# 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 the 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.

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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 a monomial having a degree of 82 = x82

An example of a binomial having a degree of 99 = x99Â + x

2.Â 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.Â Â Find the value of the polynomial 5x â€“ 4x2Â + 3 at x = 2 and x = â€“1.

Solution:

Let the polynomial be f(x) =Â 5x â€“ 4x2Â + 3

Now, for x = 2,

f(2) = 5(2)Â â€“ 4(2)2 + 3

=> f(2) = 10 â€“ 16 + 3 = â€“3

Or,Â the value of the polynomial 5x â€“ 4x2Â + 3 at x = 2 is -3.

Similarly, for x = â€“1,

f(â€“1) = 5(â€“1) – 4(â€“1)2 + 3

=>Â f(â€“1) =Â â€“5Â â€“4 + 3 = -6

The value of the polynomialÂ 5x â€“ 4x2Â + 3 at x = -1 is -6.

4. Calculate the perimeter of a rectangle whose area is 25x2Â – 35x + 12.Â

Solution:

Given,

Area of rectangle = 25x2Â – 35x + 12

We know, area of rectangle = length Ã— breadth

So, by factoring 25x2Â – 35x + 12, the length and breadth can be obtained.

25x2Â – 35x + 12 = 25x2Â – 15x – 20x + 12

=> 25x2Â – 35x + 12 = 5x(5x – 3) – 4(5x – 3)

=> 25x2Â – 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.Â Find the value of x3Â + y3Â + z3Â – 3xyz if x2Â + y2Â + z2Â = 83 and x + y + z = 15

Solution:

Consider the equation x + y + z = 15

From algebraic identities, we know that (a + b + c)2Â = a2Â + b2Â + c2Â + 2(ab + bc + ca)

So,

(x + y + z)2Â = x2Â + y2Â + z2Â + 2(xy + yz + xz)

From the question, x2Â + y2Â + z2Â = 83 and x + y + z = 15

So,

152Â = 83 + 2(xy + yz + xz)

=> 225 – 83 = 2(xy + yz + xz)

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

Using algebraic identity aÂ³ + bÂ³ + cÂ³ – 3abc = (a + b + c)(aÂ² + bÂ² + cÂ² – ab – bc – ca),

x3Â + y3Â + z3Â – 3xyz = (x + y + z)(xÂ² + yÂ² + zÂ² – (xy + yz + xz))

Now,

x + y + z = 15, xÂ² + yÂ² + zÂ² = 83 and xy + yz + xz = 71

So, x3Â + y3Â + z3Â – 3xyz = 15(83 – 71)

=> x3Â + y3Â + z3Â – 3xyz = 15 Ã— 12

Or, x3Â + y3Â + z3Â – 3xyz = 180

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

Solution:

We know that,

a3 + b3 + c3 â€“ 3abc = (a + b + c)(a2 + b2 + c2 â€“ ab â€“ bc â€“ ca) â€¦.(i)

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

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

From (ii), we have

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

â‡’ 225 â€“ 83 = 2(ab + bc + ca)

â‡’ 142/2 = ab + bc + ca

â‡’ ab + bc + ca = 71

Now, (i) can be written as

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

a3 + b3 + c3 â€“ 3abc = 15 Ã— [83 â€“ 71] = 15 Ã— 12 = 180.

7. If (x â€“ 1/x) = 4, then evaluate (x2 â€“ 1/x2) and (x4 â€“ 1/x4).

Solution:

Given, (x â€“ 1/x) = 4

Squaring both sides we get,

(x â€“ 1/x)2 = 16

â‡’ x2 â€“ 2.x.1/x + 1/x2 = 16

â‡’ x2 â€“ 2 + 1/x2 = 16

â‡’ x2 + 1/x2 = 16 + 2 = 18

âˆ´ (x2 + 1/x2) = 18 â€¦.(i)

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

(x2 + 1/x2)2 = 324

â‡’ x4 + 2.x2.1/x2 + 1/x4 = 324

â‡’ x4 + 2 + 1/x4 = 324

â‡’ x4 + 1/x4 = 324 â€“ 2 = 322

âˆ´ (x4 + 1/x4) = 322.

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

Solution:

Let p(x) = 2x3 + ax2 + x + b. Then, p( â€“2) = and p(Â½) = 0.

p(2) = 2(2)3 + a(2)2 + 2 + b = 0

â‡’ â€“16 + 4a â€“ 2 + b = 0 â‡’ 4a + b = 18 â€¦.(i)

p(Â½) = 2(Â½)3 + a(Â½)2 + (Â½) + b = 0

â‡’ a + 4b = â€“3 â€¦.(ii)

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

Hence, a = 5 and b = â€“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 â‡’ x = â€“7/3.

By the remainder theorem, we know that when p(x) is divided by g(x) then the remainder is p(â€“7/3).

Now, p(â€“7/3) = 3(â€“7/3)3 + 7(â€“7/3) = â€“490/9 â‰  0.

âˆ´ g(x) is not a factor of p(x).

10. Factorise x2 + 1/x2 + 2 â€“ 2x â€“ 2/x.

Solution:Â

x2 + 1/x2 + 2 â€“ 2x â€“ 2/x = (x2 + 1/x2 + 2) â€“ 2(x + 1/x)

= (x + 1/x)2 â€“ 2(x + 1/x)

= (x + 1/x)(x + 1/x â€“ 2).

11. Factorise x2 â€“ 1 â€“ 2a â€“ a2.

Solution:

x2 â€“ 1 â€“ 2a â€“ a2 = x2 â€“ (1 + 2a + a2)

= x2 â€“ (1 + a)2

= [x â€“ (1 â€“ a)][x + 1 + a]

= (x â€“ 1 â€“ a)(x + 1 + a)

âˆ´ x2 â€“ 1 â€“ 2a â€“ a2 = (x â€“ 1 â€“ a)(x + 1 + a).

### More Topics Related to Class 9 Polynomials

1. gauri singhal

its very nice app

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^3Â + y^3Â + z^3Â – 3xyz = (x + y + z)(xÂ² + yÂ² + zÂ² – (xy + yz + xz)) [By algebraic identities]
And
x + y + z = 15, xÂ² + yÂ² + zÂ² = 83 and xy + yz + xz = 71
So, if we substitute the values, we get:
x^3Â + y^3Â + z^3Â – 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,
xÂ² + yÂ² + zÂ² = 83
And
xy + yz + xz = 71
Now,
Acc. To the question,
x3 + y3 + z3 â€“ 3xyz which is equal to
= (x + y + z)(xÂ² + yÂ² + zÂ² â€“ (xy + yz + xz))
And these are given above
So we subtract 83 – 71

4. Smart Androidz

because the formula is x^3+y^3+z^3-3xyz = (x+y+z)(x^2 + y^2 + z^2-xy-yz-zx)
and we know x+y+z= 15 and x^2 + y^2 + z^2 = 83 and xy+yz+zx = 71 and -xy-yz-zx = -71
so we have putted the values on formula 15(83-71)

5. fenom beatz

we found the value of xy+yz+zx so in the above step we take ‘-‘ as common and then xy+yz+zx so it will -71
then it is 15*12=180

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

Nice app but difficult to do the question plz make it

15. Aishwarya

Very easy to solve

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