Question

factorise by using identities: p4-81

Answer 1

Answer:

Step-by-step explanation:

P^4 can be written as (P^2)^4 and 81 as (3^2)^2

By the formula ,

(a^2 - b^2) = (a+b) (a-b)

So,

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

= (p^2 - 9)(p^2 + 9)

Hope it helps you!

Answer 2

The factorisation of $p^4-81$ is $(p^{2}+9)(p+3)(p-3).$

Step-by-step explanation:

We have,

$p^4-81$

To find, the factorisation of $p^4-81=?$

∴ $p^4-81$

= $(p^{2})^2-(3^{2})^2$

$=(p^{2}+3^{2})(p^{2}-3^{2})$

Using identity,

$a^{2}-b^{2}=(a+b)(a-b)$

$=(p^{2}+9)(p^{2}-3^{2})$

$=(p^{2}+9)(p+3)(p-3)$

Again, using identity,

$a^{2}-b^{2}=(a+b)(a-b)$

Hence, the factorisation of $p^4-81$ is $(p^{2}+9)(p+3)(p-3).$