# RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions

RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions is part of RBSE Solutions for Class 9 Maths. Here we have given RBSE Rajasthan Board Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions.

 Board RBSE Class Class 9 Subject Maths Chapter Chapter 3 Chapter Name Polynomial Exercise Additional Questions Number of Questions Solved 36 Category RBSE Solutions

## RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions

Multiple Choice Questions

Question 1.
A polynomial (RBSESolutions.com) of degree n in x has atmost:
(A) n terms
(B) (n – 1) terms
(C) (n + 1) terms
(D) $$\frac { n }{ 2 }$$ terms
Solution
C

Question 2.
The zeroes of the (RBSESolutions.com) polynomial p(x) = x(x² – 1) are:
(A) 0, 1
(B) 0, – 1
(C) 0, – 1, 1
(D) ± 1
Solution
C Question 3.
If is equal to Solution
B

Question 4.
√7 is a polynomial (RBSESolutions.com) of degree:
(A) 0
(B) 7
(C) $$\frac { 1 }{ 2 }$$
(D) 2
Solution
A

Question 5.
The coefficient of x in (x – 3)(x – 4) is:
(A) 7
(B) 1
(C) – 7
(D) 12
Solution
C

Question 6.
If p(x) = x² -3√2x + 1, then p(3√2) is equal to:
(A) 3√2
(B) 3√2 – 1
(C) 6√2 – 1
(D) 1
Solution
D

Question 7.
Which of the following is (RBSESolutions.com) a polynomial in one variable: Solution
A

Question 8.
The degree of the zero polynomial is:
(A) 0
(B) 1
(C) any real number
(D) not exist
Solution
D Question 9.
If x91 + 91 is divided by x + 1, then (RBSESolutions.com) the remainder is:
(A) 0
(B) 90
(C) 92
(D) None of these
Solution
B

Question 10.
Zero of the polynomial p(x) = √3x + 3 is Solution
A

Question 1.
Show that $$\frac { 1 }{ 2 }$$ is zero (RBSESolutions.com) of the polynomial 2x2 + 7x – 4.
Solution.
Let p(x) = 2x2 + 7x – 4 Question 2.
Find the (RBSESolutions.com) remainder when
x4 + x3 – 2x2 + x + 1 is divided by x – 1.
Solution.
Let p(x) = x4 + x3 – 2x2 + x + 1
zero of x – 1 is 1
So, P(1) = (1)4 + (1)3 – 2(1)2 + (1) + 1 = 1 + 1 – 2 + 2 = 2
Hence, 2 is the remainder, when x4 + x3 – 2x2 + x + 1 is divided by x – 1.

Question 3.
If x – 1 is a factor of p(x) = 2x2 + kx + √2 then find (RBSESolutions.com) the value of k.
Solution.
x – 1 is a factor of p(x) = 2x2 + kx + √2
p(1) = 0 (by factor theorem)
⇒ 2(1)2 + k(1) + √2 = 0
⇒ 2 + k + √2 = 0
k = -(2 + √2)

Question 4.
Find the (RBSESolutions.com) value of m, if (x + 3) is a factor of 3x2 + mx + 6.
Solution.
Let p(x) = 3x2 + mx + 6
If x + 3 is a factor of p(x), then
p(-3) = 0
p(-3) = 3(- 3)2 + m(-3) + 6 = 0
⇒ 3 x 9 – 3m + 6 = 0
⇒ 27 – 3m + 6 = 0
⇒ 3m = 33
⇒ m = 11

Question 5.
Verify that 1 is not a zero (RBSESolutions.com) of the polynomial 4x4 – 3x3 + 2x2 – 5x + 1.
Solution.
Let p(x) = 4x4 – 3x3 + 2x2 – 5x + 1
Now
P(1) = 4(1)4 – 3(1)3 + 2(1)2 – 5(1) + 1 = 4 – 3 + 2 – 5 + 1 = -1
p(1) ≠ 0
1 is not a zero of p(x). Question 6.
Using remainder theorem, find the (RBSESolutions.com) value of k so that (4x2 + kx – 1) leaves the remainder 2 when divided by (x – 3).
Solution.
Let p(x) = 4x2 + kx – 1
Here we are given that p(3) = 2
i.e. 4(3)2 + k(3) -1 = 2
⇒ 4 x 9 + 3k – 1 = 2
⇒ 36 + 3k = 3
⇒ 3k = -33
⇒ k = -11

Question 7.
If (x – 2) is a factor of the (RBSESolutions.com) polynomial x4 – 2x3 + ax – 1, then find the value of a.
Solution.
(x – 2) is a factor of the polynomial p(x) = x4 – 2x3 + ax – 1, then according to factor theorem p(2) = 0
⇒ (2)4 – 2(2)3 + a(2) – 1 = 0
⇒ 2a – 1 = 0
⇒ a = $$\frac { 1 }{ 2 }$$

Question 8.
Find the value of
(x – y)3 + (y – z)3 + (z – x)3.
Solution.
Let a = x – y, b = y – z, c = z – x
Here a + b + c = x – y + y – z + z – x = 0
a3 + b3 + c3 = 3abc
⇒ (x – y)3 + (y – z)3 + (z – x)3 = 3(x – y)(y – z)(z – x).

Question 9.
If a, b, c are all non-zero and a + b + c = 0 then find the (RBSESolutions.com) value of $$\frac { { a }^{ 2 } }{ bc } +\frac { { b }^{ 2 } }{ ca } +\frac { { c }^{ 2 } }{ ab }$$
Solution.
We know that a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
a + b + c then a3 + b3 + c3 = 3abc …(1)
Dividing equation (1) by abc, we get Question 10. Solution.  Question 1.
Find the remainder (RBSESolutions.com) obtained on dividing p(x) = x3 + 1 by x + 1.
Solution.
By long division method Here we find that remainder is zero.
Also p(-1) = (-1)3 + 1 = -1 + 1 = 0,
which is equal to the (RBSESolutions.com) remainder obtained by actual division.

Question 2.
Find the value of
4a2 + b2 + 25c2 + 4ab – 10bc – 20ca when a = 1, b = 2 and c = 3.
Solution.
We have,
4a2 + b2 + 25c2 + 4ab – 10bc – 20ca = (2a)2 + (b)2 + (-5c)2 – 2(2a)(b) + 2(b)(-5c) + 2(-5c)(2a) = (2a + b – 5c)2
It is given that a = 1, b = 2 and c = 3,
so (2a + b – 5c)2 = (2 x 1 + 2 – 5 x 3)2 = (2 + 2 – 15)2 = (4 – 15)2 = (-11)2 = 121

Question 3.
If 3a – 2b = 11 and ab = 12, then find (RBSESolutions.com) the value of 27a3 – 8b3.
Solution.
We know that
(a – b)3 = a3 – b3 – 3ab(a – b)
Using this identity,
(3a – 2b)3 = (3a)3 – (2b)3 -3 x 3a x 2b (3a – 2b)
⇒ (3a – 2b)3 = 27a3 – 8b3 – 18ab(3a – 2b)
Now substituting 3a – 2b = 11 and ab = 12,
we get
(11)2 = 27a3 – 8b3 – 18 x 12 x 11
⇒ 1331 = 27a3 – 8b3 – 2376
⇒ 27a3 – 8b2 = 1331 + 2376
⇒ 27a3 – 8b3 = 3707

Question 4.
Use factor (RBSESolutions.com) theorem, show that (x + √2) is a factor of (2√2x2 + 5x + √2).
Solution.
Let p(x) = (2√2x2 + 5x + √2)
If x + √2 is a factor of p(x), then according to factor theorem p(√2) = 0.
p(-√2) = 2√2(-√2)2 + 5(-√2) + √2 = 2√2 x 2 – 5√2 + √2 = 5√2 – 5√2 = 0
p(-√2) = 0
(x + √2) is a factor p(x) i.e. (2√2 x2 + 5x + √2).

Question 5.
If a + b + c = 9 and ab + bc + ca = 23 then find (RBSESolutions.com) the value of a2 + b2 + c2.
Solution.
We have, a + b + c = 9 Squaring both sides, we get
(a + b + c)2 = (9)2
⇒ a2 + b2 + c2 + 2ab + 2bc + 2ca = 81
⇒ a2 + b2 + c2 + 2(ab + bc + ca) = 81
⇒ a2 + b2 + c2 + 2 x 23 = 81
⇒ a2 + b2 + c2 = 81 – 46
⇒ a2 + b2 + c2 = 35 Thus, a2 + b2 + c2 = 35. Question 6.
Simplify by using (RBSESolutions.com) suitable identity Solution. Question 7.
Factorise
(i) 216a3 – 125
(ii) a3 – b3 – a + b
Solution.
(i) we have,
216a3 – 125 = (6a)3 – (5)3
[a3 – b3 = (a – b)(a2 + ab + b2)]
= (6a – 5) {(6a)2 + 6a x 5 + (5)2}
= (6a – 5)(36a2 + 30a + 25)
(ii) We have,
a3 – b3 – a + b = a3 – b3 – (a – b)
= (a – b)(a2 + ab + b2) – (a – b)
[a3 – b3 = (a – b)(a2 + ab + b2)]
= (a – b)(a2 + ab + b2 – 1)

Question 8.
If p = 4 – q show that p3 + q3 + 12pq = 64.
Solution.
We have,
p = 4 – q ⇒ p + q = 4 …(i)
Cubing both sides, (RBSESolutions.com) we get
(p + q)3 = (4)3
⇒ p3 + 3p2q + 3pq2 + q3 = 64
⇒ p3 + q3 + 3pq(p + q) = 64
⇒ p3 + q3 + 3pq x 4 = 64 [using (i)]
⇒ p3 + q3 + 12pq = 64

Question 9. Solution. Question 1.
If (x – 3) and (x – $$\frac { 1 }{ 2 }$$) are both (RBSESolutions.com) factors of ax2 + 5x + b, show that a = b.
Solution.
Let p(x) = ax2 + 5x + b
x – 3 is a factor of p(x).
p(3) = 0  Question 2.
For what values of a and b so that (RBSESolutions.com) the polynomial x3 + 10x2 + ax – 6 is exactly divisible by (x – 1) and (x + 2).
Solution.
Let p(x) = x3 + 10x2 + ax + b
p(x) i.e. x3 + 10x2 + ax + 6 is exactly divisible by (x – 1) and (x + 2)
Therefore, p(1) and p(-2) must equal to zero.
p(1) = (1)3 + 10(1)2 + a x 1 + b = 0
⇒ 1 + 10 + a + b = 0
⇒ a + b = – 11 …(i)
Also, p(-2) = 0
(-2)3 + 10(-2)2 + a x (- 2) + b = 0
-8 + 40 – 2a + b = 0
⇒ – 2a + b = – 32 …(ii)
Solving (i) and (ii), we get
a = 7 and b = -18.

Question 3.
If ax3 + bx2 + x – 6 has x + 2 as a factor and leaves (RBSESolutions.com) a remainder 4 when divided by (x – 2), find the values of a and b.
Solution.
Let p(x) = ax3 + bx2 + x – 6
(x + 2) is a factor of p(x)
⇒ p(-2) = 0 [∴ x + 2 = 0 ⇒ x = -2]
⇒ a(-2)3 + b(-2)2 + (-2) – 6 = 0
⇒ -8a + 4b – 2 – 6 = 0
⇒ -8a + 4b = 8
⇒ -2a + b = 2 …(i)
It is given that p(x) leaves the (RBSESolutions.com) remainder 4 when it divided by (x – 2) i.e. p(2) = 4.
⇒ a(2)3 + b(2)2 + (2) – 6 = 4
⇒ 8a + 4b – 4 = 4
⇒ 8a + 46 = 8
⇒ 2a + b = 2 …(ii)
Solving (i) and (ii), we get a = 0 and b = 2.

Question 4.
Evaluate Solution.
In the given expression, we see that both numerator and denominator (RBSESolutions.com) is in the form a3 + b3 + c3 = 3abc because a + b + c = 0.
From numerator, we see that a2 – b2 + b2 – c2 + c2 – a2 = 0
⇒ (a2 – b2)3 + (b2 – c2)3 + (c2 – a2)3 = 3 (a2 – b2)(b2 – c2)(c2 – a2)
Similarly, from denominator,
a – b + b – c + c – a = 0
(a – b)3 + (b – c)3 + (c – a)3 = 3(a – b)(b – c)(c – a)
Value of the expression Question 5.
If the polynomials (3x3 + ax2 + 3x + 5) and (4x3 + x2 – 2x + a) leaves the same (RBSESolutions.com) remainder when divided by (x – 2), find the value of a. Also the find the remainder in each-case.
Solution.
Let the given polynomials are p(x) = 3x3 + ax2 + 3x + 5 and f(x) = 4x3 + x2 – 2x + a
According to question p(2) = f(2)
3(2)3 + a(2)2 + 3(2) + 5 = 4(2)3 + (2)2 – 2(2) + a
⇒ 3 x 8 + 4a + 6 + 5 = 32 + 4 – 4 + a
⇒ 24 + 4a + 11 = 32 + a
⇒ 4a – a = 32 – 35
⇒ 3a = -3
⇒ a = -1
As the remainder is same, so p(2) or f(2) would (RBSESolutions.com) be same when a = – 1
p(2) = 3(2)3 + (-1)(2)2 + 3 x 2 + 5 [∴ a = -1]
= 24 – 4 + 6 + 5
= 35 – 4 = 31
Thus, a = – 1 and p(2) or f(2) = 31

Question 6.
Prove that (a + b + c)3 – a3 – b3 – c3 = 3(a + b)(b + c)(c + a).
Solution.
We have,
L.H.S.
= (a + b + c)3 – a3 -b3 – c3
= {(a + b + c)3 – (a)3} – (b3 + c3)
= (a + b + c – a){(a + b + c)2 + a(a + b + c) + a2} – (b + c){b2 – bc + c2)
[∵ x3 – y3 = (x – y)(x2 + xy + y2) and x3 + y3 = (x + y)(x2 – xy + y2)]
= (b + c){a2 + b2 + c2 + 2ab + 2bc + 2ca + a2 + ab + ac + a2} – (b + c)(b2 – bc + c2)
= (b + c)[3a2 + b2 + c2 + 3ab + 2bc + 3ca – b2 + bc – c2]
= (b + c)[3a2 + 3ab + 3bc + 3ca]
= 3(b + c)[a2 + ab + bc + ca]
= 3(b + c)[a(a + b) + c(a + b)]
= 3(b + c)(a + b)(a + c)
= 3(a + b)(b + c)(c + a)
= R.H.S. Question 7.
(i) For what (RBSESolutions.com) value of m is x3 – 2mn2 + 16 divisible by (x + 2)?
(ii) Show that (2x – 3) is a factor of x + 2x3 – 9x + 12.
Solution.
(i) Let p(x) = x3 – 2mn2 + 16
p(x) will be divisible by (x + 2) if
p(-2)= 0
p(-2) = (-2)3 – 2m(-2)2 + 16 = -8 – 8m + 16 = – 8m + 8
p(-2) = 0
⇒ -8m + 8 = 0
⇒ 8m = 8
⇒ m = 1
(ii) We know that (2x – 3) will be a factor of x + 2x3 – 9x + 12 if p(x) on dividing by 2x – 3, leaves (RBSESolutions.com) a remainder zero So, the remainder obtained (RBSESolutions.com) on dividing p(x) by 2x – 3 is zero.
Hence, (2x – 3) is a factor of x + 2x3 – 9x + 12

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