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.

**Also Check:**

- Important 2 Marks Questions for CBSE 9th Maths
- Important 3 Marks Questions for CBSE 9th Maths
- Important 4 Marks Questions for CBSE 9th Maths

## 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 = x^{82}

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

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

**Solution:**

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

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

**Solution:**

Given,

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

We know, area of rectangle = length Ã— breadth

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

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

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

=> 25x^{2}Â – 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 x**^{3}**Â + y**^{3}**Â + z**^{3}**Â – 3xyz if x**^{2}**Â + y**^{2}**Â + z**^{2}**Â = 83 and x + y + z = 15**

**Solution:**

Consider the equation x + y + z = 15

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

So,

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

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

So,

15^{2}Â = 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),

x^{3}Â + y^{3}Â + z^{3}Â – 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, x^{3}Â + y^{3}Â + z^{3}Â – 3xyz = 15(83 – 71)

=> x^{3}Â + y^{3}Â + z^{3}Â – 3xyz = 15 Ã— 12

Or, x^{3}Â + y^{3}Â + z^{3}Â – 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}** â€“ 3abc.**

**Solution:**

We know that,

a^{3} + b^{3} + c^{3} â€“ 3abc = (a + b + c)(a^{2} + b^{2} + c^{2} â€“ ab â€“ bc â€“ ca) â€¦.(i)

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca â€¦.(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)

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

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

a^{3} + b^{3} + c^{3} â€“ 3abc = 15 Ã— [83 â€“ 71] = 15 Ã— 12 = 180.

**7. If (x â€“ 1/x) = 4, then evaluate (x**^{2}** â€“ 1/x**^{2}**) and (x**^{4}** â€“ 1/x**^{4}**).**

**Solution:**

Given, (x â€“ 1/x) = 4

Squaring both sides we get,

(x â€“ 1/x)^{2} = 16

â‡’ x^{2} â€“ 2.x.1/x + 1/x^{2} = 16

â‡’ x^{2} â€“ 2 + 1/x^{2} = 16

â‡’ x^{2} + 1/x^{2} = 16 + 2 = 18

âˆ´ (x^{2} + 1/x^{2}) = 18 â€¦.(i)

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

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

â‡’ x^{4} + 2.x^{2}.1/x^{2} + 1/x^{4} = 324

â‡’ x^{4} + 2 + 1/x^{4} = 324

â‡’ x^{4} + 1/x^{4} = 324 â€“ 2 = 322

âˆ´ (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 â€“ 1) as factors.**

**Solution:**

Let p(x) = 2x^{3 }+ ax^{2} + 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 (3x**^{3}** + 7x).**

**Solution:**

Let p(x) = 3x^{3} + 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 x**^{2}** + 1/x**^{2}** + 2 â€“ 2x â€“ 2/x.**

**Solution:**Â

x^{2} + 1/x^{2} + 2 â€“ 2x â€“ 2/x = (x^{2} + 1/x^{2} + 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 x**^{2}** â€“ 1 â€“ 2a â€“ a**^{2}**.**

**Solution:**

x^{2} â€“ 1 â€“ 2a â€“ a^{2 }= x^{2} â€“ (1 + 2a + a^{2})

= x^{2} â€“ (1 + a)^{2}

= [x â€“ (1 â€“ a)][x + 1 + a]

= (x â€“ 1 â€“ a)(x + 1 + a)

âˆ´ x^{2} â€“ 1 â€“ 2a â€“ a^{2 }= (x â€“ 1 â€“ a)(x + 1 + a).

### More Topics Related to Class 9 Polynomials

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

Why we subtracted 83-71??

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)

Please go through the complete solution for the answer.

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

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

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)

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

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