Question

121p^2-1/9p^2 factorise the following expression​

Answer 1

Answer:

$121 {p}^{2} - \frac{1}{9} {p}^{2}$

$= > {(11p})^{2} - { (\frac{1}{3} p)}^{2}$

According to the formula,

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

$= >( 11p + \frac{1}{3} p)(11p - \frac{1}{3} p)$

It is thus factorised.

However if you want to solve it,

$= > \frac{33 + 1}{3}p \times \frac{33 - 1}{3}p$

$= > \frac{34}{3}p \times \frac{32}{3}p$

$= > \frac{1088}{3}p^{2}$

Hope it helps...

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