Question

factorise:
(2p+3)(2p-3)(4p^2+9)(16p^4+81)
Answer with explanation plzzz​

Answer 1

Answer:

2 × 3[p – (2 × q)]

= 6(p – 2q)

(iii) We have, 7a2 = 7 × a × a

and , 14a = 2 × 7 × a

The two terms have 7 and a as common factors

7a2 + 14a = (7 × a × a) + (2 × 7 × a)

= 7 × a × (a + 2) = 7a(a + 2)

(iv) 16z = 2 × 2 × 2 × 2 × z

20z3 = 2 × 2 × 5 × z × z × z

The common factors are 2, 2, and z.

∴ –16z + 20z3 = –(2 × 2 × 2 × 2 × z)

+ (2 × 2 × 5 × z × z × z

= (2 × 2 × z)[–(2 × 2) + (5 × z × z)]

= 4z(– 4 + 5z2)

Step-by-step explanation:

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Answer 2

Your answer is:-

256p^8 -6561

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