 # 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.

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## 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

1. gauri singhal

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2. RISHI GAUR

tough please make the solution a bit easy

3. Byjus customer

I thank byjus for providing these important questions

4. Jyoti kr.mishra

Nice questions but little bit difficult make it easy to solve

5. Anonymous

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

1. lavanya

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)

2. Aritra pal

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. Jatin kumar

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. sheikh akmal

wow! nice app

8. Tanaya Ghosh

Thank you so much for these questions

9. Samakhya Sharma

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

10. Sakht Launda

11. maths in byjus is easy.
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12. Suhani Mahajan

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

13. chirayu jain

very good question for the exam.
thank you