Q1. Factor completely: (i) 9x2 + 24xy + 16y2 (ii) 4s2 + 20st + 25t2 (iii) 49x2 + 28xy + 4y2 (iv) 64p2 + 32/3pq + 4/9q2 *(v) 3a2 + 4ab + 4/3b2 (*vi) 9/5 s2 + 6sv + 5v2 (Hint: 2 was taken out as a common factor in Example 7. Is it possible to do something similar in Exercises (v) and (vi) above?)
Answer:
(i) 9x2 + 24xy + 16y2
= (3x)2 + 2(3x)(4y) + (4y)2 [using identity (a + b)2 = a2 + 2ab + b2]
= (3x + 4y)2
(ii) 4s2 + 20st + 25t2
= (2s)2 + 2(2s)(5t) + (5t)2
= (2s + 5t)2
(iii) 49x2 + 28xy + 4y2
= (7x)2 + 2(7x)(2y) + (2y)2 [using identity (a + b)2 = a2 + 2ab + b2]
= (7x + 2y)2
(iv) 64p2 + 32/3pq + 4/9q2
= (8p)2 + 2(8p)(2/3q) + (2/3q)2 [using identity (a + b)2 = a2 + 2ab + b2]
=(8p+2/3q)2
(v) 3a2 + 4ab + 4/3b2
Taking 3 as common
=3(a2 + 4/3.a.b + 4/9b2)
=3(a2 + 2.a.2/3b + (2/3b)2)
=3(a + 2/3b)2
(vi) 9/5 s2 + 6sv + 5v2.
Taking 5 as common
= 5(9/25s2 + 6/5sv + v2)
=5((3/5)s2 + 2.3/5s.v + (v)2)
=5(3/5 s+ v)2
Q2. Find the values of the following using the identity (a – b)2 = a2 – 2ab + b2. (i) (79)2 (ii) (193)2 (iii) (299)2
Answer:
(i) 792
we can write 79=(80-1), so a=80 , b=1.
= (80 - 1)2
= 802 - 2(80)(1) + 12
= 6400 - 160 + 1
= 6241
(ii) 1932
Write 193=(200-7), So a=200, b=7
= (200 - 7)2
= 2002 - 2(200)(7) + 72
= 40000 - 2800 + 49
= 37249
(iii) 2992
Write 299=(300-1), So a=300, b=1
= (300 - 1)2
= 3002 - 2(300)(1) + 12
= 90000 - 600 + 1
= 89401