Important Questions CBSE Class 9 Maths Chapter 2 Polynomial

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 = x⁸²

An example of a binomial having a degree of 99 = x⁹⁹ + x

2. Compute the value of 9x² + 4y² if xy = 6 and 3x + 2y = 12.

Solution:

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

3.  Find the value of the polynomial 5x – 4x² + 3 at x = 2 and x = –1.

Solution:

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.

4. Calculate the perimeter of a rectangle whose area is 25x² – 35x + 12. 

Solution:

Given,

Area of rectangle = 25x² – 35x + 12

We know, area of 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)

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 x³ + y³ + z³ – 3xyz if x² + y² + z² = 83 and x + y + z = 15

Solution:

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),

x³ + y³ + z³ – 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, x³ + y³ + z³ – 3xyz = 15(83 – 71)

=> x³ + y³ + z³ – 3xyz = 15 × 12

Or, x³ + y³ + z³ – 3xyz = 180

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

Solution:

We know that,

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.

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

Solution:

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.

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

Solution:

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.

9. Check whether (7 + 3x) is a factor of (3x³ + 7x).

Solution:

Let p(x) = 3x³ + 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 x² + 1/x² + 2 – 2x – 2/x.

Solution: 

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).

11. Factorise x² – 1 – 2a – a².

Solution:

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).