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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

• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2A
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2B
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2C
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2D
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2E
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2F
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2G
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2H
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2I
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2J
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2K

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

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

• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2A
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2B
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2C
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2D
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2E
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2F
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2G
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2H
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2I
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2J
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2K

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:

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 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.