Here we are providing Polynomials Class 9 Extra Questions Maths Chapter 2 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers.

## Extra Questions for Class 9 Maths Polynomials with Answers Solutions

**Extra Questions for Class 9 Maths Chapter 2 Polynomials with Solutions Answers**

### Polynomials Class 9 Extra Questions Very Short Answer Type

Question 1.

Factorise : 125x^{3} – 64y^{3}

Solution:

125x^{3}– 6443 = (5x)^{3} – (4y)^{3}

By using a3 – b3 = (a – b) (a^{2} + ab + b^{2}), we obtain

125x^{3}– 64y^{3} = (5x – 4y) (25x^{2} + 20xy + 16y^{2})

Question 2.

Find the value of (x + y)^{2} + (x – y)^{2}.

Solution:

(x + y)^{2} + (x – y)^{2} = x^{2} + y^{2} + 2xy + x^{2} + y^{2} – 2xy

= 2x^{2} + 2y^{2} = 21x^{2} + y^{2})

Question 3.

If p(x)= x^{2} – 2√2x+1, then find the value of p(2√2)

Solution:

Put x = 2√2 in p(x), we obtain

p(2√2) = (2√2)^{2} – 2√2(2√2) + 1 = (2√2)^{2} – (2√2)^{2} + 1 = 1

Question 4.

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

Solution:

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

Since (x + 4) or (x – (-4)} is a factor of p(x).

∴ p(-4) = 0

⇒ (-4)^{2} + 3(-4) + m = 0

⇒ 16 – 12 + m = 0

⇒ m = -4

Question 5.

Find the remainder when x^{3}+ x^{2} + x + 1 is divided by x – \(\frac{1}{2}\) using remainder theorem.

Solution:

Let p(x) = x^{3}+ x^{2} + x + 1 and q(x) = x – \(\frac{1}{2}\)

Here, p(x) is divided by q(x)

∴ By using remainder theorem, we have

Question 6.

Find the common factor in the quadratic polynomials x^{2} + 8x + 15 and x^{2} + 3x – 10.

Solution:

x^{2} + 8x + 15 = x^{2} + 5x + 3x + 15 = (x + 3) (x + 5)

x^{2} + 3x – 10 = x^{2} + 5x – 2x – 10 = (x – 2) (x + 5)

Clearly, the common factor is x + 5.

### Polynomials Class 9 Extra Questions Short Answer Type 1

Question 1.

Expand :

(i) (y – √3)^{2}

(ii) (x – 2y – 3z)^{2}

Solution: (i)

(y – √3)^{2} = y^{2} -2 × y × √3 + (√3)^{2} = y^{2} – 2√3 y + 3 (x – 2y – 3z)^{2}

= x^{2} + 1 – 2y)^{2} + (-3z)^{2} + 2 × x × (-2y) + 2 × (-2y) × (-3z) + 2 × (-3z) × x

= x^{2} + 4y^{2} + 9z^{2} – 4xy + 12yz – 6zx

Question 2.

If x + = \(\frac{1}{x}\) = 7, then find the value of x^{3} + \(\frac{1}{x^{3}}\)

Solution:

We have x + \(\frac{1}{x}\) = 7

Cubing both sides, we have

Question 3.

Show that p – 1 is a factor of p^{10} + p^{8} + p^{6} – p^{4} – p^{2} – 1.

Solution:

Let f(p) = p^{10} + p^{8} + p^{6} – p^{4} – p^{2} – 1

Put p = 1, we obtain

f(1) = 1^{10} + 1^{8} + 1^{6} – 1^{4} – 1^{2} – 1

= 1 + 1 + 1 – 1 – 1 – 1 = 0

Hence, p – 1 is a factor of p^{10} + p^{8} + p^{6} – p^{4} – p^{2} – 1.

Question 4.

If 3x + 2y = 12 and xy = 6, find the value of 27x^{3 }+ 8y^{3}

Solution:

We have 3x + 2y = 12

On cubing both sides, we have

⇒ (3x + 2y)^{3} = 12^{3}

⇒ (3x)^{3} +(2y)^{3} + 3 × 3x × 2y(3x + 2y) = √728

⇒ 27x^{3}+ 8y^{3} + 18xy(3x + 2y) = √728

⇒ 27x^{3}+ 8y^{3} + 18 × 6 × 12 = √728

⇒ 27x^{3}+ 8y^{3} + 1296 = √728

⇒ 27x^{3}+ 8y^{3} = √728 – 1296

⇒ 27x^{3}+ 8y^{3} = 432

Question 5.

Factorise : 4x^{2} + 9y^{2} + 16z^{2}2 + 12xy – 24 yz – 16xz.

Solution:

4x^{2} + 9y^{2} + 16z^{2}2 + 12xy – 24yz – 16xz

= (2x)^{2} + (3y)^{2} + (-4z)^{2} + 2(2x)(3y) + 2(3y)(= 42) + 2(- 42)(2x)

By using a^{2} + b^{2} + 2ab + 2bc + 2ca = (a + b + c)^{2}, we obtain

= (2x + 3y – 4z)^{2} = (2x + 3y – 4z) (2x + 3y – 4z)

Question 6.

Factorise : 1 – 2ab – (a^{2} + b^{2}).

Solution:

1 – 2ab – (a^{2} + b^{2}) = 1 – (a^{2} + b^{2} + 2ab)

= 1^{2} – (a + b)^{2}

= (1 + a + b) (1 – a – b) [∵ x^{2} – y^{2} = (x + y)(x – y)]

### Polynomials Class 9 Extra Questions Short Answer Type 2

Question 1.

Factorise :

Solution:

Question 2.

Factorise 64a^{3} – 27b^{3} – 144a^{2}b + 108ab^{2}.

Solution:

64a^{2} – 27b^{2} – 144a^{2}b + 108ab^{2}

= (4a)^{3} – (3b)^{3} – 36ab(4a – 3b)

= (40)^{2} – (3b)^{3} – 3 × 4a × 3b (4a – 3b)

= (4a – 3b)^{3} [∵ (x – y)^{3} = x^{3} – y^{3} – 3xy(x – y)]

= (40 – 3b) (4a – 3b) (4a – 3b)

Question 3.

What are the possible expressions for the dimensions of a cuboid whose volume is given below ?

Volume = 12ky^{2} + 8ky – 20k.

Solution:

We have, volume = 12ky^{2} + 8ky – 20k

= 4k(3y^{2} + 2y – 5) = 4k(3y^{2} + 5y – 3y – 5)

= 4k[y(3y + 5) – 1(3y + 5)] = 4k(3y + 5) (y – 1)

∴Possible expressions for the dimensions of cuboid are 4k units, (3y + 5) units and (y – 1) units.

Question 4.

If p(x) = x^{3} + 3x^{2} – 2x + 4, then find the value of p(2) + p(-2) – P(0).

Solution:

Here, p(x) = x^{3}+ 3x^{2} – 2x + 4

Now, p(2) = 2^{3} + 3(2)^{2} – 2(2) + 4

= 8 + 12 – 4 + 4 = 20

p(-2) = (-2)^{3} + 3(-2)^{2} – 21 – 2) + 4

= 8 + 12 + 4 + 4 = 12

and p(0) = 0 + 0 – 0 + 4 = 4

∴ p(2) + p(-2) – p(0) = 20 + 12 – 4 = 28.

Question 5.

If one zero of the polynomial x^{2} – √3x + 40 is 5, which is the other zero ?

Solution:

Let p(x) = x^{2} – √3x + 40

= x^{2} – 5x – 8x + 40 = x(x – 5) – 8(x – 5) = (x – 5) (x – 8)

Now, for zeroes of given polynomial, put p(x) = 0

∴ (x – 5) (x – 8) = 0

⇒ x = 5 or x = 8

Hence, other zero is 8.

Question 6.

Simplify:

Solution:

Question 7.

If one zero of the polynomial x^{2} – √3x + 40 is 5, which is the other zero ?

Solution:

Let

p(x) = x^{2} – √3x + 40

= x^{2} – 5x – 8x + 40 = x(x – 5) – 8(x – 5) = (x – 5) (x – 8)

Now, for zeroes of given polynomial, put p(x) = 0

∴ (x – 5) (x – 8) = 0

x = 5 or x = 8

⇒ Hence, other zero is 8.

### Polynomials Class 9 Extra Questions Long Answer Type

Question 1.

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

Solution:

L.H.S. = (a + b + c)^{3} – a^{3} – b^{3} – c^{3}

= {(a + b + c)^{3} – 3} – {b^{3} + c^{3}}

= (a + b + c – a) {(a + b + c)^{2} + a^{2} + a(a + b + c)} – (b + c) (b^{2} + c^{2} – bc)

= (b + c) {a^{2} + b^{2} + 2 + 2ab + 2bc + 2ca + a^{2} + a^{2} + ab + ac – b^{2} – a^{2} + bc)

= (b + c) (3a^{2} + 3ab + 3bc + 3ca}

= 3(b + c) {a^{2} + ab + bc + ca}

= 31b + c) {{a^{2} + ca) + (ab + bc)}

= 3(b + c) {a(a + c) + b(a + c)}

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

= 3(a + b)(b + c) (c + a) = R.H.S.

Question 2.

Factorise : (m + 2n)^{2} x^{2} – 22x (m + 2n) + 72.

Solution:

Let m + 2n = a

∴ (m + 2n)^{2} x^{2} – 22x (m + 2n) + 72 = a^{2}x^{2} – 22ax + 72

= a^{2}x^{2} – 18ax – 4ax + 72

= ax(ax – 18) – 4(ax – 18)

= (ax – 4) (ax – 18)

= {(m + 2n)x – 4)} {(m + 2n)x – 18)}

= (mx + 2nx – 4) (mx + 2nx – 18).

Question 3.

If x – 3 is a factor of x^{2} – 6x + 12, then find the value of k. Also, find the other factor of the – polynomial for this value of k.

Solution:

Here, x – 3 is a factor of x^{2} – kx + 12

∴ By factor theorem, putting x = 3, we have remainder 0.

⇒ (3)^{2} – k(3) + 12 = 0

⇒ 9 – 3k + 12 = 0

⇒ 3k = 21

⇒ k = 7

Now, x^{2} – 7x + 12 = x^{2} – 3x – 4x + 12

= x(x – 3) – 4(x – 3)

= (x – 3) (x – 4)

Hence, the value of k is 7 and other factor is x – 4.

Question 4.

Find a and b so that the polynomial x^{3}– 10x^{2} + ax + b is exactly divisible by the polynomials (x – 1) and (x – 2).

Solution:

Let p(x) = x^{3}– 10x^{2} + ax + b

Since p(x) is exactly divisible by the polynomials (x – 1) and (x – 2).

∴ By putting x = 1, we obtain

(1)^{3} – 10(1)^{2} + a(1) + b = 0

⇒ a + b = 9

And by putting x = 2, we obtain

(2)^{3} – 10(2)^{2} + a(2) + b = 0

8 – 40 + 2a + b = 0

⇒ 2a + b = 32

Subtracting (i) from (ii), we have

a = 23

From (i), we have 23 + b = 9 = b = -14

Hence, the values of a and b are a = 23 and b = -14

Question 5.

Factorise : x^{2} – 6x^{2} + 11x – 6.

Solution:

Let p(x) = x^{2} – 6x^{2} + 11x – 6

Here, constant term of p(x) is -6 and factors of -6 are ± 1, ± 2, ± 3 and ± 6

By putting x = 1, we have

p(1) = (1)^{3} – 6(1)^{2} + 11(1) – 6 = 1 – 6 + 11 -6 = 0

∴ (x – 1) is a factor of p(x)

By putting x = 2, we have

p(2) = (2)^{3} – 6(2)^{2} + 11(2) – 6 = 8 – 24 + 22 – 6 = 0

∴ (x – 2) is a factor of p(x)

By putting x = 3, we have

p(3) = (3)^{3} – 6(3)^{2} + 11(3) – 6 = 27 – 54 + 33 – 6 = 0

∴ (x – 3) is a factor of p(x) Since p(x) is a polynomial of degree 3, so it cannot have more than three linear factors.

∴ x^{3} – 6x^{2} + 11x – 6 = k (x – 1) (x – 2) (x – 3)

By putting x = 0, we obtain

0 – 0 + 0 – 6 = k (-1) (-2) (3)

-6 = -6k

k = 1

Hence, x^{3} – 6x^{2} + 11x – 6 = (x – 1) (x – 2)(x – 3).

Question 6.

Show that \(\frac{1}{3}\)and \(\frac{4}{3}\) are zeroes of the polynomial 9x^{3} – 6x^{2} – 11x + 4. Also, find the third zero of the polynomial.

Solution:

Let p(x) = 9x^{3}– 6x^{2} – 11x + 4

Put x = \(\frac{1}{3}\), we have

Thus, x = \(\frac{4}{3}\) is another zero of the polynomial p(x). Since x = \(\frac{1}{3}\) and x = \(\frac{4}{3}\) are the zeroes of p(x), therefore, \(\left(x-\frac{1}{3}\right)\) \(\left(x-\frac{4}{3}\right)\) (3x – 1) (3x – 4) or 9x^{2} – 15x + 4 exactly divides p(x).

⇒ 9x^{3} – 6x^{2} – 11x + 4 = (9x^{2} – 15x + 4) (x + 1)

Hence, x = -1 is its third zero.

Question 7.

Factorise : 6x^{2} – 5x^{2} – √3x + 12

Solution:

Let p(x) = 6x^{3}– 5x^{2} – √3x + 12

Here, constant term of p(x) is 12 and factors of 12 are ± 1, ± 2, ± 3, ± 4, ± 6, ± 12.

By putting x = 1, we have

p(1) = 6(1)^{3} – 5(1)^{2} – √3(1) + 12 = 6 – 5 – √3 + 12 = 0

∴ (x – 1) is a factor of p(x).

Now, by long division, we have

Thus,

p(x) = (x – 1) (6x^{2} + x – 12)

p(x) = (x – 1) (6x^{2} + 9x – 8x – 12)

p(x) = (x – 1) {3x (2x + 3) – 4(2x + 3)}

p(x) = (x – 1) (3x – 4) (2x + 3).

### Polynomials Class 9 Extra Questions HOTS

Question 1.

What must be added to polynomial f(x) = x^{4} + 2x^{2} – 2x^{2} + x – 1 so that resulting polynomial is exactly divisible by x^{2} + 2x – 3?

Solution:

Here, remainder = -x + 2

To make remainder = 0, we must add -(remainder) in the polynomial

i.e., -(-x + 2) i.e., x – 2

Hence, x4 + 2x^{3} – 2x^{2} + x – 1 + (x – 2)

Here, polynomial = x^{4} + 2x^{3}– 2x^{2} + 2x – 3 and required addition is (x – 2).

Question 2.

If x = 2 – √3, y = √3 – √7 and 2 = √7 – √4, find the value of x’ + 43 + 2?.

Solution:

Here, x + y + z = 2 – √3+ √3 – √7+√7 – 2 = 0

x^{3}+ √3 + x^{3}= 3(x)(y)(z)

= 3(2 – √3)(√3 – √7)(√7 – 2)

= 3(2√3 – 2√7 – 3 + √21)(√7 – 2)

= 3(2√21 – 14 – 3√7 + 7√3 – 4√3 + 4√7 + 6 – 2√21)

= 3(3√3 + √7 – 8)

Question 3.

If (x – a) is a factor of the polynomials x^{2} + px – q and x^{2} + rx – t, then prove that a = \(\frac{t-q}{r-p}\)

Solution:

Let f(x) = x + px -q and g(x) = x^{2} + x – t

Since x-a is factor of both f(x) and g(x)

⇒ f(a) = g(a) = 0

Now, here f(a) = a^{2} + pa – q and

g(a) = a^{2} + ra- t

⇒ a^{2} + pa – q = a + ra – t (considering f(a) = g(a)]

⇒ pa – q = ra – t

⇒ ra – pa = t – q

⇒ a(r – p) = t – q

a = \(\frac{t-q}{r-p}\)

### Polynomials Class 9 Extra Questions Value Based (VBQs)

Question 1.

If a teacher divides a material of volume 27x^{3} + 54x^{2} + 36x + 8 cubic units among three students. Is it possible to find the quantity of material ? Can you name the shape of the figure teacher obtained ? Which value is depicted by the teacher ?

Solution:

We know that, √olume = Length × Breadth × Height

Now, 27x^{3}+ 54x^{2} + 36x + 8

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

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

Thus, volume = (3x + 2) (3x + 2) (3x + 2)

Yes, it is possible to find the quantity of material. (3x + 2) units.

Cube.

Apply knowledge and use of example for clarity of subject, student friendly.

Question 2.

In a camp organised by the students of class-9 to donate amount collected to flood victims of Kashmir. At the time of payment of a juice glass at one stall of juice, stall holder asked the students to pay the remainder of x^{3}+ 3x^{2} + 3x + 1 divided by \(\left(x-\frac{1}{2}\right)\) What is the price of the juice at the stall ? Which value is depicted by class-9 students by organising such camps ?

Solution:

Let

p(x) = x^{2} + 3x^{2} + 3x + 1 and

By long division method, we have

Remainder = \(\frac{27}{8}\) or 3 \(\frac{3}{8}\)

Thus, price of the juice glass is ₹ 3 \(\frac{3}{8}\)

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