Important Questions 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.

Also Check:

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-variation of questions and develop the confidence to solve polynomial questions more efficiently.

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

Solution:

An example of 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 = 25x– 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 + c+ 2(ab + bc + ca)

So,

(x + y + z)2 = x+ 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

More Topics Related to Class 9 Polynomials

 

17 Comments

  1. its very nice app

  2. tough please make the solution a bit easy

  3. I thank byjus for providing these important questions

  4. Nice questions but little bit difficult make it easy to solve

  5. I don’t got question 5, step x^3+y^3+z^3-3xyz=15(83-71)
    Why we subtracted 83-71??

    1. We evaluated the value of xy + yz + xz, initially, which is equal to 71.
      Since, x^3 + y^3 + z^3 – 3xyz = (x + y + z)(x² + y² + z² – (xy + yz + xz)) [By algebraic identities]
      And
      x + y + z = 15, x² + y² + z² = 83 and xy + yz + xz = 71
      So, if we substitute the values, we get:
      x^3 + y^3 + z^3 – 3xyz = 15(83 – 71)
      Please go through the complete solution for the answer.

    2. The expression is x^3+y^3+z^3 = (x+y+z)(x^2+y^2+z^2-xy-yz-xz)

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

      x^2 +y^2 + z^2 = 83, xy+yz+sa = 71
      This is why 83 -71 is done

    3. x + y + z = 15,
      x² + y² + z² = 83
      And
      xy + yz + xz = 71
      Now,
      Acc. To the question,
      x3 + y3 + z3 – 3xyz which is equal to
      = (x + y + z)(x² + y² + z² – (xy + yz + xz))
      And these are given above
      So we subtract 83 – 71

  6. This is so nice we can learn easily

  7. wow! nice app

  8. Thank you so much for these questions

  9. Nice questions and its also easy to check since the solutions are given

  10. Very helpful during online education

  11. maths in byjus is easy.
    thank you for everything

  12. Please add some more questions. There should be at least 10 questions

  13. very good question for the exam.
    thank you

  14. Priyadarshini Dutta

    Nice app but difficult to do the question plz make it

Leave a Comment

Your email address will not be published. Required fields are marked *

BOOK

Free Class