RBSE Class 9 Maths Important Questions Chapter 2 Polynomials

Rajasthan Board RBSE Class 9 Maths Important Questions Chapter 2 Polynomials Important Questions and Answers.

Rajasthan Board NCERT New Syllabus RBSE Solutions for Class 9 Guide Pdf free download in Hindi Medium and English Medium are part of RBSE Solutions. Here we have given RBSE Class 9th Books Solutions.

RBSE Class 9 Maths Chapter 2 Important Questions Polynomials

I. Multiple Choice Questions:
Choose the correct answer from the given options.

Question 1.
The coefficient of x2 in (3x2 - 5) (4x + 4x2) is :
(a) 12
(b) 5
(c) - 8
(d) 8
Answer:
(c) - 8

Question 2.
Which of the following algebraic expression is a polynomial in one variable?
(a) 7x2 - \(\frac{2}{x}\)
(b) 7x2 - 3x + √2
(c) \(\frac{(x-2)(x-4)}{x}\)
(d) 7x2 - xy + 5y + 2
Answer:
(b) 7x2 - 3x + √2

Question 3.
Degree of the polynomial 4x4 + 0x3 + 0x5 + 5x + 7 is :
(a) 4
(b) 5
(c) 3
(d) 7
Answer:
(a) 4

Question 4.
If p(x) = x2 - 2√2x +1, then p(2√2) is equal to :
(a) 0
(b) 1
(c) 4√2
(d) 8√2 + 1
Answer:
(b) 1

Question 5.
If a + b + c = 0, then the value of a3 + b3 + c3 is :
(a) 6
(b) abc
(c) 2abc
(d) 3abc
Answer:
(d) 3abc

Question 6.
The factorisation of 4a2 + 8a + 3 is :
(a) (a + 1) (a + 3)
(b) (2a + 1) (2a + 3)
(c) (2a + 2) (2a + 5)
(d) (2a - 1) (2a - 3)
Answer:
(b) (2a + 1) (2a + 3)

RBSE Class 9 Maths Important Questions Chapter 2 Polynomials

Question 7.
If p(x) = x + 3, then p(x) + p(- x) is equal to :
(a) 3
(b) 2x
(c) 0
(d) 6
Answer:
(d) 6

Question 8.
The roots of the polynomial equation 3x3 - 12x = 0 are :
(a) 3, 2,-2
(b) 3, 0, 4
(c) 0, 2, -2
(d) 3, 0, 2, -2
Answer:
(c) 0, 2, -2

Question 9.
x + 1 is factor of the polynomial:
(a) x3 + x2 - x + 1
(b) x3 + x2 + x + 1
(c) x4 + x3 + x2 + 1
(d) x4 + 3x3 + 3x2 + x + 1
Answer:
(b) x3 + x2 + x + 1

Question 10.
The value of the polynomial 5x - 4x2 + 3, then x = - 1 is
(a) - 6
(b) 6
(c) 2
(d) - 2
Answer:
(a) - 6

Question 11.
If 49x2 - b = (7x + \(\frac{1}{2}\))(7x - \(\frac{1}{2}\)) , then the value of b is :
(a) 0
(b) \(\frac{1}{\sqrt{2}}\)
(c) \(\frac{1}{4}\)
(d) \(\frac{1}{2}\)
Answer:
(c) \(\frac{1}{4}\)

Question 12.
If \(\frac{x}{y}+\frac{y}{x}\) = -1 (x, y ≠ 0), then the value of x3 - y3 is :
(a) 1
(b) -1
(c) 0
(d) \(\frac{1}{2}\)
Answer:
(c) 0

Question 13.
The coefficient of x in the expansion of (x + 3)3 is :
(a) 1
(b) 9
(c) 18
(d) 27
Answer:
(d) 27

Question 14.
The factors of x2 - √2x - 4 are ;
(a) (x + 2√2)(x + √2)
(b) (x - 2√2)(x - √2)
(c) (x - 2√2) (x + √2)
(d) (x + 2)(x - 2√2)
Answer:
(c) (x - 2√2) (x + √2)

RBSE Class 9 Maths Important Questions Chapter 2 Polynomials

Question 15.
If x2 + kx + 6 = (x + 2)(x + 3) for all x, then the value of k is :
(a) 1
(b) -1
(c) 5
(d) 3
Answer:
(c) 5

II. Fill in the Blanks :

Question 1.
The coefficient of x2 in the polynomial (x - 1) (- 3x - 4) is ____________
Answer:
3

Question 2.
Degree of polynomial 4 - y2 is ____________
Answer:
2

Question 3.
If - 1 is a zero of the polynomial, x2 + 8x + k, then the value of k =
Answer:
k = 7

Question 4.
If 3x - 1 is a factor of the polynomial 6x2 + x - 1, then its other factor is
Answer:
2x + 1

Question 5.
If (x + 1) and (x + 2) are factors of x3 + 3x2 - 2αx + β, then α = ____________ and β = ____________
Answer:
-1, 0

Question 6.
If a + b + c = 9 and ab + bc + ca = 26, then a2 + b2 + c2 =
Answer:
29

RBSE Class 9 Maths Important Questions Chapter 2 Polynomials

III. True/False:
State whether the following statements are True or False.

Question 1.
If (x - 1) is a factor of polynomial 4x3 + 3x2 - 4x + k, then the value of k is 3.
Answer:
False

Question 2.
A binomial can have at most two terms.
Answer:
False

Question 3.
Every polynomial is binomial.
Answer:
False

Question 4.
A binomial may have degree 5.
Answer:
True

Question 5.
Zero of a polynomial is also the factor of given polynomial.
Answer:
True

Question 6.
A polynomial cannot have more than one zero.
Answer:
False

Question 7.
The degree of the sum of two polynomials each of degree 5 is always 5.
Answer:
False

IV. Match the Columns

Question 1.
Match the column I with the column II.

Column I

Column II

(1) Zero of zero polynomial is

(i) -3

(2) Degree of the zero polynomial is

(ii) 3

(3) Coefficient of y2 in the expression (y - l)3 is

(iii) not defined

(4) If x101 + 4 is divided by x + 1, the reminder is

(iv) -6

(5) If p(x) = 7x2 - 4 √2 is completely divisible by (x - √2), then c =

(v) any real number

Answer:

Column I

Column II

(1) Zero of zero polynomial is

(v) any real number

(2) Degree of the zero polynomial is

(iii) not defined

(3) Coefficient of y2 in the expression (y - 1)3 is

(i) -3

(4) If x101 + 4 is divided by x + 1, the reminder is

(ii) 3

(5) If p(x) = 7x2 - 4 √2 is completely divisible by (x - √2), then c =

(iv) -6

RBSE Class 9 Maths Important Questions Chapter 2 Polynomials

V. Very Short Answer Type Questions :

Question 1.
Find k, if x51 + 2x60 + 3x + k is divisible by x + 1.
Answer:
Let p(x) = x51 + 2x60 + 3x + k
Given that, p(x) is divisible by x + 1.
∴ p(-1) = 0
⇒ (- 1)51 + 2(- 1)60 + 3(- 1) + k = 0
⇒ -1 + 2 - 3 + k = 0
⇒ k - 4 + 2 = 0
⇒ k - 2 = 0
⇒ k = 2

Question 2.
Give possible expression for the length and breadth of the rectangle, whose area is given by 4a2 + 4a - 3.
Answer:
Given, area of rectangle = 4a2 + 4a - 3
= 4a2 + 6a - 2a - 3 [By splitting the middle term]
= 2a (2a + 3) - 1 (2a + 3)
= (2a - 1) (2a + 3)
Hence, possible expression for length/breadth = (2a - 1) and possible expression for breadth/length = (2a + 3).

Question 3.
What are the possible expressions for the dimensions of a cuboid, whose volume is 36fex2y - 21 kxy2 + 3fey3?
Answer:
Volume = 36kx2y - 21 kxy2 + 3ky3
= 3ky(12x2 - 7xy + y2)
= 3ky(12x2 - 4xy — 3xy + y2)
= 3ky[4x(3x - y) - y(3x - y)]
= 3ky(4x - y)(3x - y)
Hence, possible expression for length/breadth/height = Sky, possible expression for breadth/height/length = 4x - y and possible expression for height/length/breadth = 3x - y.

Question 4.
Expand (x - \(\frac{1}{2}\)y + \(\frac{1}{3}\)z)2
Answer:
 (x - \(\frac{1}{2}\)y + \(\frac{1}{3}\)z)2 = (x)2 + (- \(\frac{1}{2}\)y)2 + (\(\frac{1}{3}\)z)2 + 2x(- \(\frac{1}{2}\)y) + 2(- \(\frac{1}{2}\)y)(\(\frac{1}{3}\)z) + 2(\(\frac{1}{3}\)z)(x)
[∵ (a + b + c)2 = a2 + b2 +c2 +2ab +2bc + 2ca]
= x2 + \(\frac{1}{4}\)y2 + \(\frac{1}{9}\)z2 - xy - \(\frac{y z}{3}+\frac{2 z x}{3}\)

Question 5.
Without calculating the cubes find the value of (- 11)3 + (8)3 + (3)3.
Answer:
We have,
(-11) + 8 + 3 = 0
Therefore, (- 11)3 + (8)3 + (3)3 = 3(- 11) (8) (3) = - 792
(∵ If x + y + z = 0, then x3 + y3 + z3 = 3xyz)

RBSE Class 9 Maths Important Questions Chapter 2 Polynomials

VI. Short Answer Type Questions :

Question 1.
If x + 2y = 10, xy = 15, then find x3 + 8y3.
Answer:
We know that
(x + 2y)3 = (x)3 + (2y)3 + 3(x)(2y) (x + 2y) (Using Identity VI)
⇒ (x + 2y)3 = x3 + 8y3 + 6xy(x + 2y)
⇒ (10)3 = x3 + 8y3 + 6(15)(10)
⇒ 1000 = x3 + 8y3 + 900
⇒ x3 + 8y3 = 1000 - 900 = 100

Question 2.
Factorise : a7 + ab6.
Answer:
a7 + ab6 = a(a6 + b6) = a{(a2)3 + (b2)3}
= a(a2 + b2) {(a2)2 + (b2)2 - (a2) (b2)}
= a(a2 + b2) (a4 + b4 - a2b2)
= a(a2 + b2) (a4 + b4 + 2a2b2 - 2a2b2 - a2b2)
= a(a2 + b2) (a4 + b4 + 2a2b2 - 3a2b2)
= a(a2 + b2) {(a2 + b2)2 - (√3 ab)2}
= a(a2 + b2) (a2 + b2 - √3 3ab) (a2 + b2 + √3 ab).

Question 3.
If p(x) = x2 - 4x + 3, then evaluate p(2) - p(- 1) + p(1/2).
Answer:
p(2) = (2)2 - 4(2) + 3 = 4 - 8 + 3 = -1
p(- 1) = (- 1)2 - 4(- 1) + 3 = 1 + 4 + 3 = 8
p\(\left(\frac{1}{2}\right)\) = \(\left(\frac{1}{2}\right)^{2}\) - 4\(\left(\frac{1}{2}\right)\) + 3 = \(\frac{1}{4}\) - 2 + 3 = \(\frac{5}{4}\)
∴ p(2) - p(-1) + p\(\left(\frac{1}{2}\right)\) = -1 - 8 + \(\frac{5}{4}=\frac{-31}{4}\)

Question 4.
Use Factor Theorem to verify that x + a is a factor of xn + an for any odd positive integer n.
Answer:
Let p(x) = xn + an
The zero of x + a is - a. (x + a = 0 ⇒ x = - a)
Now, p(- a) = (- a)n + an = (- 1 )nan + an
[∵ n is an odd positive integer]
[∴ (-1)n = -1]
= (-1)an + an = 0
= - an + an = 0
∴ By Factor Theorem, x + a is a factor of xn + an for any odd positive integer n.

The remainder of the polynomial 5 + bx - 2x2 + ax3, when divided by (x - 2) is twice the remainder when it is divided by (x + 1). Show that 10a + 46 = 9.
Answer:
Let f(x) = 5 + bx - 2x2 + ax3
When f(x) is divided by (x - 2), then the remainder is f(2).
When f(x) is divided by (x + 1), then the remainder is f(-1).
Now, f(2) = 5 + 6(2) - 2(2)2 + a(2)3 = 8a + 2b - 3
and f(-1) = 5 + 6(- 1) - 2(- 1)2 + a(- 1)3 = - a - b + 3

According to the question, f(2) = 2f(- 1)
∴ 8a + 2b - 3 = 2 (- a - 6 + 3)
⇒ 8a + 2b - 3 = -2a - 2b + 6
⇒ 10a + 4b = 9
Hence proved.

RBSE Class 9 Maths Important Questions Chapter 2 Polynomials

Question 6.
Show by long division that (x - 3) is a factor of 2x4 + 3x3 - 26x2 - 5x + 6.
Answer:
We have to show that remainder is zero, while divide 2x4 + 3x3 - 26x2 - 5x + 6 by (x - 3).
RBSE Class 9 Maths Important Questions Chapter 2 Polynomials 1
Hence, (x - 3) is a factor of 2x4 + 3x3 - 26x2 - 5x + 6.

Question 7.
The polynomial 3x3 + ax2 + 3x + 5 and 4x3 + x2 - 2x + a leave remainder and R2, when divided by (x - 2) respectively. If R1 - R2 = 9, then find the value of a.
Answer:
Let f(x) = 3x3 + ax2 + 3x + 5 and g(x) = 4x3 + x2 - 2x + a Here, the zero of (x - 2) is x = 2. (∵ x - 2 = 0 ⇒ x = 2)

Where f(x) and g(x) are divided by (x - 2), then we get the remainders R1 and R2.
f(2) = 3(2 )3 + a(2)3 + 3(2) + 5
= 24 + 4a + 6 + 5
= 35 + 4a = R1 (Given)
and g(2) = 4(2)3 + (2)2 - 2(2) + a
= 32 + 4- 4 + a
= 32 + a = R1 (Given)
Also, R1 - R2 - 9
∴ 35 + 4a - (32 + a) = 9
⇒ 3 + 3a = 9
⇒ 3a = 6 ⇒ a = 3

VII. Long Answer Type Questions :

Question 1.
If both (x - 2) and (x - \(\frac{1}{2}\)) are factors of px2 + 5x + r, then show that p = r.
Answer:
Let f(x) = px2 + 5x + r
If (x - 2) is a factor of fix), then by factor theorem.
f(2) = 0 (x - 2 = 0)
⇒ x = 2
⇒ p(2)2 + 5(2) + r = 0
⇒ 4p + r + 10 = 0 ...................(1)

If (x - \(\frac{1}{2}\)) is a factor of f(x), then by factor theorem,
f(\(\frac{1}{2}\)) = 0 (x - \(\frac{1}{2}\) = 0 ⇒ x = \(\frac{1}{2}\))
⇒ p\(\left(\frac{1}{2}\right)^{2}\) + 5(\(\frac{1}{2}\)) + r = 0
\(\frac{p}{4}+\frac{5}{2}\) + r = 0
⇒ p + 4r + 10 = 0 ......(2)

Subtracting (2) from (1), we get : .
3p - 3r = 0
⇒ p = r
Hance proved.

Question 2.
Divide the polynomial 3x4 - 4x3 - 3x - 1 by x - 1 and find its quotient and remainder.
Answer:
RBSE Class 9 Maths Important Questions Chapter 2 Polynomials 2
Quotient = 3x3 - x2 - x - 4
and Remainder = - 5

RBSE Class 9 Maths Important Questions Chapter 2 Polynomials

Question 3.
If x + a is a factor of the polynomials x2 + px + q and x2 + mx + n, prove that
a = \(\frac{n-q}{m-p}\)
Answer:
Let f(x) = x2 + px + q
and g(x) = x2 + mx + n
If x + a is a factor of f(x), then by factor theorem
f(a) = 0 (x + a = 0 ⇒ x = -a)
⇒ (- a)2 + p(- a) + q = 0
⇒ a2 - pa + q = 0 ........(1)

If x + a is a factor of g(x), then by factor theorem
g(- a) = 0 (x + a = 0 ⇒ x = -a)
⇒ (- a)2 + m(- a) + n - 0
a2 - ma + n = 0 ............(2)

Subtracting (1) from (2), we get :
(p - m)a + (n - q) = 0
⇒ (m - p)a = n - q
⇒ a = \(\frac{n-q}{m-p}\)
Hence Proved.

Prasanna
Last Updated on April 29, 2022, 4:32 p.m.
Published April 29, 2022