Please see **Polynomials Exam Questions Class 9 Mathematics ** below. These important questions with solutions have been prepared based on the latest examination guidelines and syllabus issued by CBSE, NCERT, and KVS. We have provided Class 9 Mathematics Questions and answers for all chapters in your NCERT Book for Class 9 Mathematics. These solved problems for Polynomials in Class 9 Mathematics will help you to score more marks in upcoming examinations.

**Exam Questions Polynomials Class 9 Mathematics**

**Very Short Answer Type Questions:**

**Question. Write the zeroes of the polynomial p(x) = x(x – 2)(x – 3).****Ans.** For zeroes, put p(x) = 0

∴ x(x – 2)(x – 3) = 0,

Therefore, x = 0, 2, 3

**Question. Write an example of a constant polynomial.****Ans.** Constant polynomial is 7.

**Question. Find the value of m, if x + 4 is a factor of the polynomial x ^{2} + 3x + m.**

**Ans.**Given, x + 4 is a factor of x

^{2}+ 3x + m = p(x)

∴ p(– 4) = 0

or, 16 – 12 + m = 0

or, m = – 4

**Question. Write the expression which represents a polynomial.****Ans.** 3 x^{2} – x – 1.

**Question. Write the coefficient of x ^{2} in the expansion of (x – 2)^{3}.**

**Ans.**(x – 2)

^{3}= (x)

^{3}– (2)

^{3}– 3 × x × 2(x – 2)

= x

^{3}– 8 – 6x

^{2}+ 12x

Coefficient of x

^{2 }in the expansion of (x – 2)

^{3}= – 6

**Question. If x + 1/x = 4, then calculate the value of x ^{2 }+ 1/x^{2}**

**Ans.**x

^{2}+ 1/x

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

^{2}− 2(x)(1/x)

= (4)

^{2}− 2 = 16 − 2 = 14

**Question. Write any polynomial in one variable.****Ans.** √2 x^{2} + 3x or √3 y^{2} + 3y.

**Question. In the expression x ^{2} + π /2 x – 7, what is the coefficient of x ?**

**Ans.**Coefficient of x in expression x

^{2}+ π /2 x – 7 is π /2

**Question. Factorize : 20x ^{2} – 9x + 1**

**Ans.**20x

^{2}– 9x + 1 = 20x

^{2}– 5x – 4x + 1

= 5x(4x – 1) – 1(4x – 1)

= (4x – 1)(5x – 1)

**Short Answer Type Questions:**

**Question. If y = 2 and y = 0 are the zeroes of the polynomial f(y) = 2y ^{3} – 5y^{2} + ay + b, find the value of a and b.**

**Ans.**Given, f(y) = 2y

^{3}– 5y

^{2}+ ay + b

∴ f(2) = 2(2)3 – 5(2)

^{2}+ a(2) + b = 0

or, 16 – 20 + 2a + b = 0

or, 2a + b = 4 …(i)

and f(0) = b = 0

From (i), 2a + 0 = 4

or, a = 2

∴ a = 2, b = 0

**Question. If x ^{2}+ 1/x^{2} = 98. Find the value of x^{3}+ 1/x^{3}**

**Ans.**x

^{2}+ 1/x

^{2}= 98

(x + 1/x)

^{2}= 98 + 2 = 100

x + 1/x = 10

Now, x

^{3}+ 1/x

^{3}= (x + 1/x)(x

^{2}− x.1/x + 1/x

^{2})

= (10)(x

^{2}+ 1/x

^{2}− 1)

= (10)(98 − 1)

= 10 x 97 = 970

**Question. Factorize : x ^{3} – 3x^{2} – 9x – 5.**

**Ans.**Let p(x) = x

^{3}– 3x

^{2}– 9x – 5,

Since, p (– 1) = – 1 – 3 + 9 – 5 = 0

Therefore, (x + 1) is a factor of x

^{3}– 3x

^{2}– 9x – 5.

∴ (x

^{3}– 3x

^{2}– 9x – 5) = (x + 1)(x

^{2}– 4x – 5)

Now, x

^{2}– 4x – 5 = x

^{2}– 5x + x – 5

= x(x – 5) + 1(x – 5)

= (x + 1)(x – 5) 1

∴ p(x) = (x + 1)(x + 1)(x – 5)

**Question. Factorize : 6x ^{2} + 7x – 3**

**Ans.**6x

^{2}+ 7x – 3 = 6x

^{2}+ 9x – 2x – 3 (By splitting middle term)

= 3x(2x + 3) – 1 (2x + 3)

= (2x + 3) (3x – 1)

**Question. Using a suitable identity, find (98) ^{3}.**

**Ans.**(98)

^{3}= (100 – 2)

^{3}

= (100)

^{3}– (2)

^{3}– 3 × 100 × 2(100 – 2)

= 1000000 – 8 – 600 × (100 – 2)

= 1000000 – 8 – 60000 + 1200

= 1000000 – 58808 = 941192

**Question. Find the value of k, so that polynomial x ^{3} + 3x^{2} – kx – 3 has one factor as x + 3.**

**Ans.**Let f(x) = x

^{3}+ 3x

^{2}– kx – 3

Since, (x + 3) is a factor of f(x).

Then, f(– 3) = 0

or, (– 3)

^{3}+ 3(– 3)

^{2}– k(–3) – 3 = 0

or, – 27 + 27 + 3k – 3 = 0

or, 3k – 3 = 0

or, k = 1

**Question. For what value of k, is the polynomial p(x) = 2x ^{3} – kx^{2} + 3x+ 10 exactly divisible by (x + 2) ?**

**Ans.**Since, (x + 2) is a factor of p(x).

Thus, p(–2) = 0

or, 2(–2)

^{3}– k(–2)

^{2}+ 3(–2) +10 = 0

or, – 16 – 4k – 6 + 10 = 0

k = – 3

**Question. Find the value of ‘k’ if (x – 1) is a factor of p(x) = 2x ^{2} + kx + √2 .**

**Ans.**Given, p(x) = 2x

^{2}+ kx + √

Since, (x – 1) is a factor of p(x), then p(1) = 0.

∴ 2 (1)

^{2}+ k(1) + √2 = 0

or, 2 + k + √2 = 0

or, k = –2 – √2

**Question. Evaluate 111 ^{3}, using a suitable identity.**

**Ans.**111

^{3}= (100 + 11)

^{3}

= (100)

^{3}+ 3(100)

^{2}(11) + 3(100) (11)

^{2}+ (11)

^{3}

= 1367631

**Question. Factorize : (x + y) ^{3} – (x^{3}+ y^{3})**

**Ans**. (x + y)

^{3}– (x

^{3}+ y

^{3})

= (x + y)

^{3}– (x + y) (x

^{2}+ y

^{2}– xy)

= (x + y) {(x + y)

^{2}– (x

^{2}+ y

^{2}– xy)}

= (x + y) (x

^{2}+ y

^{2 }+ 2xy – x

^{2}– y

^{2}+ xy)

= (x + y) (3xy)

= 3xy(x + y)

**Long Answer Type Questions: **

**Question. Factorize : x ^{3} – 12x^{2} + 47x – 60**

**Ans.**Factor of 60 = (±1, ±2, ±3, ±4, ±5, ±6, ±10, ±12, ±15, ±30, ±60,)

p(x) = x

^{3}– 12x

^{2}+ 47x – 60

p(3) = (3)

^{3}– 12(3)

^{2}+ 47(3) – 60

= 27 – 108 + 141 – 60

= 168 – 168 = 0

∴ x = 3 is a zero of p(x) or (x – 3) is a factor of p(x).

x

^{3}– 12x

^{2}+ 47x – 60

= x

^{2}(x – 3) – 9x(x – 3) + 20(x – 3)

= (x – 3)(x

^{2}– 9x + 20)

= (x – 3)(x

^{2}– 5x – 4x + 20)

= (x – 3)[x(x – 5) – 4(x – 5)]

= (x – 3)(x – 4)(x – 5)

**Question. If x + 1/x = 5, evaluate x ^{2} + 1/x^{2}.**

**Ans.**x + 1/x = 5

On squaring both sides, we get

(x + 1/x)

^{2}= 5

^{2}

or, x

^{2}+( 1/x)

^{2 }+ 2 × x + 1/x = 25 [(a + b)

^{2}= a

^{2 }+ b

^{2}+ 2ab]

or, x

^{2}+ 1/x

^{2}+ 2 = 25

or, x

^{2}+ 1/x

^{2}= 25 − 2

or x

^{2}+ 1/x

^{2}= 23

**Question. If x + 1/x = √3 , evaluate x ^{3} + 1/x^{3}.**

**Ans.**x + 1/x = √3

Cubing both sides, we get

(x + 1/x)

^{3}= (√3)

^{3}

or, x

^{3}+ 1/x

^{3 }+ 3x. 1/x(x + 1/x) = 3√3

or, x

^{3}+ 1/x

^{3 }+ 3(√3) = 3√3

or, x

^{3}+ 1/x

^{3 }= 0

**Question. Prove that x ^{3}+ y^{3} + z^{3 }– 3xyz = 1/2(x + y + z) [(x – y)^{2} + (y – z)^{2} + (z – x)^{2}]**

**Ans.**RHS = 1/2(x + y + z) [(x – y)

^{2}+ (y – z)

^{2}+ (z – x)

^{2}]

= 1/2(x + y + z) [ x

^{2}+ y

^{2}− 2xy + y

^{2}+ z

^{2}− 2yz + z

^{2}+ x

^{2}− 2zx]

= 1/2(x + y + z)[2x

^{2}+ 2y

^{2}+ 2z

^{2}– 2xy – 2yz – 2zx]

= 1/2(x + y + z). 2[x

^{2}+ y

^{2 }+ z

^{2}– xy – yz – zx]

= x

^{3}+ y

^{3}+ z

^{3 }– 3xyz

[Using identity x

^{3}+ y

^{3}+ z

^{3 }– 3xyz = (x + y + z) (x

^{2}+ y

^{2 }+ z

^{2}– xy – yz – zx)]

= LHS

**Question. If x − 1/x =2 , find x ^{4} + 1/x^{4} **

**Ans.**x

^{2}+ 1/x

^{2}= (x − 1/x)

^{2}+ 2

= 4 + 2 = 6

(x

^{2}+ 1/x

^{2})

^{2}= (x

^{4}+ 1/x

^{4}) + 2

or, (6)

^{2}= x

^{4}+ 1/x

^{4}+ 2

or, 36 − 2 = x

^{4}+ 1/x

^{4}

or, x

^{4}+ 1/x

^{4}= 34

**Question. Prove that (a ^{2}– b^{2} )^{3} + (b^{2} – c^{2} )^{3} + (c^{2} – a^{2})^{3} = 3(a + b)(b + c)(c + a)(a – b)(b – c)(c – a).**

**Ans.**Let x = a

^{2}– b

^{2}, y = b

^{2}– c

^{2}, z = c

^{2}– a

^{2}

Now, x + y + z = a

^{2}– b

^{2}+ b

^{2}– c

^{2}+ c

^{2}– a

^{2}= 0

∴ x + y + z = 0

or, x

^{3 }+ y

^{3}+ z3 = 3xyz

i.e., (a

^{2}– b

^{2})

^{3}+ (b

^{2}– c

^{2})

^{3}+ (c

^{2}– a

^{2})

^{3}

= 3(a

^{2}– b

^{2})(b

^{2}– c

^{2})(c

^{2}– a

^{2})

= 3(a + b)(a – b)(b + c)(b – c)(c + a)(c – a)

= 3(a + b)(b + c)(c + a)(a – b)(b – c)(c – a)