Using a2 – b2 = (a + b) (a – b), find
(i) 512– 492
(ii) (1.02)2– (0.98)2
(iii) 1532– 1472
(iv) 12.12– 7.92
Using identities, evaluate.
(i) 71²
(ii) 99²
(iii) 1022
(iv) 998²
(v) 5.2²
(vi) 297 x 303
(vii) 78 x 82
(viii) 8.92
(ix) 10.5 x 9.5
Show that.
(i) (3x + 7)2 – 84x = (3x – 7)2
(ii) (9p – 5q)2+ 180pq = (9p + 5q)2
(iii) (4/3m – 3/4n)2 + 2mn = 16/9 m2 + 9/16 n2
(iv) (4pq + 3q)2– (4pq – 3q)2 = 48pq2
(v) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
Simplify.
(i) (a2 – b2)2
(ii) (2x + 5)2 – (2x – 5)2
(iii) (7m – 8n)2 + (7m + 8n)2
(iv) (4m + 5n)2 + (5m + 4n)2
(v) (2.5p – 1.5q)2 – (1.5p – 2.5q)2
(vi) (ab + bc)2– 2ab²c
(vii) (m2 – n2m)2 + 2m3n2
Find the following squares by using the identities.
(i) (b – 7)2
(ii) (xy + 3z)2
(iii) (6x2 – 5y)2
(iv) [(2m/3) + (3n/2)]2
(v) (0.4p – 0.5q)2
(vi) (2xy + 5y)2
Use the identity (x + a) (x + b) = x2 + (a + b) x + ab to find the following products.
(i) (x + 3) (x + 7)
(ii) (4x + 5) (4x + 1)
(iii) (4x – 5) (4x – 1)
(iv) (4x + 5) (4x – 1)
(v) (2x + 5y) (2x + 3y)
(vi) (2a2 + 9) (2a2 + 5)
(vii) (xyz – 4) (xyz – 2)
Use a suitable identity to get each of the following products.
(i) (x + 3) (x + 3)
(ii) (2y + 5) (2y + 5)
(iii) (2a – 7) (2a – 7)
(iv) (3a – 1/2)(3a – 1/2)
(v) (1.1m – 0.4) (1.1m + 0.4)
(vi) (a2+ b2) (- a2+ b2)
(vii) (6x – 7) (6x + 7)
(viii) (- a + c) (- a + c)
(ix) (1/2x + 3/4y) (1/2x + 3/4y)
(x) (7a – 9b) (7a – 9b)
Simplify.
(i) (x2– 5) (x + 5) + 25
(ii) (a2+ 5) (b3+ 3) + 5
(iii)(t + s2)(t2 – s)
(iv) (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)
(v) (x + y)(2x + y) + (x + 2y)(x – y)
(vi) (x + y)(x2– xy + y2)
(vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y
(viii) (a + b + c)(a + b – c)
Find the product.
(i) (5 – 2x) (3 + x)
(ii) (x + 7y) (7x – y)
(iii) (a2+ b) (a + b2)
(iv) (p2 – q2) (2p + q)
Multiply the binomials.
(i) (2x + 5) and (4x – 3)
(ii) (y – 8) and (3y – 4)
(iii) (2.5l – 0.5m) and (2.5l + 0.5m)
(iv) (a + 3b) and (x + 5)
(v) (2pq + 3q2) and (3pq – 2q2)
(vi) (3/4 a2 + 3b2) and 4( a2 – 2/3 b2)