Factors of Algebraic Expressions – Maharashtra Board Class 7 Solutions for Mathematics (English Medium)
MathematicsGeneral ScienceMaharashtra Board Solutions
Exercise 76:
Solution 1:
- 7p = 7 × p
∴ Factors of 7p are 7 and p. - 6m = 6 × m = 2 × 3 × m
∴ Factors of 6m are 2, 3, and m. - 9xy = 9 × x × y = 3 × 3 × x × y
∴ Factors of 9xy are 3, 3, x, and y. - 22ab = 22 × a × b = 2 × 11 × a × b
∴ Factors of 22ab are 2, 11, a, and b. - p2q = p × p × q
∴ Factors of p2q are p, p, and q. - 10xy2 = 2 × 5 × x × y × y
∴ Factors of 10xy2 are 2, 5, x, y, and y. - 5a2 = 5 × a × a
∴ Factors of 5a2 are 5, a, and a. - 15m2n = 3 × 5 × m × m × n
∴ Factors of 15m2n are 3, 5, m, m, and n. - 30a2b2 = 2 × 3 × 5 × a × a × b × b
∴ Factors of 30a2b2 are 2, 3, 5, a, a, b, and b. - 12x3 = 2 × 2 × 3 × x × x × x
∴ Factors of 12x3 are 2, 2, 3, x, x, and x.
Exercise 77:
Solution 1:
- 8m, 4m2n
8m = 2 × 2 × 2 × m
4m2n = 2 × 2 × m × m × n - 3x2y, 12xy2
3x2y = 3 × x × x × y
12xy2 = 2 × 2 × 3 × x × y × y - 15a2bc, 5ab, 20abc2
15a2bc = 3 × 5 × a × a × b × c
5ab = 5 × a × b
20abc2 = 2 × 2 × 5 × a × b × c × c
Solution 2:
- 4p2q, 16pq2
4p2q = 2 × 2 × p × p × q
16pq2 = 2 × 2 × 2 × 2 × p × q × q
∴ Common factors = 2 × 2 × p × q = 4pq - 18x3y2, 12x3y
18x3y2 = 2 × 3 × 3 × x × x × x × y × y
12x3y = 2 × 2 × 3 × x × x × x × y
∴ Common factors = 2 × 3 × x × x × x × y = 6x3y - 7a3b2c, 28a2bc2
7a3b2c = 7 × a × a × a × b × b × c
28a2bc2 = 2 × 2 × 7 × a × a × b × c × c
∴ Common factors = 7 × a × a × b × c = 7a2bc - 8x3y2, 10x2y3, 6x2y
8x3y2 = 2 × 2 × 2 × x × x × x × y × y
10x2y3 = 2 × 5 × x × x × y × y × y
6x2y = 2 × 3 × x × x × y
∴ Common factors = 2 × x × x × y = 2x2y - 24mnp2, 22m2p2, 30m2n2p
24mnp2 = 2 × 2 × 2 × 3 × m × n × p × p
22m2p2 = 2 × 11 × m × m × p × p
30m2n2p = 2 × 3 × 5 × m × m × n × n × p
∴ Common factors = 2 × m × p = 2mp
Solution 3:
- 6m2n2, 10m2n
Common factors: 2m2n - 38a3b2, 57ab2
Common factors: 19ab2 - 11x2y3, xy2
Common factors: xy2 - 35p2q2r, 40q3r2, 50pq2r
Common factors: 5q2r - 15x3y3, 39x2z2, 48xy2z3
Common factors: 3x
Exercise 78:
Solution 1:
- 4a + 8b = 4 × a + 4 × 2 × b
= 4(a + 2b) - 5m + 15n = 5 × m + 5 × 3 × n
= 5(m + 3n) - abp – abq = ab × p – ab × q
= ab(p – q) - x2 + x3 = x2 + x2 × x
= x2(1 + x) - mnx + mny = mn × x + mn × y
= mn(x + y) - 4x2y + 3xy2 = 4 × x × x × y + 3 × x × y × y
= xy(4x + 3y) - 15p2q – 20q = 3 × 5 × p × p × q – 4 × 5 × q
= 5q(3p2 – 4) - a2bc + abc2 = a × a × b × c + a × b × c × c
= abc(a + c) - 18m2n – 27m3 = 2 × 9 × m2 × n – 3 × 9 × m2 × m
= 9m2(2n – 3m) - 24p3q2 + 28p2q3 = 4 × 6 × p2 × p × q2 + 4 × 7 × p2 × q2 × q
= 4p2q2(6p + 7q)
Exercise 79:
Solution 1:
- ab + cd + ac + bd
= ab + ac + bd + cd
= a(b + c) + d(b + c)
= (a + d)(b + c) - 2x2 + 4x3 + 2x + 1
= 4x3 + 2x2 + 2x + 1
= 2x2(2x + 1) + 1(2x + 1)
= (2x + 1)(2x2 + 1) - ax + bx – ay – by
= ax – ay + bx – by
= a(x – y) + b(x – y)
= (x – y)(a + b) - y – 1 + y3 – y2
= y – 1 + y3 – y2
= 1(y – 1) + y2(y – 1)
= (y – 1)(1 + y2) - b2 + bc + ab + ac
= b2 + bc + ab + ac
= b(b + c) + a(b + c)
= (b + c)(b + a) - 2x2 + xy – 2xy2 – y3
= 2x2 – 2xy2 + xy – y3
= 2x(x – y2) + y(x – y2)
= (x – y2)(2x + y) - 12pm + 18qm + 6pn + 9nq
= 12pm + 6pn + 18qm + 9nq
= 6p(2m + n) + 9q(2m + n)
= (2m + n)(6p + 9q)
= (2m + n)3(2p + 3q)
= 3(2p + 3q)(2m + 3n) - m3 + m2 + m + 1
= m3 + m2 + m + 1
= m2(m + 1) + 1(m + 1)
= (m2 + 1)(m + 1) - am + an + al + bm + bl + bn
= am + an + al + bm + bl + bn
= a(m + n + l) + b(m + n + l)
= (a + b)(m + n + l) - 3y3 – 6y2 + 4y – 8
= 3y3 – 6y2 + 4y – 8
= 3y2(y – 2) + 4(y – 2)
= (3y2 + 4)(y – 2)
Exercise 80:
Solution 1(1):
Solution 1(2):
Solution 1(3)
Solution 1(4):
Solution 1(5):
Solution 1(6):
Solution 1(7):
Solution 1(8):
Solution 1(9):
Solution 1(10):
Solution 1(11):
Solution 1(12):
Solution 1(13):
Solution 1(14):
Solution 1(15):
Solution 1(16):
Solution 1(17):
Solution 1(18):
Exercise 81:
Solution 1(1):
Solution 1(2):
Solution 1(3):
Solution 1(4):
Solution 1(5):
Solution 1(6):
Solution 1(7):
Solution 1(8):
Solution 1(9):
Solution 1(10):
Solution 1(11):
Solution 1(12):
Leave a Reply