RS Aggarwal Maths Solutions Class 9

RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2F

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RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2F

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2F.

Other Exercises

Factorize :

Question 1.
Solution:
25x2 – 64y2
= (5x)2 – (8y)2
= (5x + 8y) (5x – 8y) Ans.

Question 2.
Solution:
100 – 9x2
(10)2 – (3x)2
= (10 + 3x) (10 – 3x) Ans.

Question 3.
Solution:
5x2 – 7y2
=(√5x)2 +(√7y)2
=(√5x – √7y)(√5x – √7y) Ans

Question 4.
Solution:
(3x + 5y)2 – 4z2
= (3x + 5y)2 – (2z)2
= (3x + 5y + 2z) (3x + 5y – 2z) Ans.

Question 5.
Solution:
150 – 6x2
= 6(25 – x2)
= 6{(5)2 – (x2)}
= 6 (5 + x) (5 – x) Ans.

Question 6.
Solution:
20x2 – 45
= 5 (4x2 – 9)
= 5{(2x)2 – (3)2}
= 5 (2x + 3) (2x – 3) Ans.

Question 7.
Solution:
3x3 – 48x
= 3x (x2 – 16)
= 3x {(x)2 – (4)2}
= 3x (x + 4) (x – 4) Ans.

Question 8.
Solution:
2 – 50x2
= 2(1 – 25x2) = 2 {(1)2 – (5x)2}
= 2 (1 + 5x) (1 – 5x) Ans.

Question 9.
Solution:
27a2 – 48b2
= 3(9a2 – 16b2)
= 3 {(3a)2 – (4b)2}
= 3(3a + 4b) (3a – 4b) Ans.

Question 10.
Solution:
x – 64x3
= x (1 – 64x2)
= x{(1)2 – (8x)2}
= x (1 + 8x) (1 – 8x) Ans.

Question 11.
Solution:
8ab2 – 18a3
= 2a (4b2 – 9a2)
= 2a {(2b)2 – (3a)2}
= 2a (2b + 3a) (2b – 3a) Ans

Question 12.
Solution:
3a3b – 243ab3
= 2ab (a2 -81 b2)
= 3ab {(a)2 – (9b)2}
= 3ab (a + 9b) (a – 9b) Ans.

Question 13.
Solution:
(a + b)3 – a – b
= (a + b)3 – 1 (a + b)
= (a + b) {(a + b)2 – 1}
= (a + b) {(a + b)2 – (1)2}
= (a + b) (a + b + 1) (a + b – 1) Ans.

Question 14.
Solution:
108 a2 – 3(b – c)2
= 3 {36a2 – (b – c)2}
= 3 {(6a)2 – (b – c)2}
= 3 {(6a + (b – c)} {6a – (b – c)}
= 3 (6a + b – c) (6a – b + c) Ans.

Question 15.
Solution:
x3 – 5x2 – x + 5
= x2(x – 5) -1(x – 5)
= (x – 5) (x2 – 1)
= (x – 5) {(x)2 – (1)2}
= (x – 5) (x + 1) (x – 1) Ans.

Question 16.
Solution:
a2 + 2ab + b2 – 9c2
= (a + b)2 – (3c)2
= {a2 + b2 + 2ab – (a + b)2}
= (a + b + 3c) (a + b – 3c) Ans.

Question 17.
Solution:
9 – a2 + 2ab – b2
= 9 – (a2 – 2ab + b2)
= (3)2 -(a- b)2
{ ∴a2 + b2 – 2ab = (a – b)2}
= (3 + a – b) (3 – a + b) Ans.

Question 18.
Solution:
a2 – b2 – 4ac + 4c2
= a2 – 4ac + 4c2 – b2
= (a)2 – 2 x a x 2c + (2c)2 – (b)2
= (a – 2c)2 – (b)2
= (a – 2c + b) (a – 2c – b)
= (a + b – 2c) (a – b – 2c) Ans.

Question 19.
Solution:
9a2 + 3a – 8B – 64b2
= 9a2 – 64b2 + 3a – 8b
= (3a)2 – (8b)2 + 1(3a – 8b)
= (3a + 8b) (3a – 8b) + 1 (3a – 8b)
= (3a – 8b) (3a + 8b + 1) Ans.

Question 20.
Solution:
x2 – y2 + 6y _ 9
= (x)2 – (y2 – 6y + 9)
= (x)2 – {(y)2 – 2 x y x 3 + (3)2}
= (x)2 – (y – 3)2
= (x + y – 3) (x – y + 3) Ans.

Question 21.
Solution:
4x2 – 9y2 – 2x – 3y
= (2x)2 – (3y)2 – 1(2x + 3y)
= (2x + 3y) (2x – 3y) – 1( 2x + 3y)
= (2x + 3y) (2x – 3y – 1) Ans.

Question 22.
Solution:
x4 – 1
= (x2)2 – (1)2
= (x2 + 1) (x2 – 1)
= (x2 + 1) {(x)2 – (1)2}
= (x2 + 1) (x + 1) (x – 1) Ans.

Question 23.
Solution:
a – b – a2+ b2
= 1 (a – b) – (a2 – b2)
= 1 (a – b) – (a + b) (a – b)
= (a – b) (1 – a – b) Ans.

Question 24.
Solution:
x4 – 625
= (x2)2 – (25)2
= (x2 + 25) (x2 – 25)
= (x2 + 25) {(x)2 – (5)2}
= (x2 + 25) (x + 5) (x – 5) Ans.

Hope given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2F 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 9 Solutions Chapter 2 Polynomials Ex 2E

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RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2E

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2E.

Other Exercises

Factorize:

Question 1.
Solution:
9x2 + 12xy
= 3x (3x + 4y) Ans.

Question 2.
Solution:
18x2y – 24xyz
= 6xy (3x – 4z) Ans.

Question 3.
Solution:
27a3b3 – 45a4b2
= 9a3b2 (3b – 5a) Ans.

Question 4.
Solution:
2a (x + y) – 3b(x + y)
= (x + y) (2a – 3b) Ans.

Question 5.
Solution:
2x (p2 + q2) + 4y (p2 + q2)
= 2(p2 + q2) (x + 2y) Ans

Question 6.
Solution:
x (a – 5) + y (5 – a)
= x (a – 5) -y (a – 5)
= (a – 5) (x – y) Ans.

Question 7.
Solution:
4(a + b) – 6 (a + b)2
= 2(a + b) {2 – 3 (a + b)}
= 2(a + b) (2 – 3a – 3b) Ans.

Question 8.
Solution:
8(3a – 2b)2 – 10 (3a – 2b)
= 2(3a – 2b) {4 (3a – 2b) – 5}
= 2(3a – 2b) (12a – 8b – 5) Ans.

Question 9.
Solution:
x (x + x)3 – 3x2 y (x + y)
= x (x + y) {(x + y)2 – 3xy}
= x (x + y) [x2 + y2 + 2xy – 3xy)
= x (x + y) (x2 + y2 – xy) Ans.

Question 10.
Solution:
x3 + 2x2 + 5x + 10
= x2 (x + 2) + 5 (x + 2)
= (x + 2) (x2 + 5) Ans.

Question 11.
Solution:
x2 + xy – 2xz – 2yz
= x (x + y) -2z (x + y)
= (x + y) (x – 2z) Ans.

Question 12.
Solution:
a3 b – a2b + 5ab – 5b.
= b (a3 – a2 + 5a – 5)
= b {(a2 (a – 1) + 5 (a – 1)}
= b (a – 1) (a2 + 5) Ans.

Question 13.
Solution:
8 – 4a – 2a3 + a4
= 4 (2 – a) – a3 (2 – a)
= (2 – a) (4 – a3) Ans.

Question 14.
Solution:
x3 – 2x2y + 3xy2 – 6y3
= x2 (x – 2y) + 3y2 (x – 2y)
= (x – 2y) (x2 + 3y2) Ans

Question 15.
Solution:
px – 5q + pq – 5x
= px – 5x + pq – 5q
= x(p – 5) + q(p – 5)
= {p – 5) (x + q) Ans.

Question 16.
Solution:
x2 + y – xy – x
= x2 – x – xy + y
= x (x – 1) – y (x – 1)
= (x – 1) (x – y) Ans.

Question 17.
Solution:
(3a – 1)2 – 6a + 2
= (3a – 1)2 – 2 (3a – 1)
= (3a – 1) (3a – 1 – 2)
= (3a – 1) (3a – 3)
= 3(3a – 1) (a – 1) Ans.

Question 18.
Solution:
(2x – 3)2 – 8x + 12
= (2x – 3)2 – 4(2x – 3)
= (2x – 3) (2x – 3 – 4)
= (2x – 3) (2x – 7) Ans.

Question 19.
Solution:
a3 + a – 3a2 – 3
= a3 – 3a2 + a – 3
= a2 (a – 3) + 1 (a – 3)
= (a – 3) (a2 + 1) Ans.

Question 20.
Solution:
3ax – 6ay – 8by + 4bx
= 3ax – 6ay + 4bx – 8by
= 3a (x – 2y) + 4b (x – 2y)
= (x – 2y) (3a + 4b) Ans

Question 21.
Solution:
abx2 + a2x + b2x + ab
= ax (bx + a) + b (bx + a)
= (bx + a) (ax + b) Ans.

Question 22.
Solution:
x3 – x2 + ax + x – a – 1
= x3 – x2 + ax – a + x – 1
= x2 (x – 1) + a (x – 1) + 1 (x – 1)
= (x – 1) (x2 + a + 1) Ans.

Question 23.
Solution:
2x + 4y – 8xy – 1
= 2x – 8xy -1+4y
= 2x (1 – 4y) -1 (1 – 4y)
= (1 – 4y) (2x – 1) Ans.

Question 24.
Solution:
ab (x2 +y2) – xy (a2 + b2)
= abx2 + aby2 – a2xy – b2xy
= abx2 – a2xy – b2xy + aby2
= ax (bx – ay) – by (bx – ay)
= (bx – ay) (ax – by) Ans.

Question 25.
Solution:
a2 + ab (b + 1) + b3
= a2 + ab2 + ab + b3
= a (a + b2) + b (a + b2)
= (a + b2) (a + b) Ans

Question 26.
Solution:
a3 + ab (1 – 2a) – 2b2
= a3 + ab – 2a2b – 2b2
= a3 – 2a2b + ab – 2b2
= a2 (a – 2b) + b (a – 2b)
= (a – 2b) (a2 + b) Ans.

Question 27.
Solution:
2a2 + bc – 2ab – ac
= 2a2 – 2ab – ac + bc
= 2a (a – b) – c (a – b)
= (a – b) (2a – c) Ans.

Question 28.
Solution:
(ax + by)2 + (bx – ay)2
= a2x2 + b2y2 + 2abxy + b2x2 + a2y2 – 2bxy
= a2x2 + b2y2 + b2x2 + a2y2
= a2x2 + b2x2 + a2y2 + b2y2
= x2(a2 + b2) + y2(a2 + b2)
= (a2 + b2) (x2 + y2) Ans.

Question 29.
Solution:
a (a + b – c) – bc
= a2 + ab – ac – bc
= a (a + b) – c (a + b)
(a + b) (a – c) Ans.

Question 30.
Solution:
a(a – 2b – c) + 2bc
= a2 – 2ab – ac +2bc
= a2 – ac – 2ab + 2bc
= a (a – c) – 2b (a – c)
= (a – c) (a – 2b) Ans.

Question 31.
Solution:
a2x2 + (ax2 + 1) x + a
= a2x2 + ax3 + x + a
= a2x2 + ax3 + a + x
= ax2 (a + x) + 1 (a + x)
– (a + x) (ax2 + 1) Ans

Question 32.
Solution:
ab (x2 + 1) + x (a2 + b2)
= abx2 + ab + a2x + b2x
= abx2 + a2x + b2x + ab
= ax (bx + a) + b (bx + a)
= (bx + a) (ax + b) Ans.

Question 33.
Solution:
x2 – (a + b) x + ab
= x2 – ax – bx + ab
= x (x – a) – b (x – a)
= (x – a) (x – b) Ans.

Question 34.
Solution:
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2E 34

Hope given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2E 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 9 Solutions Chapter 2 Polynomials Ex 2D

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RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2D

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2D.

Other Exercises

Using factor theorem, show that :

Question 1.
Solution:
By factor theorem, x – 2 will be a factor of f(x) = x3 – 8 if f(2) = 0
(∴ x-2 = 0=>x = 2)
Now f(2) = (2)3 – 8 = 8- 8 = 0
Hence (x – 2) is a factor of f(x) Ans.

Question 2.
Solution:
By factor theorem,
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2D Q2.1

Question 3.
Solution:
By factor theorem,
(x – 1) is a factor of f(x)=(2x4 + 9x3 + 6x2– 11x – 6)
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2D Q3.1

Question 4.
Solution:
By factor theorem, (x + 2) will
a factor of f (x) = x4 – x4 + 2
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2D Q4.1

Question 5.
Solution:
By factor theorem, (x + 5) will be a factor of f(x) = 2x3 + 9x2 – 11x – 30 if f(-5) = 0
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2D Q5.1

Question 6.
Solution:
By factor theorem, (2x – 3) is a factor of f(x) = 2x4 + x3 – 8x2 – x + 6
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2D Q6.1

Question 7.
Solution:
By factor theorem, (x – √2 ) will be a factor of f(x) = 7x2 – 4√2x – 6
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2D Q7.1

Question 8.
Solution:
By factor theorem, (x + √2) will be a factor of f(x) = 2√2 x3 + 5x + √2
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2D Q8.1

Question 9.
Solution:
Let f(x) = 2x3 + 9x2 + x + k and x – 1 is a factor of f(x)
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2D Q9.1

Question 10.
Solution:
Let f(x) = 2x3 – 3x2 – 18x + a and x – 4 is its factor
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2D Q10.1

Question 11.
Solution:
Let f(x) – x4 – x3 – 11x2 – x + a
f(x) is divisible by (x + 3)
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2D Q11.1

Question 12.
Solution:
Let f(x) = 2x3 + ax2 + 11x + a + 3
and (2x – 1) is its factor
Let 2x – 1 = 0 then 2x = 1
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2D Q12.1
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2D Q12.2

Question 13.
Solution:
Let f(x) = x3 – 10x2 + ax + b and (x – 1) and (x – 2) are its factors
∴ x – 1 = 0 =>x=1
and x – 2 = 2 =>x=2
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2D Q13.1

Question 14.
Solution:
Let f(x) = x4 + ax3 – 7x2 – 8x + b ,
and (x + 2) and (x + 3) are its factors
∴x + 2 = 0 => x = -2
and x + 3= 0 => x = -3
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2D Q14.1
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2D Q14.2

Question 15.
Solution:
Let f(x) = x3 – 3x2 – 13x + 15
Now x2 + 2x – 3 = x2 + 3x – x – 3
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2D Q15.1

Question 16.
Solution:
Let f(x) = x3 + ax2 + bx + 6 and (x – 2) is its factor
Let x – 2 = 0 then x = 2
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2D Q16.1
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2D Q16.2

Hope given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2D are helpful to complete your math homework.

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RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2C

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RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2C

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2C.

Other Exercises

Using remainder theorem, find the remainder when :

Question 1.
Solution:
f(x) = (x3 – 6x2 + 9x + 3)
Let x-1 = 0, then x = 1
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2C Q1

Question 2.
Solution:
f(x) = 2x3 – 5x2 + 9x – 8)
Let x-3 = 0, then x = 3
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2C Q2.1

Question 3.
Solution:
f(x) = (3x4 – 6x2 – 8x + 2)
Let x-2 = 0, the x = 2
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2C Q2.3.1

Question 4.
Solution:
f(x) = (x3 – 7x2 + 6x + 4)
Let x-6 = 0,then x=6
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2C Q2.4.1

Question 5.
Solution:
f(x)=(x3 – 6x2 + 13x + 60)
Let x+2=0,then x=-2
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2C Q5.1

Question 6.
Solution:
f(x)=(2x4 + 6x3 + 2x2 + x – 8)
Let x+3=0,then x=-3
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2C Q6.1

Question 7.
Solution:
f(x)=(4x3 – 12x2 + 11x – 5)
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2C Q7.1

Question 8.
Solution:
f(x)=(81x4 + 54x3 – 9x2 – 3x + 2)
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2C Q8.1

Question 9.
Solution:
f(x)=(x3 – ax2 + 2x – a)
let x-a=0, then x=a
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2C Q9.1

Question 10.
Solution:
f(x)=(ax3+ 3x2 – 3)
g(x)=(2x3 -5x + a)
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2C Q10.1

Question 11.
Solution:
f(x)=x4 – 2x3 + 3x2 – ax + b
let x-1=0, then x=1
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2C Q11.1
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2C Q11.2

Hope given RS Aggarwal Solutions Class 9 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.

RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2B

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RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2B

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2B.

Other Exercises

Question 1.
Solution:
p(x) = 5 – 4x + 2x2
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2B Q1.1

Question 2.
Solution:
p(y) = 4 + 3y – y2 + 5y3
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2B Q2.1

Question 3.
Solution:
f(t)=4t2-3t+6
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2B Q3.1

Question 4.
Solution:
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2B Q4.1
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2B Q4.2

Question 5.
Solution:
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2B Q5.1
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2B Q5.2
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2B Q5.3

Hope given RS Aggarwal Solutions Class 9 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.