Question

${a}^{2} + 1 \div {a}^{2} - 18 \: \: factorize$

Answer 1

Given equation : $a^2+\dfrac{1}{a^2}-18$

Splitting 18 in two numbers,

18 = 2 + 16

$\implies a^2 + \dfrac{1}{a^2} - 2 - 16 \\\\\\\implies a^2 + \dfrac{1}{a^2} - 2(1) - 16\\\\\\\implies a^2 + \dfrac{1}{a^2} - 2\bigg(a \times \dfrac{1}{a} \bigg)- 4^2$

From factorization, we know that the value of a^2 + b^2 - 2ab is ( a - b )^2

Therefore, $a^2 + \dfrac{1}{a^2} - 2\bigg(a \times \dfrac{1}{a} \bigg)= \bigg( a - \dfrac{1}{a} \bigg)^2$

$\implies \bigg( a- \dfrac{1}{a} \bigg)^2 - 4^2$

From factorization, we know that the value of a^2 - b^2  is ( a + b )( a - b )

$\implies \bigg( a- \dfrac{1}{a} + 4\bigg)\bigg( a - \dfrac{1}{a} - 4 \bigg)$

Therefore, a^2 + 1 / a^2 - 18 in factorized form is ( a - 1 / a - 4 )( a - 1 / a + 4 )