6. Factorisation of Algebraic Expressions
Solutions to Practice Sets 6.1, 6.2, 6.3 and 6.4, with every step shown. Maharashtra State Board (Balbharati), Standard Eight Mathematics.
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Practice Set 6.1
Method: multiply the coefficient of the square term by the constant term. Find two factors of this product whose sum is the coefficient of the middle term. Split the middle term using these two numbers, then group in pairs.
(1) x2 + 9x + 18
x2 + 9x + 18
Product = 18; (6) × (3) = 18 and (6) + (3) = 9
= x2 + 6x + 3x + 18
= x(x + 6) + 3(x + 6)
Answer: (x + 6)(x + 3)
(2) x2 − 10x + 9
x2 − 10x + 9
Product = 9; (−9) × (−1) = 9 and (−9) + (−1) = −10
= x2 − 9x − x + 9
= x(x − 9) − 1(x − 9)
Answer: (x − 9)(x − 1)
(3) y2 + 24y + 144
y2 + 24y + 144
Product = 144; (12) × (12) = 144 and (12) + (12) = 24
= y2 + 12y + 12y + 144
= y(y + 12) + 12(y + 12)
This can also be written as (y + 12)².
Answer: (y + 12)(y + 12)
(4) 5y2 + 5y − 10
5y2 + 5y − 10
Take out the common factor 5 first.
= 5(y2 + y − 2)
Product = −2; (2) × (−1) = −2 and (2) + (−1) = 1
= 5(y2 + 2y − y − 2)
= 5[y(y + 2) − 1(y + 2)]
Answer: 5(y + 2)(y − 1)
(5) p2 − 2p − 35
p2 − 2p − 35
Product = −35; (−7) × (5) = −35 and (−7) + (5) = −2
= p2 − 7p + 5p − 35
= p(p − 7) + 5(p − 7)
Answer: (p − 7)(p + 5)
(6) p2 − 7p − 44
p2 − 7p − 44
Product = −44; (−11) × (4) = −44 and (−11) + (4) = −7
= p2 − 11p + 4p − 44
= p(p − 11) + 4(p − 11)
Answer: (p − 11)(p + 4)
(7) m2 − 23m + 120
m2 − 23m + 120
Product = 120; (−15) × (−8) = 120 and (−15) + (−8) = −23
= m2 − 15m − 8m + 120
= m(m − 15) − 8(m − 15)
Answer: (m − 15)(m − 8)
(8) m2 − 25m + 100
m2 − 25m + 100
Product = 100; (−20) × (−5) = 100 and (−20) + (−5) = −25
= m2 − 20m − 5m + 100
= m(m − 20) − 5(m − 20)
Answer: (m − 20)(m − 5)
(9) 3x2 + 14x + 15
3x2 + 14x + 15
Product = 3 × 15 = 45; (9) × (5) = 45 and (9) + (5) = 14
= 3x2 + 9x + 5x + 15
= 3x(x + 3) + 5(x + 3)
Answer: (x + 3)(3x + 5)
(10) 2x2 + x − 45
2x2 + x − 45
Product = 2 × (−45) = −90; (10) × (−9) = −90 and (10) + (−9) = 1
= 2x2 + 10x − 9x − 45
= 2x(x + 5) − 9(x + 5)
Answer: (x + 5)(2x − 9)
(11) 20x2 − 26x + 8
20x2 − 26x + 8
Take out the common factor 2 first.
= 2(10x2 − 13x + 4)
Product = 10 × 4 = 40; (−8) × (−5) = 40 and (−8) + (−5) = −13
= 2(10x2 − 8x − 5x + 4)
= 2[2x(5x − 4) − 1(5x − 4)]
Answer: 2(5x − 4)(2x − 1)
(12) 44x2 − x − 3
44x2 − x − 3
Product = 44 × (−3) = −132; (−12) × (11) = −132 and (−12) + (11) = −1
= 44x2 − 12x + 11x − 3
= 4x(11x − 3) + 1(11x − 3)
Answer: (11x − 3)(4x + 1)
Practice Set 6.2
Formula: a3 + b3 = (a + b)(a2 − ab + b2)
(1) x3 + 64y3
x3 + 64y3
= x3 + (4y)3
= (x + 4y)[x2 − x(4y) + (4y)2]
Answer: (x + 4y)(x2 − 4xy + 16y2)
(2) 125p3 + q3
125p3 + q3
= (5p)3 + q3
= (5p + q)[(5p)2 − 5p × q + q2]
Answer: (5p + q)(25p2 − 5pq + q2)
(3) 125k3 + 27m3
125k3 + 27m3
= (5k)3 + (3m)3
= (5k + 3m)[(5k)2 − 5k × 3m + (3m)2]
Answer: (5k + 3m)(25k2 − 15km + 9m2)
(4) 2l3 + 432m3
2l3 + 432m3
= 2(l3 + 216m3)
= 2[l3 + (6m)3]
= 2(l + 6m)[l2 − l × 6m + (6m)2]
Answer: 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]
Answer: 3(2a + 3b)(4a2 − 6ab + 9b2)
(6) y3 + 18y3
y3 + 18y3
= y3 + (12y)3
= (y + 12y)[y2 − y × 12y + (12y)2]
Answer: (y + 12y)(y2 − 12 + 14y2)
(7) a3 + 8a3
a3 + 8a3
= a3 + (2a)3
= (a + 2a)[a2 − a × 2a + (2a)2]
Answer: (a + 2a)(a2 − 2 + 4a2)
(8) 1 + q3125
1 + q3125
= (1)3 + (q5)3
= (1 + q5)[12 − 1 × q5 + (q5)2]
Answer: (1 + q5)(1 − q5 + q225)
In (4) and (5), the numbers are not perfect cubes, so first take out the common factor (2 and 3) to get perfect cubes.
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Practice Set 6.3
Formula: a3 − b3 = (a − b)(a2 + ab + b2)
(1) y3 − 27
y3 − 27
= y3 − (3)3
= (y − 3)[y2 + y × 3 + (3)2]
Answer: (y − 3)(y2 + 3y + 9)
(2) x3 − 64y3
x3 − 64y3
= x3 − (4y)3
= (x − 4y)[x2 + x × 4y + (4y)2]
Answer: (x − 4y)(x2 + 4xy + 16y2)
(3) 27m3 − 216n3
27m3 − 216n3
= 27(m3 − 8n3)
= 27[m3 − (2n)3]
= 27(m − 2n)[m2 + m × 2n + (2n)2]
Answer: 27(m − 2n)(m2 + 2mn + 4n2)
(4) 125y3 − 1
125y3 − 1
= (5y)3 − (1)3
= (5y − 1)[(5y)2 + 5y × 1 + 12]
Answer: (5y − 1)(25y2 + 5y + 1)
(5) 8p3 − 27p3
8p3 − 27p3
= (2p)3 − (3p)3
= (2p − 3p)[(2p)2 + 2p × 3p + (3p)2]
Answer: (2p − 3p)(4p2 + 6 + 9p2)
(6) 343a3 − 512b3
343a3 − 512b3
= (7a)3 − (8b)3
= (7a − 8b)[(7a)2 + 7a × 8b + (8b)2]
Answer: (7a − 8b)(49a2 + 56ab + 64b2)
(7) 64x3 − 729y3
64x3 − 729y3
= (4x)3 − (9y)3
= (4x − 9y)[(4x)2 + 4x × 9y + (9y)2]
Answer: (4x − 9y)(16x2 + 36xy + 81y2)
(8) 16a3 − 128b3
16a3 − 128b3
= 16(a3 − 8b3)
= 16[a3 − (2b)3]
= 16(a − 2b)[a2 + a × 2b + (2b)2]
Answer: 16(a − 2b)(a2 + 2ab + 4b2)
(1) (x + y)3 − (x − y)3
(x + y)3 − (x − y)3
Take a = x + y and b = x − y, and use a³ − b³ = (a − b)(a² + ab + b²).
= [(x + y) − (x − y)][(x + y)2 + (x + y)(x − y) + (x − y)2]
= [2y][x2 + 2xy + y2 + x2 − y2 + x2 − 2xy + y2]
= 2y(3x2 + y2)
Answer: 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]
= [10b][9a2 + 30ab + 25b2 + 9a2 − 25b2 + 9a2 − 30ab + 25b2]
= 10b(27a2 + 25b2)
Answer: 270a2b + 250b3
(3) (a + b)3 − a3 − b3
(a + b)3 − a3 − b3
= a3 + 3a2b + 3ab2 + b3 − a3 − b3
Answer: 3a2b + 3ab2
(4) p3 − (p + 1)3
p3 − (p + 1)3
= p3 − (p3 + 3p2 + 3p + 1)
= p3 − p3 − 3p2 − 3p − 1
Answer: −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]
= [−4ab][9x2y2 − 12xyab + 4a2b2 + 9x2y2 − 4a2b2 + 9x2y2 + 12xyab + 4a2b2]
= −4ab(27x2y2 + 4a2b2)
Answer: −108x2y2ab − 16a3b3
Practice Set 6.4
Method: factorise every numerator and denominator, then cancel the factors common to a numerator and a denominator. To divide, multiply by the reciprocal of the second expression.
(1) m2 − n2(m + n)2 × m2 + mn + n2m3 − n3
m2 − n2(m + n)2 × m2 + mn + n2m3 − n3
= (m + n)(m − n)(m + n)(m + n) × m2 + mn + n2(m − n)(m2 + mn + n2)
Answer: 1m + n
(2) a2 + 10a + 21a2 + 6a − 7 × a2 − 1a + 3
a2 + 10a + 21a2 + 6a − 7 × a2 − 1a + 3
= (a + 7)(a + 3)(a + 7)(a − 1) × (a + 1)(a − 1)a + 3
Answer: a + 1
(3) 8x3 − 27y34x2 − 9y2
8x3 − 27y34x2 − 9y2
= (2x)3 − (3y)3(2x)2 − (3y)2
= (2x − 3y)(4x2 + 6xy + 9y2)(2x + 3y)(2x − 3y)
Answer: 4x2 + 6xy + 9y22x + 3y
(4) x2 − 5x − 24(x + 3)(x + 8) × x2 − 64(x − 8)2
x2 − 5x − 24(x + 3)(x + 8) × x2 − 64(x − 8)2
= (x − 8)(x + 3)(x + 3)(x + 8) × (x + 8)(x − 8)(x − 8)(x − 8)
Answer: 1
(5) 3x2 − x − 2x2 − 7x + 12 ÷ 3x2 − 7x − 6x2 − 4
3x2 − x − 2x2 − 7x + 12 ÷ 3x2 − 7x − 6x2 − 4
= 3x2 − x − 2x2 − 7x + 12 × x2 − 43x2 − 7x − 6
= (3x + 2)(x − 1)(x − 3)(x − 4) × (x + 2)(x − 2)(3x + 2)(x − 3)
Answer: (x − 1)(x − 2)(x + 2)(x − 3)2(x − 4)
(6) 4x2 − 11x + 616x2 − 9
4x2 − 11x + 616x2 − 9
= 4x2 − 8x − 3x + 6(4x)2 − (3)2
= (4x − 3)(x − 2)(4x + 3)(4x − 3)
Answer: x − 24x + 3
(7) a3 − 275a2 − 16a + 3 ÷ a2 + 3a + 925a2 − 1
a3 − 275a2 − 16a + 3 ÷ a2 + 3a + 925a2 − 1
= a3 − 275a2 − 16a + 3 × 25a2 − 1a2 + 3a + 9
= (a − 3)(a2 + 3a + 9)(5a − 1)(a − 3) × (5a + 1)(5a − 1)a2 + 3a + 9
Answer: 5a + 1
(8) 1 − 2x + x21 − x3 × 1 + x + x21 + x
1 − 2x + x21 − x3 × 1 + x + x21 + x
= (1 − x)2(1 − x)(1 + x + x2) × 1 + x + x21 + x
= (1 − x)(1 − x)(1 − x)(1 + x + x2) × 1 + x + x21 + x
Answer: 1 − x1 + x
Factors used: in (5), 3x² − x − 2 = (3x + 2)(x − 1) and 3x² − 7x − 6 = (3x + 2)(x − 3); in (7), 5a² − 16a + 3 = (5a − 1)(a − 3). Each is found by splitting the middle term as in Practice Set 6.1.