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2
Identity 4: (x + a)(x + b) = x + (a + b)x + ab
Proof: We have,
2
2
(x + a)(x + b) = x(x + b) + a(x + b) = x + bx + ax + ab = x + (b + a)x + ab
Geometrical Representation of (x + a)(x + b)
1. Draw a rectangle ABCD of length (x + a) units and breadth (x + b) units x + b
and divide it into four parts i.e., U, V, X, and W, as shown alongside. A b x B
E
2. Area of rectangle ABCD = Area of rectangle AEFI

+ Area of rectangle EBHF a U V a
+ Area of rectangle DGFI ab ax (x + a)
+ Area of square FHCG F H
I
fi (x + a)(x + b) = ab + xa + bx + x 2 X W x x
bx x 2
2
Thus, (x + a)(x + b) = x + (a + b)x + ab D G C
Example 19: Expand the following using suitable identities:

(a) (xy + yz) 2 (b)    3m − 1 2 (c) (3x + 5y)(3x – 5y)

4
2
2
2
2
2
2
Solutions: (a) We have, (xy + yz) = (xy) + 2(xy)(yz) + (yz) [Using identity (a + b) = a + 2ab + b ]
2 2
2
2 2
= x y + 2xy z + y z
 1 2  1  1 2
2
(b) We have, 3m − 4 = (3m) – 2 × 3m ×   +  



4

4

2
2
2
[Using identity (a – b) = a – 2ab + b ]
3 m 1
2
= 9m − +
2 16
2
2
(c) We have, (3x + 5y)(3x – 5y) = (3x) – (5y) 2 [Using (a + b)(a – b) = a – b ]
2
2
= 9x – 25y 2
Example 20: Simplify the following using the identity: (x + a)(x + b) = x + (a + b)x + ab
2
(a) (2z + 9)(2z – 7) (b) (pqr – 1)(pqr – 3)
2
Solutions: (a) We have, (2z + 9)(2z – 7) = (2z) + (9 – 7)2z + (9)(–7) [Here, x = 2z, a = 9, b = –7]
2
= 4z + 4z – 63
(b) We have, (pqr – 1)(pqr – 3) = [pqr + (–1)][pqr + (–3)] [Here, x = pqr, a = –1, b = –3]
2
2 2
= (pqr) + [(–1) + (–3)] pqr + (–1)(–3) = p q r – 4pqr + 3
2
Example 21: Using the identities, evaluate the following:
(a) (4.9) 2 (b) 399 × 402
2
2
2
2
Solutions: (a) We have, (4.9) = (5.0 – 0.1) [Using (a – b) = a – 2ab + b ]
2
2
2
= (5.0) – 2 × (5.0) × (0.1) + (0.1) = 25 – 1 + 0.01 = 24.01
(b) We have, 399 × 402 = (400 – 1)(400 + 2) = (400) + (–1 + 2) 400 + (–1)(2)
2
2
[Using (x + a)(x + b) = x + (a + b)x + ab]
= 160000 + 400 – 2 = 160398
Mathematics-8 208