Question

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Answer 1

Answer:

(i) 4p2–9q2

= (2p)2-(3q)2

= (2p-3q)(2p+3q)

Using identity: x2-y2 = (x+y)(x-y)

(ii) 63a2–112b2

= 7(9a2 –16b2)

= 7((3a)2–(4b)2)

= 7(3a+4b)(3a-4b)

Using identity: x2-y2 = (x+y)(x-y)

(iii) 49x2–36

= (7a)2 -62

= (7a+6)(7a–6)

Using identity: x2-y2 = (x+y)(x-y)

(iv) 16x5–144x3

= 16x3(x2–9)

= 16x3(x2–9)

= 16x3(x–3)(x+3)

Using identity: x2-y2 = (x+y)(x-y)

(v) (l+m) 2-(l-m) 2

= {(l+m)-(l–m)}{(l +m)+(l–m)}

Using Identity: x2-y2 = (x+y)(x-y)

= (l+m–l+m)(l+m+l–m)

= (2m)(2l)

= 4 ml

(vi) 9x2y2–16

= (3xy)2-42

= (3xy–4)(3xy+4)

Using Identity: x2-y2 = (x+y)(x-y)

(vii) (x2–2xy+y2)–z2

= (x–y)2–z2

Using Identity: (x-y)2 = x2-2xy+y2

= {(x–y)–z}{(x–y)+z}

= (x–y–z)(x–y+z)

Using Identity: x2-y2 = (x+y)(x-y)

(viii) 25a2–4b2+28bc–49c2

= 25a2–(4b2-28bc+49c2 )

= (5a)2-{(2b)2-2(2b)(7c)+(7c)2}

= (5a)2-(2b-7c)2

Using Identity: x2-y2 = (x+y)(x-y) ,

we have

= (5a+2b-7c)(5a-2b-7c)

hope it helps you sorry if worng