Question

Factorise this question​

Answer 1

Step-by-step explanation:

$\frac{ {a}^{2} }{9} - 16 {b}^{2} \\ \\ = ( \frac{a}{3} ) {}^{2} - {(4b)}^{2} \\ \\ = ( \frac{a}{3} - 4b)( \frac{a}{3} + 4b) \\ \\ \\ \\ using \: formula \: \: \: \: {x}^{2} - {y}^{2} = (x - y)(x + y)$

Answer 2

Answer:

$(\frac{a}{3} + 4b)(\frac{a}{3} - 4b)$

Step-by-step explanation:

To factorise:

$\frac{1}{9}a^{2} - 16b^{2}$

Method to factorise:

Since both the terms are perfect squares and one term is subtracted from the other, we can use the following identity to factorise this expression:

a² - b² = (a + b)(a - b)

√$\frac{1}{9}a^{2}$ = $\frac{1}{3}a$ = $\frac{a}{3}$

√$16b^{2}$ = $4b$

So, the factorised expression will be:

$(\frac{a}{3} + 4b)(\frac{a}{3} - 4b)$

Other identities:

• (a + b)² = a² + 2ab + b²

• (a - b)² = a² - 2ab + b²

• (x + a)(x + b) = x² + x(a + b) + ab

Hope it helps!