CBSE Class 9 – Page 133

RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2E

January 5, 2022June 18, 2018 by Prasanna

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

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RD Sharma Class 9 Solutions Chapter 3 Rationalisation VSAQS

January 5, 2022June 18, 2018 by Prasanna

RD Sharma Class 9 Solutions Chapter 3 Rationalisation VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 3 Rationalisation VSAQS

Question 1.
Write the value of (2 + \(\sqrt { 3 } \) ) (2 – \(\sqrt { 3 } \)).
Solution:
(2+ \(\sqrt { 3 } \) )(2- \(\sqrt { 3 } \) ) = (2)²-(\(\sqrt { 3 } \) )²{∵ (a + b) (a – b) = a² – b²}
= 4-3=1

Question 2.
Write the reciprocal of 5 + \(\sqrt { 2 } \).
Solution:
RD Sharma Class 9 Solutions Chapter 3 Rationalisation VSAQS Q2.1

Question 3.
Write the rationalisation factor of 7 – 3\(\sqrt { 5 } \) .
Solution:
Rationalising factor of 7 – 3\(\sqrt { 5 } \) is 7 + 3\(\sqrt { 5 } \)
{∵ (\(\sqrt { a } \) + \(\sqrt { b } \) ) (\(\sqrt { a } \) – \(\sqrt { b } \)) = a-b}

Question 4.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation VSAQS Q4.1
Solution:

Question 5.
If x =\(\sqrt { 2 } \) – 1 then write the value of \(\frac { 1 }{ x }\).
Solution:
RD Sharma Class 9 Solutions Chapter 3 Rationalisation VSAQS Q5.1

Question 6.
If a = \(\sqrt { 2 } \) + h then find the value of a –\(\frac { 1 }{ a }\)
Solution:
RD Sharma Class 9 Solutions Chapter 3 Rationalisation VSAQS Q6.1

Question 7.
If x = 2 + \(\sqrt { 3 } \), find the value of x + \(\frac { 1 }{ x }\).
Solution:
RD Sharma Class 9 Solutions Chapter 3 Rationalisation VSAQS Q7.1

Question 8.
Write the rationalisation factor of \(\sqrt { 5 } \) – 2.
Solution:
Rationalisation factor of \(\sqrt { 5 } \) – 2 is \(\sqrt { 5 } \) + 2 as
(\(\sqrt { a } \) + \(\sqrt { b } \))(\(\sqrt { a } \) – \(\sqrt { b } \)) = a – b

Question 9.
Simplify : \(\sqrt { 3+2\sqrt { 2 } }\).
Solution:
RD Sharma Class 9 Solutions Chapter 3 Rationalisation VSAQS Q9.1

Question 10.
Simplify : \(\sqrt { 3-2\sqrt { 2 } }\).
Solution:
RD Sharma Class 9 Solutions Chapter 3 Rationalisation VSAQS Q10.1

Question 11.
If x = 3 + 2 \(\sqrt { 2 } \), then find the value of \(\sqrt { x } \) – \(\frac { 1 }{ \sqrt { x } }\).
Solution:
RD Sharma Class 9 Solutions Chapter 3 Rationalisation VSAQS Q11.1

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RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.2

January 5, 2022June 18, 2018 by Prasanna

RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.2

Question 1.
Rationalise the denominators of each of the following(i – vii):
RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.2 Q1.1 >
Solution:

Question 2.
Find the value to three places of decimals of each of the following. It is given that
RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.2 Q2.1
Solution:

Question 3.
Express each one of the following with rational denominator:
RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.2 Q3.1
Solution:

Question 4.
Rationales the denominator and simplify:
RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.2 Q4.1
Solution:

Question 5.
Simplify:
RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.2 Q5.1
Solution:

Question 6.
In each of the following determine rational numbers a and b:
RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.2 Q6.1
Solution:

Question 7.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.2 Q7.1
Solution:

Question 8.
Find the values of each of the following correct to three places of decimals, it being given that \(\sqrt { 2 } \) = 1.4142, \(\sqrt { 3 } \) = 1-732, \(\sqrt { 5 } \) = 2.2360, \(\sqrt { 6 } \) = 2.4495 and \(\sqrt { 10 } \) = 3.162.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.2 Q8.1
Solution:

Question 9.
Simplify:
RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.2 Q9.1
Solution:

Question 10.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.2 Q10.1
Solution:

Question 11.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.2 Q11.1
Solution:

Question 12.
RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.2 Q12.1
Solution:

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RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2D

January 5, 2022June 18, 2018 by Prasanna

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

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

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

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

January 5, 2022June 18, 2018 by Prasanna

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

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

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