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 : 125x3 – 64y3
Solution:
125x3– 6443 = (5x)3 – (4y)3
By using a3 – b3 = (a – b) (a2 + ab + b2), we obtain
125x3– 64y3 = (5x – 4y) (25x2 + 20xy + 16y2)

Question 2.
Find the value of (x + y)2 + (x – y)2.
Solution:
(x + y)2 + (x – y)2 = x2 + y2 + 2xy + x2 + y2 – 2xy
= 2x2 + 2y2 = 21x2 + y2)

Question 3.
If p(x)= x2 – 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 x2 + 3x + m.
Solution:
Let p(x) = x2 + 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 x3+ x2 + x + 1 is divided by x – \(\frac{1}{2}\) using remainder theorem.
Solution:
Let p(x) = x3+ x2 + x + 1 and q(x) = x – \(\frac{1}{2}\)
Here, p(x) is divided by q(x)
∴ By using remainder theorem, we have
Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers 1

Question 6.
Find the common factor in the quadratic polynomials x2 + 8x + 15 and x2 + 3x – 10.
Solution:
x2 + 8x + 15 = x2 + 5x + 3x + 15 = (x + 3) (x + 5)
x2 + 3x – 10 = x2 + 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 = y2 -2 × y × √3 + (√3)2 = y2 – 2√3 y + 3 (x – 2y – 3z)2
= x2 + 1 – 2y)2 + (-3z)2 + 2 × x × (-2y) + 2 × (-2y) × (-3z) + 2 × (-3z) × x
= x2 + 4y2 + 9z2 – 4xy + 12yz – 6zx

Question 2.
If x + = \(\frac{1}{x}\) = 7, then find the value of x3 + \(\frac{1}{x^{3}}\)
Solution:
We have x + \(\frac{1}{x}\) = 7
Cubing both sides, we have
Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers 2

Question 3.
Show that p – 1 is a factor of p10 + p8 + p6 – p4 – p2 – 1.
Solution:
Let f(p) = p10 + p8 + p6 – p4 – p2 – 1
Put p = 1, we obtain
f(1) = 110 + 18 + 16 – 14 – 12 – 1
= 1 + 1 + 1 – 1 – 1 – 1 = 0
Hence, p – 1 is a factor of p10 + p8 + p6 – p4 – p2 – 1.

Question 4.
If 3x + 2y = 12 and xy = 6, find the value of 27x3 + 8y3
Solution:
We have 3x + 2y = 12
On cubing both sides, we have
⇒ (3x + 2y)3 = 123
⇒ (3x)3 +(2y)3 + 3 × 3x × 2y(3x + 2y) = √728
⇒ 27x3+ 8y3 + 18xy(3x + 2y) = √728
⇒ 27x3+ 8y3 + 18 × 6 × 12 = √728
⇒ 27x3+ 8y3 + 1296 = √728
⇒ 27x3+ 8y3 = √728 – 1296
⇒ 27x3+ 8y3 = 432

Question 5.
Factorise : 4x2 + 9y2 + 16z22 + 12xy – 24 yz – 16xz.
Solution:
4x2 + 9y2 + 16z22 + 12xy – 24yz – 16xz
= (2x)2 + (3y)2 + (-4z)2 + 2(2x)(3y) + 2(3y)(= 42) + 2(- 42)(2x)
By using a2 + b2 + 2ab + 2bc + 2ca = (a + b + c)2, we obtain
= (2x + 3y – 4z)2 = (2x + 3y – 4z) (2x + 3y – 4z)

Question 6.
Factorise : 1 – 2ab – (a2 + b2).
Solution:
1 – 2ab – (a2 + b2) = 1 – (a2 + b2 + 2ab)
= 12 – (a + b)2
= (1 + a + b) (1 – a – b) [∵ x2 – y2 = (x + y)(x – y)]

Polynomials Class 9 Extra Questions Short Answer Type 2

Question 1.
Factorise :
Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers 3
Solution:
Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers 4

Question 2.
Factorise 64a3 – 27b3 – 144a2b + 108ab2.
Solution:
64a2 – 27b2 – 144a2b + 108ab2
= (4a)3 – (3b)3 – 36ab(4a – 3b)
= (40)2 – (3b)3 – 3 × 4a × 3b (4a – 3b)
= (4a – 3b)3 [∵ (x – y)3 = x3 – y3 – 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 = 12ky2 + 8ky – 20k.
Solution:
We have, volume = 12ky2 + 8ky – 20k
= 4k(3y2 + 2y – 5) = 4k(3y2 + 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) = x3 + 3x2 – 2x + 4, then find the value of p(2) + p(-2) – P(0).
Solution:
Here, p(x) = x3+ 3x2 – 2x + 4
Now, p(2) = 23 + 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 x2 – √3x + 40 is 5, which is the other zero ?
Solution:
Let p(x) = x2 – √3x + 40
= x2 – 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:
Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers 5
Solution:
Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers 6

Question 7.
If one zero of the polynomial x2 – √3x + 40 is 5, which is the other zero ?
Solution:
Let
p(x) = x2 – √3x + 40
= x2 – 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 – a3 – b3 – c3 = 3(a + b) (b + c) (c + a).
Solution:
L.H.S. = (a + b + c)3 – a3 – b3 – c3
= {(a + b + c)3 – 3} – {b3 + c3}
= (a + b + c – a) {(a + b + c)2 + a2 + a(a + b + c)} – (b + c) (b2 + c2 – bc)
= (b + c) {a2 + b2 + 2 + 2ab + 2bc + 2ca + a2 + a2 + ab + ac – b2 – a2 + bc)
= (b + c) (3a2 + 3ab + 3bc + 3ca}
= 3(b + c) {a2 + ab + bc + ca}
= 31b + c) {{a2 + 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 x2 – 22x (m + 2n) + 72.
Solution:
Let m + 2n = a
∴ (m + 2n)2 x2 – 22x (m + 2n) + 72 = a2x2 – 22ax + 72
= a2x2 – 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 x2 – 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 x2 – 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, x2 – 7x + 12 = x2 – 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 x3– 10x2 + ax + b is exactly divisible by the polynomials (x – 1) and (x – 2).
Solution:
Let p(x) = x3– 10x2 + 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 : x2 – 6x2 + 11x – 6.
Solution:
Let p(x) = x2 – 6x2 + 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.
∴ x3 – 6x2 + 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, x3 – 6x2 + 11x – 6 = (x – 1) (x – 2)(x – 3).

Question 6.
Show that \(\frac{1}{3}\)and \(\frac{4}{3}\) are zeroes of the polynomial 9x3 – 6x2 – 11x + 4. Also, find the third zero of the polynomial.
Solution:
Let p(x) = 9x3– 6x2 – 11x + 4
Put x = \(\frac{1}{3}\), we have
Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers 7
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 9x2 – 15x + 4 exactly divides p(x).
⇒ 9x3 – 6x2 – 11x + 4 = (9x2 – 15x + 4) (x + 1)
Hence, x = -1 is its third zero.

Question 7.
Factorise : 6x2 – 5x2 – √3x + 12
Solution:
Let p(x) = 6x3– 5x2 – √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
Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers 8
Thus,
p(x) = (x – 1) (6x2 + x – 12)
p(x) = (x – 1) (6x2 + 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) = x4 + 2x2 – 2x2 + x – 1 so that resulting polynomial is exactly divisible by x2 + 2x – 3?
Solution:
Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers 9
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 + 2x3 – 2x2 + x – 1 + (x – 2)
Here, polynomial = x4 + 2x3– 2x2 + 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
x3+ √3 + x3= 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 x2 + px – q and x2 + rx – t, then prove that a = \(\frac{t-q}{r-p}\)
Solution:
Let f(x) = x + px -q and g(x) = x2 + x – t
Since x-a is factor of both f(x) and g(x)
⇒ f(a) = g(a) = 0
Now, here f(a) = a2 + pa – q and
g(a) = a2 + ra- t
⇒ a2 + 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 27x3 + 54x2 + 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, 27x3+ 54x2 + 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 x3+ 3x2 + 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) = x2 + 3x2 + 3x + 1 and
Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers 10
By long division method, we have
Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers 11
Remainder = \(\frac{27}{8}\) or 3 \(\frac{3}{8}\)
Thus, price of the juice glass is ₹ 3 \(\frac{3}{8}\)
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