Class 10 Maths RS Aggarwal Solutions

RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4A

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4A.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4A
• RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4B
• RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4C
• RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4D
• RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4E
• RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQS
• RS Aggarwal Solutions Class 10 Chapter 4 Triangles Test Yourself

Question 1.
Solution:
In ∆ABC, points D and E are on the sides AB and AC respectively such that DE || BC.



⇒ 7x = 28 ⇒ x = 4

Question 2.
Solution:
D and E are the points on the sides AB and AC of ∆ABC respectively and DE || BC,


By cross multiplication,
⇒ (7x – 4) (3x) = (5x – 2) (3x + 4)
⇒ 21x² – 12x = 15x² + 20x – 6x – 8
⇒ 21x² – 12x – 15x² – 20x + 6x = -8
⇒ 6x² – 26x + 8 = 0
⇒ 3x² – 13x + 4 = 0
⇒ 3x² – 12x – x + 4 = 0
⇒ 3x (x – 4) – 1 (x – 4) = 0
⇒ (x – 4) (3x – 1) = 0
Either x – 4 = 0, then x = 4
or 3x – 1 = 0 then 3x = 1 ⇒ x = \(\frac { 1 }{ 3 }\) which is not possible
x = 4

Question 3.
Solution:
D and E are the points on the sides AB and AC of ∆ABC
(i) AD = 5.7 cm, DB = 9.5 cm, AE = 4.8 cm and EC = 8 cm

Question 4.
Solution:
In ∆ABC, AD is the bisector of ∠A

Question 5.
Solution:
Given : M is a point on the side BC of a parallelogram ABCD.
DM when produced meets AB produced at N.

Question 6.
Solution:
Given : In trapezium AB || DC.
M and N are the mid points of sides AD and BC respectively.
MN is joined.
To prove : MN || AB or DC.
Construction: Produce AD and BC to meet at P.

Question 7.
Solution:
ABCD is a trapezium and its diagonals AC and BD intersect each other at O.

Question 8.
Solution:
Given : In ∆ABC, M and N are points on the sides AB and AC respectively such that
BM = CN and if ∠B = ∠C.

Question 9.
Solution:
Given : ∆ABC and ∆DBC are on the same side of BC.
P is any point on BC.
PQ || AB and PR || BD are drawn, which meet AC at Q and CD at R.
AD is joined.
To prove : QR || AD.
Proof:
In ∆ABC, PQ||AB


Hence proved.

Question 10.
Solution:
Given : In the given figure, D is the mid point of BC. AD is joined. O is any point on AD. BO and CO are produced to meet AC at F and AB at E respectively. AD is produced to X such that D is the mid point of OX.
To prove: AO : AX = AF : AB and EF || BC
Construction. Join BX and CX.

D is the midpoint of BC.
Proof: BD = DC and OD = DX. (given)
OBXC is a parallelogram. (Diagonals of a ||gm bisect each other)

Question 11.
Solution:
Given : In ||gm, ABCD, P is mid point of DC and Q is the mid point of AC. Such that
CQ = \(\frac { 1 }{ 4 }\) AC.
PQ is produced to meet BC at R.
To prove : R is the mid point of BC.
Construction : Join BD which intersects AC at O.

Proof: Diagonals of a parallelogram bisect each other
CO = \(\frac { 1 }{ 2 }\) AC.
But CQ = \(\frac { 1 }{ 2 }\) CO (given)
Q is the mid point of CO.
PQ || DO
⇒ PR || DB and P is mid point of CD
R is the midpoint BC.
Hence proved.

Question 12.
Solution:
Given : In ∆ABC, AB = AC
D and E are the points on AB and AC such that
AD = AE
To prove : BCED is a cyclic quadrilateral.
Proof: In ∆ABC,
AB = AC,
D and E are the mid points of AB and AC respectively.

But these are opposite angles of quad. BCED
BCED is a cyclic quadrilateral
Hence, B, C, E and D are concyclic.

Question 13.
Solution:
Given : In ∆ABC, BD is the bisector of AB which meets AC at D.
A line PQ || AC which meets AB, BC, BD at P, Q, R respectively.
To prove : PR x BQ = QR x BP
Proof: In ∆ABC, PQ || AC
and BD is the bisector of ∠B

Hope given RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4A are helpful to complete your math homework.

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RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Test Yourself

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Test Yourself

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 2 Polynomials Ex 2A
• RS Aggarwal Solutions Class 10 Chapter 2 Polynomials Ex 2B
• RS Aggarwal Solutions Class 10 Chapter 2 Polynomials Ex 2C
• RS Aggarwal Solutions Class 10 Chapter 2 Polynomials MCQS
• RS Aggarwal Solutions Class 10 Chapter 2 Polynomials Test Yourself

MCQ
Question 1.
Solution:
(b) Polynomial is x² – 2x – 3
=> x² – 3x + x – 3
= x(x – 3) + 1(x – 3)
= (x – 3) (x + 1)
Either x – 3 = 0, then x = 3
or x + 1 = 0, then x = -1
Zeros are 3, -1

Question 2.
Solution:
(a) α, β, γ are the zeros of Polynomial is x³ – 6x² – x + 30
Here, a = 1, b = -6, c = -1, d = 30
αβ + βγ + γα = \(\frac { c }{ a }\) = \(\frac { -1 }{ 1 }\) = -1

Question 3.
Solution:
(c)

Question 4.
Solution:
(c) Let α and β be the zeros of the polynomial 4x² – 8kx + 9

Short-Answer Questions
Question 5.
Solution:

Either x + 15 = 0, then x = -15
or x – 13 = 0, then x = 13
Zeros are 13, -15

Question 6.
Solution:
The polynomial is (a² + 9) x² + 13x + 6a

Question 7.
Solution:
Zeros are 2 and -5
Sum of zeros = 2 + (-5) = 2 – 5 = -3
and product of zeros = 2 x (-5) = -10
Now polynomial will be
x² – (Sum of zeros) x + Product of zeros
= x² – (-3)x + (-10)
= x² + 3x – 10

Question 8.
Solution:
(a – b), a, (a + b) are the zeros of the polynomial x³ – 3x² + x + 1
Here, a = 1, b = -3, c = 1, d = 1
Now, sum of zeros = \(\frac { -b }{ a }\) = \(\frac { -(-3) }{ 1 }\) = 3
a – b + a + a + b = 3

Question 9.
Solution:
Let f(x) = x³ + 4x² – 3x – 18
If 2 is its zero, then it will satisfy it
Now, (x – 2) is a factor Dividing by (x – 2)
Hence, x = 2 is a zero of f(x)

Question 10.
Solution:
Sum of zeros = -5
and product of zeros = 6
Quadratic polynomial will be
x² – (Sum of zeros) x + Product of zeros
= x² – (-5) x + 6
= x² + 5x + 6

Question 11.
Solution:

Question 12.
Solution:
p(x) = x³ + 3x² – 5x + 4
g(x) = x – 2
Let x – 2 = 0, then x = 2
Remainder = p(2) = (2)³ + 3(2)² – 5(2) + 4 = 8 + 12 – 10 + 4 = 14

Question 13.
Solution:
f(x) = x³ + 4x² + x – 6
and g(x) = x + 2
Let x + 2 = 0, then x = -2
f(-2) = (-2)³ + 4(-2)² + (-2) – 6 = -8 + 16 – 2 – 6 = 0
Remainder is zero, x + 2 is a factor of f(x)

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:
f(x) = x⁴ + 4x² + 6
=> (x²)² + 4(x²) + 6 = y² + 4y + 6 (Let x² = y)
Let α, β be the zeros of y² + 4y + 6
Sum of zeros = -4
and product of zeros = 6
But there are no factors of 6 whose sum is -4 {Factors of 6 = 1 x 6 and 2 x 3}
Hence, f(x) Has no zero (real).

Long-Answer Questions
Question 17.
Solution:
3 is one zero of p(x) = x³ – 6x² + 11x – 6
(x – 3) is a factor of p(x)
Dividing, we get

Question 18.
Solution:

Question 19.
Solution:
p(x) = 3x⁴ + 5x³ – 7x² + 2x + 2
Dividing by x² + 3x + 1,
we get,
Quotient = 3x² – 4x + 2

Question 20.
Solution:
Let p(x) = x³ + 2x² + kx + 3
g(x) = x – 3
and r(x) = 21
Dividing p(x) by g(x), we get

But remainder = 21
3 + 3k + 45 = 21
3k = 21 – 45 – 3
=> 3k = 21 – 48 = -27
k = -9
Second method:
x – 3 is a factor of p(x) : x = 3
Substituting the value of x in p(x)
p(3) = 3³ + 2 x 3² + k x 3 + 3
= 27 + 18 + 3k + 3
48 + 3k = 21
=> 3k = -48 + 21 = -27
k = -9
Hence, k = -9

Hope given RS Aggarwal Solutions Class 10 Chapter 2 Polynomials Test Yourself are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 10 Solutions Chapter 2 Polynomials MCQS

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Class 10 Solutions Chapter 2 Polynomials MCQS

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 2 Polynomials Ex 2A
• RS Aggarwal Solutions Class 10 Chapter 2 Polynomials Ex 2B
• RS Aggarwal Solutions Class 10 Chapter 2 Polynomials Ex 2C
• RS Aggarwal Solutions Class 10 Chapter 2 Polynomials MCQS
• RS Aggarwal Solutions Class 10 Chapter 2 Polynomials Test Yourself

Choose the correct answer in each of the following questions.
Question 1.
Solution:
(d) √2 x² – 3√3 x + √6 is polynomial, others are not polynomial.

Question 2.
Solution:
(d) x + \(\frac { 3 }{ x }\) is not a polynomial, other are polynomial.

Question 3.
Solution:
(c) Let f(x) = x² – 2x – 3
= x² – 3x + x – 3
= x(x – 3) + 1(x – 3)
= (x – 3)(x + 1)
If x – 3 = 0, then x – 3
and if x + 1 = 0, then x = -1
Zeros are 3, -1

Question 4.
Solution:
(b)

Question 5.
Solution:
(c)

Question 6.
Solution:
(b) Polynomial is x² + \(\frac { 1 }{ 6 }\) x – 2

Question 7.
Solution:
(a)

Question 8.
Solution:
(c) Sum of zeros = 3
Product of zeros = -10
Polynomial : x² – (Sum of zeros) x + Product of zeros
= x² – 3x – 10

Question 9.
Solution:
(c) Zeros are 5 and -3
Sum of zeros = 5 – 3 = 2
Product of zeros = 5 x (-3) = -15
Polynomial: x² – (Sum of zeros) x + Product of zeros
= x² – 2x – 15

Question 10.
Solution:
(d)

Question 11.
Solution:
(b) Let f(x) = x² + 88x +125
Here, sum of roots = \(\frac { -b }{ a }\) = -88
and product = \(\frac { c }{ a }\) = 125
Product is positive,
Both zeros can be both positive or both negative.
Sum is negative.
Both zeros are negative.

Question 12.
Solution:
(b) α and β are the zeros of x² + 5x + 8
Then sum of zeros (α + β) = \(\frac { -b }{ a }\) = \(\frac { -5 }{ 1 }\) = -5

Question 13.
Solution:
(c) α and β are the zeros of 2x² + 5x – 9
Product of zeros (αβ) = \(\frac { c }{ a }\) = \(\frac { -9 }{ 2 }\)

Question 14.
Solution:
(d) 2 is a zero of kx² + 3x + k
It will satisfy the quadratic equation kx² + 3x + k = 0
k(2)² + 3x² + 1 = 0
4k + 6 + k = 0
=> 5k = -6
k = \(\frac { -6 }{ 5 }\)

Question 15.
Solution:
(b) -4 is a zero of (k – 1) x² + 4x + 1
-4 will satisfy the equation (k – 1) x² + kx + 1 = 0
=> (k – 1)(-4)² + k(-4) + 1 =0
=> 16k – 16 – 4k + 1 = 0
=> 12k – 15 = 0
=> 12k = 15
=> k = \(\frac { 15 }{ 12 }\) = \(\frac { 5 }{ 4 }\)

Question 16.
Solution:
(c)

Question 17.
Solution:
(a)

Question 18.
Solution:
(d) Polynomial: kx² + 2x + 3k

Question 19.
Solution:
(b) α, β are the zeros of the polynomial x² + 6x + 2

Question 20.
Solution:
(a) α, β, γ are the zeros of x³ – 6x² – x + 30
Then αβ + βγ + γα = \(\frac { c }{ a }\) = \(\frac { -1 }{ 1 }\) = -1

Question 21.
Solution:
(a) α, β, γ are the zeros of 2x³ + x² – 13x + 6, then
αβγ = \(\frac { -d }{ a }\) = \(\frac { -6 }{ 2 }\) = -3

Question 22.
Solution:
(c) α, β, γ are the zeros of p(x) such that

Question 23.
Solution:
(a)

Question 24.
Solution:
(b) If one zero of cubic polynomial ax³ + bx² + cx + d = 0
Let a be zero, then

Question 25.
Solution:
(c)

Question 26.
Solution:
(d)

Question 27.
Solution:
(c) p(x) is divided by q(x), then
p(x) = q(x) x g(x) + r(x)
Either r(x) = 0
Degree of r(x) < deg of g(x)

Question 28.
Solution:
(d) (a) is not a linear polynomial.
(b) is trinomial not binomial.
(c) is not a monomial.
(d) 5x² is monomial is true.

Hope given RS Aggarwal Solutions Class 10 Chapter 2 Polynomials MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2C

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2C. You must go through NCERT Solutions for Class 10 Maths to get better score in CBSE Board exams along with RS Aggarwal Class 10 Solutions.

Very-Short-Answer Questions
Question 1.
Solution:
Let other zero of x² – 4x + 1 be a, then
Sum of zeros = \(\frac { -b }{ a }\) = \(\frac { -(-4) }{ 1 }\) = 4
But one zero is 2 + √3
Second zero = 4 – (2 + √3) =4 – 2 – √3 = 2 – √3

Question 2.
Solution:
Let f(x) = x² + x – p(p + 1)
= x² + (p + 1) x – px – p(p + 1)
= x(x + p + 1) – p(x + p + 1)
= (x + p + 1) (x – p)
Either x + p + 1 = 0, then x = -(p + 1)
or x – p = 0, then x = p
Hence, zeros are p and -(p + 1)

Question 3.
Solution:
p(x) = x² – 3x – m(m + 3)
= x² – (m + 3)x + mx – m(m + 3)
= x(x – m – 3) + m(x – m – 3)
= (x – m – 3)(x + m)
Either x – m – 3 = 0, then x = m + 3
or x + m = 0, then x = -m
Zeros are (m + 3), -m

Question 4.
Solution:
a and p are the zeros of a polynomial
and α + β = 6, αβ = 4
Polynomial = x² – (α + β)x + αβ = x² – (6)x + 4 = x² – 6x + 4

Question 5.
Solution:
One zero of kx² + 3x + k is 2
x = 2 will satisfy it
⇒ k(2)² + 3 x 2 + k = 0
⇒ 4k + 6 + k= 0
⇒5k + 6 = 0
⇒ 5k = -6
⇒ k = \(\frac { -6 }{ 5 }\)
Hence, k = \(\frac { -6 }{ 5 }\)

Question 6.
Solution:
3 is a zero of the polynomial 2x² + x + k
Then 3 will satisfy it
2x² + x + k = 0
⇒ 2(3)² + 3 + k = 0
⇒ 18 + 3+ k = 0
⇒ 21 + k = 0
⇒ k = -21
Hence, k = -21

Question 7.
Solution:
-4 is a zero of polynomial x² – x – (2k + 2)
Then it will satisfy the equation
x² – x – (2k + 2) = 0
⇒ (-4)2 – (-4) – 2k – 2 = 0
⇒ 16 + 4 – 2k – 2 = 0
⇒ -2k + 18 = 0
⇒ 2k = 18
k = 9

Question 8.
Solution:
1 is a zero of the polynomial ax² – 3(a – 1)x – 1
Then 1 will satisfy the equation ax² – 3(a – 1) x – 1 = 0
a(1)² – 3(a – 1) x 1 – 1 = 0
⇒ a x 1 – 3a + 3 – 1 = 0
⇒ a – 3a + 2 = 0
⇒ -2a + 2 = 0
⇒ 2a = 2
⇒ a = 1

Question 9.
Solution:
-2 is a zero of 3x² + 4x + 2k
It will satisfy the equation 3x² + 4x + 2k = 5
3(-2)² + 4(-2) + 2k = 0
⇒ 3 x 4 + 4(-2) + 2k = 0
⇒ 12 – 8 + 2k = 0
⇒ 4 + 2k=0
⇒ 2k = -4
⇒ k = -2
k = -2

Question 10.
Solution:
Let f(x) = x² – x – 6
= x² – 3x + 2x – 6
= x(x – 3) + 2(x – 3)
= (x – 3)(x + 2)
(x – 3)(x + 2) = 0
Either x – 3 = 0, then x = 3
or x + 2 = 0, then x = -2
Zeros are 3, -2

Question 11.
Solution:
Sum of zeros = 1
and polynomial is kx² – 3x + 5
Sum of zeros = \(\frac { -b }{ a }\) = \(\frac { -(-3) }{ k }\) = \(\frac { 3 }{ k }\)
\(\frac { 3 }{ k }\) = 1
⇒ k = 3
Hence, k = 3

Question 12.
Solution:
Product of zeros of polynomial x² – 4x + k is 3
Product of zeros = \(\frac { c }{ a }\)
⇒ \(\frac { k }{ 1 }\) = 3
⇒ k = 3

Question 13.
Solution:
x + a is a factor of
f(x) = 2x² + (2a + 5) x + 10
Let x + a = 0, then
Zero of f(x) = -a
Now f(-a) = 2 (-a)² + (2a + 5)(-a) + 10 = 0
2a² – 2a² – 5a + 10 = 0
⇒ 5a = 10
⇒ a = 2

Question 14.
Solution:
(a – b), a, (a + b) are the zeros of 2x³ – 6x² + 5x – 7
Sum of zeros = \(\frac { -b }{ a }\)
⇒ a – b + a + a + b = \(\frac { -(-6) }{ 2 }\)
⇒ 3a = \(\frac { 6 }{ 2 }\)
⇒ 3a = 3
⇒ a = 1

Question 15.
Solution:
f(x) = x³ + x² – ax + 6 is divisible by x² – x
Remainder will be zero
Now dividing f(x) by x² – x
Remainder = (2 – a) x + b

(2 – a) x + b = 0
2 – a = 0
⇒ a = 2 and b = 0
Hence, a = 2, b = 0

Question 16.
Solution:
α and β are the zeros of polynomial f(x) = 2x² + 7x + 5

Question 17.
Solution:
Division algorithm for polynomials:
If f(x) and g(x) are any two polynomials with g(x) ≠ 0, then we can find polynomial q(x) and r(x).
f(x) = q(x) x g(x) + r(x)
where r (x) = 0
or [degree of r(x) < degree of g(x)]
or Dividend=Quotient x Division + Remainder

Question 18.
Solution:
Sum of zeros = \(\frac { -1 }{ 2 }\)
Product of zeros = -3
Polynomial: x² – (Sum of zeros) x + product of zeros

Short-Answer Questions
Question 19.
Solution:

Question 20.
Solution:

Question 21.
Solution:
α and β are the zeros of polynomial f(x) = x² – 5x + k

(1)² = (5)² – 4 k
1 ⇒ 25 – 4k
⇒ 4k = 25 – 1 = 24
Hence, k = 6

Question 22.
Solution:

Question 23.
Solution:
α and β are the zeros of polynomial
f(x) = 5x² – 7x + 1

Question 24.
Solution:

Question 25.
Solution:
(a – b), a and (a + b) are the zeros of the polynomial
f(x) = x³ – 3x² + x + 1

Hope given RS Aggarwal Solutions Class 10 Chapter 2 Polynomials Ex 2C are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

NCERT Maths Solutions for Ex 2.2 class 10 Polynomials is the perfect guide to boost up your preparation during CBSE 10th Class Maths Examination.

RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2B

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2B. You must go through NCERT Solutions for Class 10 Maths to get better score in CBSE Board exams along with RS Aggarwal Class 10 Solutions.

Question 1.
Solution:
Zeros are 3, -2, 1 of
p(x) = x³ – 2x² – 5x + 6
Here, a = 1, b = -2, c = -5, d = 6
We know that if α, β and γ are the roots of
f(x) = ax³ + bx² + cx + d, then

Question 2.
Solution:
Zeros are 5, -2 and \(\frac { 1 }{ 3 }\) of

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Find the quotient and the remainder when:
Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:
f(x) = 2x⁴ + 3x³ – 2x² – 9x – 12
g(x) = x² – 3
Quotient [q(x)] = 2x² + 3x + 4
Remainder [r(x)] = 0
Remainder is zero.
x² – 3 is a factor of f(x)

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

= (x + 1) (x + 4) (x – 3)
If x + 4 = 0, then x = -4
If x – 3 = 0, then x = 3
Zeros are -1, -4, 3

Question 13.
Solution:

Question 14.
Solution:

=> (x – 3) (x + 3) [x (x + 2) – 1 (x + 2)]
=> (x – 3) (x + 3) (x + 2) (x – 1)
Other zeros will be
If x + 2 = 0 then x = -2
and if x – 1 =0, then x = 1
Zeros are 3, -3, -2, 1

Question 15.
Solution:
2 and -2 are the two zeros of the polynomial
f(x) = x⁴ + x³ – 34x² – 4x + 120,
Then (x – 2) (x + 2) or x² – 4 will its the factor of f(x)
Now dividing f(x) by x² – 4, we get

f(x) = (x – 2) (x + 2) (x² + x – 30)
= (x – 2)(x + 2)[x² + 6x – 5x – 30]
= (x – 2) (x + 2)[x(x + 6) – 5(x + 6)]
= (x – 2) (x + 2) (x + 6) (x – 5)
Other two zeros are
If x + 6 = 0, then x = -6 and
if x – 5 = 0, then x = 5
Roots of f(x) are 2, -2, -6, 5

Question 16.
Solution:

Other two zeros are :
if x + 5 = 0, then x = -5
and if x – 4 = 0, then x = 4
Hence, all the zeros of f(x) are : √3, – √3, 4, -5

Question 17.
Solution:
√3 and – √3 are the zeros of the polynomial
f(x) = 2x⁴ – 3x³ – 5x² + 9x – 3
=> (x – √3) (x + √3) or (x² – 3) is a factor of f(x)
Now, dividing f(x) by x² – 3, we get

Question 18.
Solution:

Question 19.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 2 Polynomials Ex 2B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.