Question

Factorise the following by grouping the terms:
O) $p^2+\left(\frac{1}{p^2}\right)-2-5p+\left(\frac{5}{p}\right)$

Answer 1

Given: $p^2+\left(\frac{1}{p^2}\right)-2-5p+\left(\frac{5}{p}\right)$

On grouping the terms, we get,

$=(p^2+\frac{1}{p^2})-2-(5p-\frac{5}{p})$

$=(p^2+\frac{1}{p^2}-2p\left(\frac{1}{p}\right))-(5p-\frac{5}{p})$

$={(p-\frac{1}{p})}^2-5(p-\frac{1}{p})[\because (a-b{)}^2=a^2+b^2-2ab]$

$=(p-\frac{1}{p})(p-\frac{1}{p}-5)$

Hence, the answer is $(p-\frac{1}{p})(p-\frac{1}{p}-5)$.