Question

Factorise:

$p^4-81$

• A. $(p^2+9)(p^2-9)$
• B. $(p^4-{27}^4)$
• C. $(p^3-3^3)$
• D. $(p^4-3)$

Answer 1

The correct option is A $(p^2+9)(p^2-9)$

$p^4-81$
$=p^4-(3{)}^4$
$=(p^2{)}^2-(3^2{)}^2$ Using,$[a^2-b^2=(a+b\left)\right(a-b\left)\right]$
$=(p^2+3^2)(p^2-3^2)$ Using,$[a^2-b^2=(a+b\left)\right(a-b\left)\right]$
$=(p^2+9)(p^2-9)$