Question

factorise the following

a² + 1 / a² - 18

ans : (a - 1/a + 4) ( a - 1/a - 4)

Answer 1

$a^2+1/a^2-18=a^2+1/a^2-2*a*1/a-16=(a+1/a)^2-(4)^2=(a-1/a+4)(a-1/a-4)$

In the first step,2*a*1/a=2

and -2-16=-18 so I splitted them

Second:

$(a-1/a)^2=a^2+1/a^2-2$

Last step:

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

Answer 2

Answer:  The required factored form of the given expression is

$\left(a-\dfrac{1}{a}+4\right)\left(a-\dfrac{1}{a}-4\right).$

Step-by-step explanation:  We are given to factorize the following expression  :

$E=a^2+\dfrac{1}{a^2}-18~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

We will be using the following two formulas :

$(i)~(x-y)^2=x^2+y^2-2xy,\\\\(ii)~x^2-y^2=(x+y)(x-y).$

The expression (i) can be factorized as follows :

$E\\\\\\=a^2+\dfrac{1}{a^2}-18\\\\\\=a^2+\left(\dfrac{1}{a}\right)^2-2\times a\times \dfrac{1}{a}-18+2\\\\\\=\left(a-\dfrac{1}{a}\right)^2-16\\\\\\=\left(a-\dfrac{1}{a}\right)^2-4^2\\\\\\=\left(a-\dfrac{1}{a}+4\right)\left(a-\dfrac{1}{a}-4\right).$

Thus, the required factored form of the given expression is

$\left(a-\dfrac{1}{a}+4\right)\left(a-\dfrac{1}{a}-4\right).$