DAV Class 8 Maths Solution Chapter 7 Algebraic Identities
DAV School Books Class 8 Maths Solution Chapter 7 Algebraic Identities all Question Answer.
DAV Class 8 7th chapter Algebraic Identities full Chapter explanation is provided by expert teacher. DAV class 8 maths chapter wise solution can be found here.
DAV School Books Class 8 Maths Solution Chapter 7 Algebraic Identities:
Worksheet 1
1) Find the following by using Identity 1: (a+b)2 = a2+ 2ab + b2
i) (2x+5)2
ii) (8x+3y)2
iii) [(3a/5)+(2b/3)]2
iv) (7pq+4ab)2
v) (0.2x+1.5y)2
vi) (2m2+3n2)2
Ans:
i) (2x+5)2
= (2x)2+ 2(2x)(5) + 52
= 4x2+20x+25
ii) (8x+3y)2
= (8x)2+ 2(8x)(3y) + (3y)2
= 64x2+48xy+9y2
iii) [(3a/5)+(2b/3)]2
(3a/5)2+ 2(3a/5)(2b/3) + (2b/3)2
= (9a2/25) + (12ab/15) + (4b2/9)
iv) (7pq+4ab)2
= (7pq)2+ 2(7pq)(4ab) + (4ab)2
= 49a2b2+56pqab+16a2b2
v) (0.2x+1.5y)2
= (0.2x)2+ 2(0.2x)(1.5y) + (1.5y)2
= 0.04x2+0.6xy+2.25y2
vi) (2m2+3n2)2
= (2m2)2+ 2(2m2)(3n2) + (3n2)2
= 4m4+12m2n2+9n4
2) Evaluate the following by using Identity 1: (a+b)2 = a2+ 2ab + b2
i) 1012
ii) 522
iii) 8.12
iv) 2032
v) 4102
vi) 10.22
Ans:
i) 1012
= (100+1)2
= 1002+2×100×1+12
= 10000+200+1
= 10201
ii) 522
= (50+2)2
= 502+2×50×2+22
= 2500+200+4
= 2704
iii) 8.12
= (8+0.1)2
= 82+2×8×0.1+ (0.1)2
= 64+1.6+0.01
= 65.61
iv) 2032
= (200+3)2
= 2002+2×200×3+32
= 40000+1200+9
= 41209
v) 4102
= (400+10)2
= 4002+2×400×10+102
= 160000+8000+100
= 168100
vi) 10.22
= (10+0.2)2
= 102+2×10×0.2+0.22
= 100+4+0.04
= 104.04
Worksheet 2
1) Find the following by using Identity 2: (a-b)2 = a2– 2ab + b2
i) (x – 7)2
ii) (5a – 4b)2
iii) [(7x/4) – (2y/3)]2
iv) (8mn – 3pq)2
v) (0.1x – 0.5y)2
vi) (a2 – b2)2
Ans:
i) (x – 7)2
= x2–2(x)(7)+72
= x2-14x+49
ii) (5a – 4b)2
= (5a)2–2(5a)(4b)+(4b)2
= 25a2– 40ab+16b2
iii) [(7x/4) – (2y/3)]2
= (7x/4)2–2(7x/4)(2y/3)+(2y/3)2
= (49x2/16) – (28xy/12) + (4y2/9)
iv) (8mn – 3pq)2
= (8mn)2–2(8mn)(3pq)+(3pq)2
= 64m2n2– 48mnpq+9p2q2
v) (0.1x – 0.5y)2
= (0.1x)2–2(0.1x)(0.5y)+(0.5y)2
= 0.01x2– 0.1xy+0.25y2
vi) (a2 – b2)2
= (a2)2–2(a2)(b2)+(b2)2
= a4– 2a2b2 + b4
2) Evaluate the following by using Identity 2: (a-b)2 = a2– 2ab + b2
i) 482
ii) 9.92
iii) 2992
iv) 982
v) 872
vi) 19.92
Ans:
i) 482
= (50–2)2
= 502 – 2×50×2+22
= 2500–200+4
= 2304
ii) 9.92
= (10–0.1)2
= 102 – 2×10×0.1+0.12
= 100–2+0.01
= 98.01
iii) 2992
= (300–1)2
= 3002 – 2×300×1+12
= 90000–600+1
= 89401
iv) 982
= (100–2)2
= 1002–2×100×2+22
= 10000–400+4
= 9604
v) 872
= (90–3)2
= 902 – 2×90×3+32
= 8100–540+9
= 7569
vi) 19.92
= (20–0.1)2
= 202 – 2×20×0.1+0.12
= 400–4+0.01
= 396.01
Worksheet 3
1) Find the following products using Identity 3: (a+b)(a-b) = a2-b2
i) (2x + 7y) × (2x-7y)
ii) (5ab + 8c) × (5ab-8c)
iii) (4p2 + q2) × (4p2-q2)
iv) [(x/3) + (y/2)] × [(x/3) × (y/2)]
v) (0.1m + 0.2n) × (0.1m-0.2n)
vi) (a3 + b3) × (a3-b3)
Ans :
i) (2x + 7y) × (2x-7y)
= (2x)2–(7y)2
= 4x2 – 49y2
ii) (5ab + 8c) × (5ab-8c)
= (5ab)2–(8c)2
= 25a2b2 – 64c2
iii) (4p2 + q2) × (4p2-q2)
= (4p)2–(q)2
= 16p2 – q2
iv) [(x/3) + (y/2)] × [(x/3) × (y/2)]
= (x/3)2–(y/2)2
= (x2/9) – (y2/4)
v) (0.1m + 0.2n) × (0.1m-0.2n)
= (0.1m)2–(0.2n)2
= 0.01m2 – 0.04n2
vi) (a3 + b3) × (a3-b3)
= (a3)2–(b3)2
= a6 – b6
2) Evaluate the following using Identity 3: (a+b)(a-b) = a2-b2
i) 812 – 192
ii) 2902 – 2102
iii) 582 – 122
iv) 1762 – 242
v) 3672 – 332
vi) 5452 – 4452
Ans:
i) 812 – 192
= (81+19) × (81–19)
= 100×62
= 6200
ii) 2902 – 2102
= (290+210) × (290–210)
= 100×62
= 6200
iii) 582 – 122
= (58+12) × (58–12)
= 70×46
= 3200
iv) 1762 – 242
= (176+24) × (176–24)
= 200×152
= 30400
v) 3672 – 332
= (367+33) × (367–33)
= 400×334
= 133600
vi) 5452 – 4452
= (545+445) × (545–445)
= 990×100
= 99000
3) Simplify the following products by expressing these as difference of two squares:
i) 107 × 93
ii) 211 × 289
iii) 195 × 205
iv) 308 × 292
v) 12.4 × 11.6
vi) 30.9 × 29.1
Ans:
i) 107 × 93
= (100+7) × (100–7)
= 1002–72
= 10000–49
= 9951
ii) 211 × 189
= (2001) × (200–11)
= 2002–112
= 40000–121
= 39879
iii) 205 × 195
= (200+5) × (100–5)
= 2002–52
= 40000–25
= 39975
iv) 308 × 292
= (300+8) × (300–8)
= 3002–82
= 90000–64
= 89936
v) 12.4 × 11.6
= (12+0.4) × (12–0.4)
= 122–0.42
= 144–0.16
= 143.84
vi) 30.9 × 29.1
= (30+0.9) × (30–0.9)
= 302–0.92
= 900–0.81
= 899.19
Worksheet 4
Expand the following:
1) (x–2y+3z)2
2) (–5x+2y+z)2
3) (4x+y–3z)2
4) (3a–5b–7c)2
5) (–2a–b+3c)2
6) (–a+6b–2c)2
7) (1+2x–3y)2
8) (2x–4y–1)2
9) (6p–5q–4r)2
10) (p+5q+2)2
11) (3x–4y–5)2
12) (–2x+6y+4)2
Ans:
1) (x–2y+3z)2
=[x+(–2y)+3z]2
= x2+(2y)2+(3z)2+2(x)(-2y)+2(–2y)(3z)+2(3z)(x)
= x2+4y2+9z2–4xy–12yz+6zx
2) (–5x+2y+z)2
=[(–5x)+2y+z]2
= (–5x)2+(2y)2+(z)2+2(–5x)(2y)+2(2y)(z)+2(z)(–2x)
= 25x2+4y2+z2–20xy+4yz–4zx
3) (4x+y–3z)2
=[4x+y+(–3z)]2
= (4x)2+(y)2+(–3z)2+2(4x)(y)+2(y)( –3z)+2(–3z)(4x)
= 16x2+y2+9z2+8xy–6yz–24zx
4) (3a–5b–7c)2
=[3a+(–5b)+( –7c)]2
= (3a)2+(–5b)2+(–7c)2+2(3a)( –5b)+2(–5b)( –7c)+2(–7c)(3a)
= 9a2+25b2+49c2–30ab+70bc–42ca
5) (–2a–b+3c)2
=[(–2a)+(–b)+(c)]2
= (–2a)2+(–b)2+(c)2+2(–2a)(–b)+2(–b)(c)+2(c)(–2a)
= 4a2+b2+c2+4ab–2bc–4ca
6) (–a+6b–2c)2
= [(–a)+(6b)+(–2c)]2
= (–a)2+(6b)2+(–2c)2+2(–a)(6b)+2(6b)(–2c)+2(–2c)(–a)
= a2+36b2+4c2–12ab–24bc+4ca
7) (1+2x–3y)2
= [(1)+(2x)+(–3y)]2
= (1)2+(2x)2+(–3y)2+2(1)(2x)+2(2x)(–3y)+2(–3y)(1)
= 1+4x2+9y2+4x–12xy–6y
8) (2x–4y–1)2
= [(2x)+(–4y)+(–1)]2
= (2x)2+(–4y)2+(–1)2+2(2x)( –4y)+2(–4y)(–1)+2(–1)(2x)
= 4x2+16y2+1–16xy+8y–4x
9) (6p–5q–4r)2
=[6p+(–5q)+( –4r)]2
= (6p)2+(–5q)2+(–4r)2+2(6p)( –5q)+2(–5q)( –4r)+2(–4r)(6p)
= 36p2+25q2+16r2–60pq+40qr–48rq
10) (p+5q+2)2
= (p)2+(5q)2+(2)2+2(p)( 5q)+2(5q)( 2)+2(2)(p)
= p2+25q2+4+10pq+20q+4p
11) (3x–4y–5)2
= [(3x)+(–4y)+(–5)]2
= (3x)2+(–4y)2+(–5)2+2(3x)( –4y)+2(–4y)(–5)+2(–5)(3x)
= 9x2+16y2+25–24xy+40y–30x
12) (–2x+6y+4)2
= [(–2x)+(6y)+(4)]2
= (–2x)2+(6y)2+(4)2+2(–2x)( 6y)+2(6y)(4)+2(4)( –2x)
= 4x2+36y2+16–24xy+48y–16x
Worksheet 5
1) Find the product by using suitable identity:
i) (x+5) × (x+4)
ii) (a+3) × (a+6)
iii) (x – 9) × (x+7)
iv) (x+8) × (x – 7)
v) (z – 3) × (z – 1)
vi) (p – 5) × (p – 4)
vii) (y – 1) × (y+2)
viii) (z+3) × (z – 7)
ix) (p+8) × (p – 3)
x) (z+6) × (z – 5)
xi) (x – 6) × (x – 9)
xii) (x – 10) × (x+9)
xiii) (y – 4) × (y+4)
xiv) (x – 4) × (x – 14)
xv) (x – 8) × (x – 2)
Ans:
i) (x+5) × (x+4)
= x2+ (5+4)x + 20
= x2+9x+20
ii) (a+3) × (a+6)
= a2+ (3+6)a + 18
= a2+9a+18
iii) (x – 9) × (x+7)
= [x+ (–9)] × (x+7)
= x2+ (–9+7)x + (–9) ×7
= x2–2x–63
iv) (x+8) × (x–7)
= (x+8) × [x+ (–7)]
= x2+ (8–7)x + 8(–7)
= x2+x–56
v) (z – 3) × (z – 1)
= [z+(–3)] × [z+ (–1)]
= z2+ (–3–1)z + (–3)(–1)
= z2–4z+3
vi) (p – 5) × (p – 4)
= [p+ (–5)] × [p+ (–4)]
= p2+ (–5–4)p + (–5)(–4)
= p2–9p+20
vii) (y – 1) × (y+2)
= [y+ (–1)] × [y+ (2)]
= y2+ (–1+2)y + (–1)(2)
= y2+y–2
viii) (z+3) × (z – 7)
= [z+(3)] × [z+ (–7)]
= z2+ (3–7)z + (3)(–7)
= z2–4z–21
ix) (p+8) × (p – 3)
= [p+ (8)] × [p+ (–3)]
= p2+ (8–3)p + (8)(–3)
= p2+5p–24
x) (z+6) × (z – 5)
= [z+(6)] × [z+ (–5)]
= z2+ (6–5)z + (6)(–5)
= z2+z–30
xi) (x – 6) × (x – 9)
= [x+ (–6)] × [x+ (–9)]
= x2+ (–6–9)x + (–6)(–9)
= x2–15x–54
xii) (x – 10) × (x+9)
= [x+ (–10)] × [x+ (9)]
= x2+ (–10+9)x + (–10)(9)
= x2–x–90
xiii) (y – 4) × (y+4)
= [y+ (–4)] × [y+ (4)]
= y2+ (–4+4) y + (–4)(4)
= y2–16
xiv) (x – 4) × (x – 14)
= [x+ (–4)] × [x+ (–14)]
= x2+ (–4–14)x + (–4)( –14)
= x2–18x+56
xv) (x – 8) × (x – 2)
= [x+ (–8)] × [x+ (–2)]
= x2+ (–8–2)x + (–8)( –2)
= x2–10x–16
2) By using a suitable identity, evaluate the following:
i) 102 × 104
ii) 105 × 103
iii) 206 × 205
iv) 98 × 96
v) 87 × 85
vi) 104 × 95
vii) 97 × 102
viii) 203 × 198
ix) 35 × 37
x) 106 × 93
Ans:
i) 102 × 104
= (100+2) × (100+4)
= 1002+ (2+4)100+8
= 10000+600+8
= 10608
ii) 105 × 103
= (100+5) × (100+3)
= 1002+ (5+3)100+15
= 10000+800+15
= 10815
iii) 206 × 205
= (200+6) × (200+5)
= 2002+ (6+5)200+30
= 40000+2200+30
= 42230
iv) 98 × 96
= (100–2) × (100–4)
= 1002+ (–2–4)100+ (–2)( –4)
= 10000–600+8
= 9408
v) 87 × 85
= (100–13) × (100–15)
= 1002+ (–13–15)100+ (–13)( –15)
= 10000–2800+195
= 7395
vi) 104 × 95
= (100+4) × (100–5)
= 1002+ (4–5)100+ (4)( –5)
= 10000–100–20
= 9880
vii) 97 × 102
= (100–3) × (100+2)
= 1002+ (–3+2)100+ (–3)( 2)
= 10000–100–6
= 9894
viii) 203 × 198
= (200+3) × (200–2)
= 2002+ (3–2)200+ (3)(–2)
= 40000+200–6
= 40194
ix) 35 × 37
= (40–5) × (40–3)
= 402+ (–5–3)40+ (–5)(–3)
= 1600–320+15
= 1295
x) 106 × 93
= (100+6) × (100–7)
= 1002+ (6–7)100+ (6)(–7)
= 10000–100–42
= 9858
3) Evaluate the following products:
i) (x2+3) × (x2+4)
ii) [x+(4/3)] × [x+(1/3)]
iii) [x– (3/5)] × [x– (1/2)]
iv) (y2–6) × (y2+7)
v) (z2+4) × [z2–(1/4)]
vi) (y2–3) × (y2–1)
vii) (x3+5) × (x3+2)
viii) [p2–(1/4)] × [p2+(1/8)]
ix) [z+ (1/6)] × (z+6)
Ans:
i) (x2+3) × (x2+4)
= [x2+ (3)] × [x2+ (4)]
= (x2)2+ (3+4)x2 + (3)(4)
= x4+7x2+12
ii) [x+(4/3)] × [x+(1/3)]
= [x+ (4/3)] × [x+ (1/3)]
= x2+ [(4/3) + (1/3)]x + (4/3)(1/3)
= x2+ (5/3)x + (4/9)
iii) [x– (3/5)] × [x– (1/2)]
= [x+ (–3/5)] × [x+ (–1/2)]
= x2+ [(–3/5) + (–1/2)]x + (–3/5)( –1/2)
= x2– (11/10)x + (3/10)
iv) (y2–6) × (y2+7)
= [y2+ (–6)] × [y2+ (7)]
= (y2)2+ (–6+7)y2 + (–6)(-7)
= y4+y2+42
v) (z2+4) × [z2–(1/4)]
= [z2+ (4)] × [z2+ (–1/4)]
= (z2)2+ [(4) – (1/4)]z2 + (4)(–1/4)
= z4+ (15/4)z2–1
vi) (y2–3) × (y2–1)
= [y2+ (–3)] × [y2+ (–1)]
= (y2)2+ (–3–1)y2 + (–3)( –1)
= y4–4y2+3
vii) (x3+5) × (x3+2)
= [x3+ (5)] × [x3+ (2)]
= (x3)2+ (5+2)x3 + (5)(2)
= x6+7x3+10
viii) [p2–(1/4)] × [p2+(1/8)]
= [p2+ (–1/4)] × [p2+ (1/8)]
= (p2)2+ [(–1/4) + (1/8)]p2 + (–1/4)( 1/8)
= p4– (1/8)p2 – (1/32)
ix) [z+ (1/6)] × (z+6)
= [z+ (1/6)] × [z+ (6)]
= (z)2+ [(1/6) + 6]z2 + (1/6)( 6)
= z2+ (37/6)z2 + 1
Worksheet 6
Factorise:
1) x2+z2–2xz
2) 4x2 +9y2 –12xy
3) 64a2 +49b2 + 112ab
4) 121p2 + 16q2 – 88pq
5) 9x2y – 24xy2 + 16y3
6) 2a3 + 4a2b + 2ab2
7) 50a2 + 98b2 + 140ab
8) 64x2 – 81y2
9) 25p2 – 9q2
10) 16a2b – 64b3
11) 25x3y3 – 49xy
12) p4 – 256
13) a2 – (b–c)2
14) 25m2–(4n+3l)2
15) (2a+3b)2 – 4c2
16) (64m2–144mn+81n2)
17) 16x2+9y2+4z2+24xy+12yz+16zx
18) x2+4y2+9z2–4xy+12yz–6zx
19) 4a2+b2+25c2–4ab–10bc+20ca
20) a2+ (b2/4) + (c2/9) +ab+ (bc/3) + (2/3)ca
Ans:
1) x2+z2–2xz
= X2–2xz+z2
= (x–z)2
= (x–z)(x–z)
2) 4x2 +9y2 –12xy
= (2x)2 –2(2x)(3y)+(3y)2
= (2x–3y)2
= (2x–3y)(2x–3y)
3) 64a2 +49b2 + 112ab
= (8a)2 –2(8a)(7b)+(7b)2
= (8a–7b)2
= (8a–7b)(8a–7b)
4) 121p2 + 16q2 – 88pq
= (11p)2 –2(11p)(4q)+(4q)2
= (11p–4q)2
= (11p–4q)(11p–4q)
5) 9x2y – 24xy2 + 16y3
= y[9x2–24xy+16y2]
= y[(3x)2–2(3x)(4y)+(4y)2]
= y[3x–4y]2
= y(3x–4y)(3x–4y)
6) 2a3 + 4a2b + 2ab2
= 2a[a2+2ab+b2]
= 2a[(a)2 + 2(a)(b) +(b)2]
= 2a[a+b]2
= 2a(a+b)(a+b)
7) 50a2 + 98b2 + 140ab
= 2[25a2+49b2+ 70ab]
= 2[(5a)2+2(5a)(7b)+(7b)2]
= 2(5a+7b)2
= 2(5a+7b)(5a+7b)
8) 64x2 – 81y2
= (8x)2–(9y)2
= (8x+9y)(8x–9y)
9) 25p2 – 9q2
= (5p)2–(3q)2
= (5p+3q)(5p–3q)
10) 16a2b – 64b3
= b[16a2–64b2]
= b[(4a)2– (8b)2]
= b(4a+8b)(4a–8b)
11) 25x3y3 – 49xy
= xy[25x2y2–49]
= xy[(5xy)2 –72]
= xy(5xy+7)(5xy–7)
12) p4 – 256
= (p2)2–162
= (p2+16)(p2–16)
= (p2+16)(p+4)(p–4)
13) a2 – (b–c)2
= [a+(b–c)] × [a–(b–c)]
= (a+b–c) × (a–b+c)
14) 25m2– (4n+3l)2
= (5m)2 – (4n+3l)2
= (5m+4n+3l) × [5m– (4n+3l)]
= (5m+4n+3l) × (5m–4n–3l)
15) (2a+3b)2 – 4c2
= (2a+3b)2–(2c)2
= (2a+3b+2c) × (2a+3b–2c)
16) (64m2–144mn+81n2)
= (8m)2–2(8m)(9n) + (9n)2
= (8m–9n)2
= (8m–9n)(8m–9n)
17) 16x2+9y2+4z2+24xy+12yz+16zx
= (4x)2+(3y)2+(2z)2+2(4x)(3y) +2(3y)(2z) +2(2z)(4x)
= (4x+3y+2z)2
= (4x+3y+2z) × (4x+3y+2z)
18) x2+4y2+9z2–4xy+12yz–6zx
= (–x)2+ (2y)2+(3z)2+ 2(–x)(2y) +2(2y)(3z) +2(2z)(–x)
= [–x+2y+3z]2
= (–x+2y+3z) × (–x+2y+3z)
19) 4a2+b2+25c2–4ab–10bc+20ca
= (2a)2+(–b)2+(5c)2+ 2(2a)( –b) + 2(–b)(5c) +2(5c)(2a)
= (2a–b+5c)2
= (2a–b+5c) × (2a–b+5c)
20) a2+ (b2/4) + (c2/9) +ab+ (bc/3) + (2/3)ca
= a2+ (b/2)2+ (c/3)2 + 2(a)(b/2) + 2(b/2)(c/3) + 2(c/3)(a)
= [a+ (b/2) + (c/3)]2
= [a+ (b/2) + (c/3)] × [a+ (b/2) + (c/3)]
Worksheet 7
Factorise the following expressions:
1) x2+14x+33
2) y2 – y – 6
3) x2 – x – 72
4) x2+12x+27
5) y2+y – 132
6) x2+11x+30
7) x2–11x–42
8) z2–12z+27
9) p2-5p-6
10) P2+P-56
11) x2-8x-65
12) x2+8x+15
13) x2+2x-24
14) x2-3x-54
15) a2-7a+12
16) p2-3pq+2q2
Ans:
1) x2+14x+33
=> To find two numbers whose product is 33 and sum is 14
As the product is positive so both the numbers are either positive or negative.
Since the sum is positive we can say that both the numbers are positive.
Required factors are 11 and 3
∴ x2+14x+33
= x2+ (11+3)x+ (11)(3)
= x2+11x+3x+(11)(3)
= x(x+11) + 3(x+11)
= (x+11)(x+3)
2) y2 – y – 6
=> To find two numbers whose product is (-6) and sum is (-1)
As the product is negative so one number must be negative and the other number is positive.
Since the sum is negative we can say that of the two numerically greater number is negative.
Required factors are (-3) and 2
∴ y2–y–6
= y2– (3–2)y+ (2)( –3)
= y2–3y+2y– (2)(3)
= y(y–3) + 2(y–3)
= (y–3)(y+2)
3) x2 – x – 72
=> To find two numbers whose product is (-72) and sum is (-1)
As the product is negative so one number must be negative and the other number is positive.
Since the sum is negative we can say that of the two numerically greater number is negative.
Required factors are (–9) and 8
∴ x2–x–72
= x2– (9–8)x+ (–9)(8)
= x2–9x+8x–(9)(8)
= x(x–9) + 8(x–9)
= (x–9)(x+8)
4) x2+12x+27
=> To find two numbers whose product is 27 and sum is 12
As the product is positive so both the numbers are either positive or negative.
Since the sum is positive we can say that both the numbers are positive.
Required factors are 9 and 3
∴ x2+12x+27
= x2+ (9+3)x+ (9)(3)
= x2+9x+3x+(9)(3)
= x(x+9) + 3(x+9)
= (x+9)(x+3)
5) y2+y–132
=> To find two numbers whose product is (-132) and sum is (1)
As the product is negative so one number must be negative and the other number is positive.
Since the sum is positive we can say that of the two numerically greater number is positive.
Required factors are (12) and (-11)
∴ y2+y–132
= y2+ (12–11)y+ (12)( –11)
= y2+12y–11y– (12)(11)
= y(y+12) –11(y+12)
= (y+12)(y–11)
6) x2+11x+30
=> To find two numbers whose product is 30 and sum is 11.
As the product is positive so both the numbers are either positive or negative.
Since the sum is positive we can say that both the numbers are positive.
Required factors are 6 and 5
∴ x2+11x+30
= x2+ (6+5)x+ (6)(5)
= x2+6x+5x+(6)(5)
= x(x+6) + 5(x+6)
= (x+6)(x+5)
7) x2–11x–42
=> To find two numbers whose product is (-42) and sum is (-11)
As the product is negative so one number must be negative and the other number is positive.
Since the sum is negative we can say that of the two numerically greater number is negative.
Required factors are (–14) and 3
∴ x2–11x–42
= x2– (14–3)x+ (–14)(3)
= x2–14x+3x+(–14)(3)
= x(x–14) + 3(x–14)
= (x–14)(x+3)
8) z2–12z+27
=> To find two numbers whose product is 27 and sum is (-12).
As the product is positive so both the numbers are either positive or negative.
Since the sum is negative we can say that both the numbers are negative.
Required factors are (-9) and (-3)
∴ z2–12z+27
= z2– (9+3) z+ (–9)( –3)
= z2–9z–3z+ (–9)( –3)
= z(z–9) – 3(z–9)
= (z–9)(z–3)
9) p2-5p-6
=> To find two numbers whose product is (-6) and sum is (-5)
As the product is negative so one number must be negative and the other number is positive.
Since the sum is negative we can say that of the two numerically greater number is negative.
Required factors are (–6) and 1
∴ p2–5p–6
= p2– (6–1)p+ (–6)(1)
= p2–6p+p+(–6)(1)
= p(p–6) + 1(p–6)
= (p–6)(p+1)
10) P2+P-56
=> To find two numbers whose product is (-56) and sum is (1)
As the product is negative so one number must be negative and the other number is positive.
Since the sum is positive we can say that of the two numerically greater number is positive.
Required factors are (8) and (-7)
∴ p2+p–56
= p2+ (8–7)p+ (8)( -7)
= p2+8p–7p– (8)(7)
= p(p+8) –7(p+8)
= (p+8)(p–7)
11) x2-8x-65
=> To find two numbers whose product is (-65) and sum is (-8)
As the product is negative so one number must be negative and the other number is positive.
Since the sum is negative we can say that of the two numerically greater number is negative.
Required factors are (–13) and 5
∴ x2–8x–65
= x2– (13–5)x+ (–13)(5)
= x2–13x+5x+(–13)(5)
= x(x–13) + 5(x–13)
= (x–13)(x+5)
12) x2+8x+15
=> To find two numbers whose product is 15 and sum is 8.
As the product is positive so both the numbers are either positive or negative.
Since the sum is positive we can say that both the numbers are positive.
Required factors are 5 and 3
∴ x2+8x+15
= x2+ (5+3)x+ (5)(3)
= x2+5x+3x+(5)(3)
= x(x+5) + 3(x+5)
= (x+5)(x+3)
13) x2+2x-24
=> To find two numbers whose product is (-24) and sum is (2)
As the product is negative so one number must be negative and the other number is positive.
Since the sum is positive we can say that of the two numerically greater number is positive.
Required factors are (6) and (-4)
∴ x2+2x–24
= x2+ (6–4)x+ (6)( -4)
= x2+6x–4x– (6)(4)
= x(x+6) –4(x+6)
= (x+6)(x–4)
14) x2-3x-54
=> To find two numbers whose product is (-54) and sum is (-3)
As the product is negative so one number must be negative and the other number is positive.
Since the sum is negative we can say that of the two numerically greater number is negative.
Required factors are (–9) and 6
∴ x2–3x–54
= x2– (9–6)x+ (–9)(6)
= x2–9x+6x+ (–9)(6)
= x(x–9) + 6(x–9)
= (x–9)(x+6)
15) a2-7a+12
=> To find two numbers whose product is 12 and sum is (-7).
As the product is positive so both the numbers are either positive or negative.
Since the sum is negative we can say that both the numbers are negative.
Required factors are (-4) and (-3)
∴ a2–7a+12
= a2– (4+3) a+ (–4)( –3)
= a2–4a–3a+ (–4)( –3)
= a(a–4) – 3(a–4)
= (a–4)(a–3)
16) p2-3pq+2q2
= p2–2pq–pq+2q2
= p(p–2q) –q(p–2q)
= (p-q)(p-2q)
Previous Chapter (Compound Interest) Next Chapter (Polynomials)