Exercise 6.7
1. Find the following products:
(i) (x + 4) (x + 7)
(ii) (x – 11) (x + 4)
(iii) (x + 7) (x – 5)
(iv) (x – 3) (x – 2)
(v) (y2 – 4) (y2 – 3)
(vii) (3x + 5) (3x + 11)
(viii) (2×2 – 3) (2×2 + 5)
(ix) (z2 + 2) (z2 – 3)
(x) (3x – 4y) (2x – 4y)
(xi) (3×2 – 4xy) (3×2 – 3xy)
(xiv) (x2 + 4) (x2 + 9)
(xv) (y2 + 12) (y2 + 6)
Solution:
(i) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab.
(x + 4) (x + 7)
= x2+(4+7)x+4×7
= x2+11x+28
(ii) Here, we will use the identity (x – a) (x + b) = x2 + (b – a)x – ab.
(x – 11) (x + 4)
= x2+(4−11)x–11×4
= x2–7x–44
(iii) Here, we will use the identity (x + a) (x – b) = x2 + (a – b)x – ab.
(x + 7) (x – 5)
= x2+(7−5)x–7×5
= x2+2x–35
(iv) Here, we will use the identity (x – a) (x – b) = x2 – (a + b)x + ab.
(x – 3) (x – 2)
= x2–(3+2)x+3×2
= x2–5x+6
(v) Here, we will use the identity (x – a) (x – b) = x2 – (a + b)x – ab.
(y2 – 4) (y2 – 3)
= (y2)2–(4+3)(y2)+4×3
= y4–7y2+12
(vi) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab.
(x+43)(x+34)
= x2+(43+34)x+43×34)
= x2+2512x+1
(vii) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab.
(3x + 5) (3x + 11)
= (3x)2+(5+11)(3x)+5×11
= 9x2+48x+55
(viii) Here, we will use the identity (x – a) (x + b) = x2 + (b – a)x – ab.
(2x2 – 3) (2x2 + 5)
= (2x2)2+(5–3)(2x2)–3×5
= 4x4+4x2–15
(ix) Here, we will use the identity (x + a) (x – b) = x2 + (a – b)x – ab.
(z2 + 2) (z2 – 3)
= (z2)2+(2–3)(z2)–2×3
= z4–z2–6
(x) Here, we will use the identity (x – a) (x – b) = x2 – (a + b)x – ab.
(3x – 4y) (2x – 4y)
= (4y – 3x) (4y – 2x) (Taking common -1 from both parentheses)
= (4y)2–(3x+2x)(4y)+3x×2x
= 16y2–(12xy+8xy)+6x2
= 16y2–20xy+6x2
(xi) Here, we will use the identity (x – a) (x – b) = x2 – (a + b)x – ab.
(3x2 – 4xy) (3x2 – 3xy)
= (3x2)2–(4xy+3xy)(3x2)+4xy×3xy
= 9x4–(12x3y+9x3y)+12x2y2
= 9x4–21x3y+12x2y2
(xiv) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab.
(x2 + 4) (x2 + 9)
= (x2)2 + (4+9) (x2) + 4 × 9
= x4+13x2 + 36
(xv) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab.
(y2 + 12) (y2 + 6)
= (y2)2 + (12+6)(y2) + 12 × 6
= y4+18xy2+72
(xvi) Here, we will use the identify (x +a) (x + b) = x2 (a + b) x- ab.
(xvii) Here, we will use the identify (x +a) (x + b) = x2 (a + b) x- ab.
2. Evaluate the following:
(i) 102 x 106
(ii) 109 x 107
(iii) 35 x 37
(iv) 53 x 55
(v) 103 x 96
(vi) 34 x 36
(vii) 994 x 1006
Solution:
(i) Here, we will use the identity (x + a) (x + b) = x2 + (a + b) x + ab
102 x 106
= (100 + 2) (100 + 6)
= 1002 + (2 + 6)100 + 2 x 6
=10000 + 800 +12 = 10812
(ii) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab
109 x 107
= (100 + 9) (100 + 7)
= 1002 + (9 + 7)100 + 9 x 7
=10000 + 1600 + 63 = 11663
(iii) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab
35 x 37
= (30 + 5) (30 + 7)
= 302 + (5 + 7)30 + 5 x 7
= 900 + 360 + 35 = 1295
(iv) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab
53 x 55
= (50 + 3) (50 + 5)
= 502 + (3 + 5)50 + 3 x 5
= 2500 + 400 + 15 = 2915
(v) Here, we will use the identity (x + a) (x – b) = x2 + (a – b)x – ab
103 x 96
= (100 + 3) (100 – 4)
= 1002 + (3 – 4)100 – 3 x 4
= 10000 – 100 – 12 = 9888
(vi) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab
34 x 36
= (30 + 4) (30 + 6)
= 302 + (4 + 6)30 + 4 x 6
= 900 + 300 + 24 = 1224
(vii) Here, we will use the identity (x – a) (x + b) = x2 + (b – a)x – ab
994 x 1006
= (1000 – 6) x (1000 + 6)
= 10002 + (6 – 6) x 1000 – 6 x 6
= 1000000 – 36 = 999964