Factorisation of Algebraic Expressions
Class-8-Mathematics-Chapter-6-Maharashtra Board
Solutions
Practice Set 6.1
Question 1.1. Factorise.
(1) x2 + 9x + 18
x2 + 9x + 18
= x2 + 6x + 3x + 18
= x(x + 6) + 3(x + 6)
= (x + 6)(x + 3).
(2) x2 − 10x + 9
x2 − 10x + 9
= x2 – x − 9x + 9
= x(x − 1) − 9(x − 1).
= (x − 1)(x − 9).
(3) y2 + 24y + 144
y2 + 24y + 144
= y2 + 12y + 12y + 144
= y(y + 12) + 12(y + 12)
= (y + 12)(y + 12).
(4) 5y2 + 5y − 10
5y2 + 5y − 10
= 5(y2 + y − 2) ... (Taking 5 as a common factor)
= 5(y2 + 2y – y − 2)
= 5[y(y + 2) − 1(y + 2)]
= 5(y + 2) (y − 1).
(5) p2 − 2p − 35
p2 − 2p − 35
= p2 − 7p + 5p − 35
= p(p − 7) + 5(p − 7)
= (p − 7)(p + 5).
(6) p2 − 7p − 44
p2 − 7p − 44
= p2 − 11p + 4p − 44
= p(p − 11) + 4(p − 11)
= (p − 11)(p + 4).
(7) m2 − 23m + 120
| m2 − 23m + 120
= m2 − 8m − 15m + 120 = m(m − 8) − 15(m − 8) = (m − 8)(m − 15). |
120
= 4 × 30 = 4 × 2 × 15 = 8 × 15 |
(8) m2 − 25m + 100
| m2 − 25m + 100
m2 − 20m − 5m + 100 = m(m − 20) − 5(m − 20) = (m − 20) (m − 5). |
100
= (−20) × (−5)
|
(9) 3x2 + 14x + 15
| 3x2 + 14x + 15
=3x2 + 9x + 5x + 15 =3x(x+3)+5(x+3) =(x+3) (3x+5). |
15 × 3 = 45
= 5 × 9 |
(10) 2x2 + x − 45
| 2x2 + x − 45
= 2x2 + 10x − 9x − 45 = 2x(x + 5) − 9(x + 5) = (x + 5)(2x − 9). |
2 × (− 45) = − 90
= 10 × (−9) |
(11) 20x2 − 26x + 8
| 20x2 − 26x + 8
= 2(10x2 − 13x + 4) ... (Taking 2 as a common factor) = 2(10x2 − 5x − 8x + 4) = 2[5x(2x − 1) − 4(2x − 1)] = 2(2x − 1) (5x − 4). |
10 × 4 = 40 = (−5) × (−8) |
(12) 44x2 − x − 3
| 44x2 − x – 3
= 44x2 + 11x − 12x −3 = 11x(4x + 1) − 3(4x + 1) = (4x + 1) (11x − 3). |
44 × −3 = −132
= 11 × (−12) |
Practice Set 6.2
Question 2.1. Factorise.
(1) x3 + 64y3
x3 + 64y3
= (x)3 + (4y)3
= (x + 4y) [x2 − x(4y) + (4y)2]
= (x + 4y) (x2 – 4xy + 16y2).
(2) 125p3 + q3
125p3 + q3
= (5p)3 + (q)3
= (5p + q)[(5p)2 − 5p × q + (q)2]
= (5p − q)(25p2 − 5pq + q2).
(3) 125k3 + 27m3
125k3 + 27m3
= (5k)3 + (3m)3
= (5k + 3m)[(5k)2 − 5k × 3m + (3m)2]
= (5k + 3m)(25k2 − 15km + 9m2).
(4) 2l3 + 432m3
2l3 + 432m3
= 2(l3 + 216m3)
= 2[(l)3+(6m)3]
= 2(l + 6m)[(l)2 − l x 6m + (6m)2]
= 2(l + 6m) (l2 − 6lm + 36m2).
(5) 24a3 + 81b3
24a3 + 81b3
= 3(8a3 + 27b3)
= 3[(2a)3 + (3b)3]
= 3(2a + 3b) [(2a)2 − 2a × 3b + (3b)2]
= 3(2a +3b) (4a2 − 6ab + 9b2).
(6) y3 + \(\frac{1}{8y^3}\)
y3 + \(\frac{1}{8y^3}\)
= y3 + \((\frac{1}{2y})^3\)
= (y + \(\frac{1}{2y}\))[y2 – y × \(\frac{1}{2y}\) + \((\frac{1}{2y})^2\) ]
= (y +)[y2 – \(\frac{1}{2}\) + \(\frac{1}{4y^2}\)]
(7) a3 + \(\frac{8}{a^3}\)
a3 + \(\frac{8}{a^3}\)
= a3 + \((\frac{2}{a})^3\)
= (a + \(\frac{2}{a}\))[a2 – a × \(\frac{2}{a}\) + ((\frac{2}{a})^2\)]
= (a + \(\frac{2}{a}\))[a2 – 2 + \(\frac{4}{a^2}\)]
(8) 1 + \(\frac{q^3}{125}\)
1 + \(\frac{q^3}{125}\)
= 13 + \((\frac{q}{5})^3\)
= (1 + \(\frac{q}{5}\))[12 – 1 × \(\frac{q}{5}\) + \((\frac{q}{5})^2\)]
= (1 + \(\frac{q}{5}\))[12 – \(\frac{q}{5}\) + \(\frac{q^2}{25}\)]
Practice Set 6.3
Question 3.1. Factorise :
(1) y3 − 27
y3 − 27
= (y)3 − (3)3
= (y − 3)(y2 + 3y + 9).
(2) x3 − 64y3
x3 − 64y3
= (x)3 − (4y)3
= (x − 4y)(x2 + 4xy + 16y2).
(3) 27m3 − 216n3
27m3 − 216n3
= 27(m3 − 8n3)
= 27[(m)3 − (2n)3]
= 27(m − 2n) (m2 + 2mn + 4n2).
(4) 125y3 − 1
125y3 − 1
= (5y)3 − (1)3
= (5y − 1)(25y2 + 5y + 1).
(5) 8p3 − \(\frac{27}{p^3}\)
8p3 − \(\frac{27}{p^3}\)
= (2p)3 − \((\frac{3}{p})^3\)
= (2p − \(\frac{3}{p}\))[(2p)2 + 2p × \(\frac{3}{p}\)+\(\frac{9}{p^2}\)]
= (2p − \(\frac{3}{p}\))[4p2 + 6 + \(\frac{9}{p^2}\)]
(6) 343a3 − 512b3
343a3 − 512b3
= (7a)3 − (8b)3
= (7a − 8b)(49a2 + 56ab + 64b2).
(7) 64x3 − 729y3
64x3 − 729y3
= (4x)3 − (9y)3
= (4x − 9y)(16x2 + 36xy + 81y2).
(8) 16a3 − \(\frac{128}{b^3}\)
16a3 − \(\frac{128}{b^3}\)
= 16(a3 − \(\frac{8}{b^3}\))
= 16[(a)3 − \((\frac{2}{b})^3\)]
= 16(a − \(\frac{2}{b}\))[a2 + a × \(\frac{2}{b}\) + \((\frac{2}{b})^2\)]
= 16(a − \(\frac{2}{b}\))[a2 + \(\frac{2a}{b}\) + \(\frac{4}{b^2}\)]
Question 3.2. Simplify :
(1) (x + y)3 − (x − y)3
(x + y)3 − (x − y)3
= [(x + y) − (x − y)] [(x + y)2 + (x + y)(x − y) + (x − y)2]
= (x + y – x + y) (x2 + 2xy + y2 + x2 − y2 + x2 − 2xy + y2)
= 2y(3x2 + y2)
= 6x2y + 2y3.
(2) (3a + 5b)3 − (3a − 5b)3
(3a + 5b)3 −(3a − 5b)3
= [(3a + 5b) − (3a − 5b)] [(3a + 5b)2 + (3a + 5b) (3a − 5b) + (3a − 5b)2]
= (3a + 5b − 3a + 5b) (9a2 + 30ab + 25b2 + 9a2 − 25b2 + 9a2 − 30ab + 25b2)
= 10b(27a2 + 25b2)
= 270a2b + 250b3.
(3) (a + b)3 − a3 − b3
(a+b)3 − a3 − b3
= a3 + 3a2b + 3ab2 + b3 − a3 − b3
= 3a2b + 3ab2.
(4) p3 − (p + 1)3
p3 − (p + 1)3
= [p − (p + 1)][p2 + p(p + 1) + (p + 1)2]
= (p – p − 1)(p2 + p2 + p + p2 + 2p + 1)
= −1(3p2 + 3p + 1)
= − 3p2 −3p − 1.
(5) (3xy − 2ab)3 − (3xy + 2ab)3
(3xy − 2ab)3 − (3xy + 2ab)3
= [3xy − 2ab − (3xy + 2ab)] [(3xy − 2ab)2 + (3xy − 2ab)(3xy + 2ab) + (3xy + 2ab)2]
= (3xy − 2ab − 3xy − 2ab) (9x2y2 – 12xyab + 4a2b2 + 9x2y2 − 4a2b2 + 9x2y2 + 12xyab + 4a2b2)
= − 4ab (27x2y2 +4a2b2)
= − 108x2y2ab − 16a3b3.
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