.**

**Solution:**

Consider the equation 3x + 2y = 12

Now, square both sides:

(3x + 2y)^{2} = 12^{2}

=> 9x^{2} + 12xy + 4y^{2} = 144

=>9x^{2} + 4y^{2} = 144 – 12xy

From the questions, xy = 6

So,

9x^{2} + 4y^{2} = 144 – 72

Thus, the value of 9x^{2} + 4y^{2 }= 72

**3. ** **Find the value of the polynomial 5x – 4x ^{2} + 3 at x = 2 and x = –1.**

**Solution:**

Let the polynomial be f(x) = 5x – 4x^{2} + 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 – 4x^{2} + 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 – 4x^{2} + 3 at x = -1 is -6.

**4. Calculate the perimeter of a rectangle whose area is 25x**^{2}** – 35x + 12. **

**Solution:**

Given,

Area of rectangle = 25x^{2} – 35x + 12

We know, area of rectangle = length × breadth

So, by factoring 25x^{2} – 35x + 12, the length and breadth can be obtained.

25x^{2} – 35x + 12 = 25x^{2 }– 15x – 20x + 12

=> 25x^{2} – 35x + 12 = 5x(5x – 3) – 4(5x – 3)

=> 25x^{2} – 35x + 12 = (5x – 3)(5x – 4)

So, the length and breadth are (5x – 3)(5x – 4).

Now, perimeter = 2(length + breadth)

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

= 2(5x – 3 + 5x – 4) = 2(10x – 7) = 20x – 14

So, the perimeter = 20x – 14

**5. Find the value of x**^{3}** + y**^{3}** + z**^{3}** – 3xyz if x**^{2}** + y**^{2}** + z**^{2}** = 83 and x + y + z = 15**

**Solution:**

Consider the equation x + y + z = 15

From algebraic identities, we know that (a + b + c)^{2} = a^{2} + b^{2} + c^{2 }+ 2(ab + bc + ca)

So,

(x + y + z)^{2} = x^{2 }+ y^{2} + z^{2} + 2(xy + yz + xz)

From the question, x^{2} + y^{2} + z^{2} = 83 and x + y + z = 15

So,

15^{2} = 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

**6. If a + b + c = 15 and a**^{2}** + b**^{2}** + c**^{2}** = 83, find the value of a**^{3}** + b**^{3}** + c**^{3}** – 3abc.**

**Solution:**

We know that,

a^{3} + b^{3} + c^{3} – 3abc = (a + b + c)(a^{2} + b^{2} + c^{2} – ab – bc – ca) ….(i)

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca ….(ii)

Given, a + b + c = 15 and a^{2} + b^{2} + c^{2} = 83

From (ii), we have

15^{2} = 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

a^{3} + b^{3} + c^{3} – 3abc = (a + b + c)[(a^{2} + b^{2} + c^{2} ) – (ab + bc + ca)]

a^{3} + b^{3} + c^{3} – 3abc = 15 × [83 – 71] = 15 × 12 = 180.

**7. If (x – 1/x) = 4, then evaluate (x**^{2}** + 1/x**^{2}**) and (x**^{4}** + 1/x**^{4}**).**

**Solution:**

Given, (x – 1/x) = 4

Squaring both sides we get,

(x – 1/x)^{2} = 16

⇒ x^{2} – 2.x.1/x + 1/x^{2} = 16

⇒ x^{2} – 2 + 1/x^{2} = 16

⇒ x^{2} + 1/x^{2} = 16 + 2 = 18

∴ (x^{2} + 1/x^{2}) = 18 ….(i)

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

(x^{2} + 1/x^{2})^{2} = 324

⇒ x^{4} + 2.x^{2}.1/x^{2} + 1/x^{4} = 324

⇒ x^{4} + 2 + 1/x^{4} = 324

⇒ x^{4} + 1/x^{4} = 324 – 2 = 322

∴ (x^{4} + 1/x^{4}) = 322.

**8. Find the values of a and b so that (2x**^{3}** + ax**^{2}** + x + b) has (x + 2) and (2x – 1) as factors.**

**Solution:**

Let p(x) = 2x^{3 }+ ax^{2} + 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.

**9. Check whether (7 + 3x) is a factor of (3x**^{3}** + 7x).**

**Solution:**

Let p(x) = 3x^{3} + 7x and g(x) = 7 + 3x. Now g(x) = 0 ⇒ x = –7/3.

By the remainder theorem, we know that when p(x) is divided by g(x) 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).

**10. Factorise x**^{2}** + 1/x**^{2}** + 2 – 2x – 2/x.**

**Solution:**

x^{2} + 1/x^{2} + 2 – 2x – 2/x = (x^{2} + 1/x^{2} + 2) – 2(x + 1/x)

= (x + 1/x)^{2} – 2(x + 1/x)

= (x + 1/x)(x + 1/x – 2).

**11. Factorise x**^{2}** – 1 – 2a – a**^{2}**.**

**Solution:**

x^{2} – 1 – 2a – a^{2 }= x^{2} – (1 + 2a + a^{2})

= x^{2} – (1 + a)^{2}

= [x – (1 – a)][x + 1 + a]

= (x – 1 – a)(x + 1 + a)

∴ x^{2} – 1 – 2a – a^{2 }= (x – 1 – a)(x + 1 + a).