RS Aggarwal Class 9 Math Chapter 2 Polynomials Exercise 2E Solution

EXERCISE 2E

Question 1:
9x² + 12xy = 3x (3x + 4y)

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

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

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

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

Question 6:
x (a – 5) + y (5 – a)
= x (a – 5) + y (-1) (a – 5)
= (x – y) (a – 5)

Question 7:
4 (a + b) – 6 (a + b)²
= (a + b) [4 – 6 (a + b)]
= 2 (a + b) (2 – 3a – 3b)
= 2 (a + b) (2 – 3a – 3b)

Question 8:
8 (3a – 2b)² – 10 (3a – 2b)
= (3a – 2b) [8(3a – 2b) – 10]
= (3a – 2b) 2[4 (3a – 2b) – 5]
= 2 (3a – 2b) (12 a – 8b – 5)

Question 9:
x (x + y)³ – 3x²y (x + y)
= x (x + y) [(x + y)² – 3xy]
= x (x + y) (x² + y² + 2xy – 3xy)
= x (x + y) (x² + y² – xy)

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

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

Question 12:
a³b – a²b + 5ab – 5b
= a²b (a – 1) + 5b (a – 1)
= (a – 1) (a²b + 5b)
= (a – 1) b (a² + 5)
= b (a – 1) (a² + 5)

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

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

Question 15:
px + pq – 5q – 5x
= p(x + q) – 5 (q + x)
= (x + q) (p – 5)

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

Question 17:
(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)

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

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

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

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

Question 22:
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)

Question 23:
2x + 4y – 8xy – 1
= 2x – 1 – 8xy + 4y
= (2x – 1) – 4y (2x – 1)
= (2x – 1) (1 – 4y)

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

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

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

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

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

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

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

Question 31:
a²x² + (ax² + 1)x + a
= a²x² + ax³ + x + a
= ax² (a + x) + 1 (x + a)
= (ax² + 1) (a + x)

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

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

Question 34: