Question

Factor the expression completely over the complex numbers. x^4−81

Answer 1

Consider the expression
`x^4-81.`

First, note that
`x^4=(x^2)^2` and
`81=9^2.` Then use formula

`a^2-b^2=(a-b)(a+b).`

Thus,

`x^4-81=(x^2)^2-9^2=(x^2-9)(x^2+9).`

Now, use the same formula for

`x^2-9=x^2-3^2=(x-3)(x+3)`

and

`x^2+9=x^2-(-9)=x^2-9i^2=x^2-(3i)^2=(x-3i)(x+3i).`

At last

`x^4-81=(x-3)(x+3)(x-3i)(x+3i).`

Answer 2

Apply difference of two squares,


`{x}^(4) - 8 1 = ({x}^(2) ) {}^(2) - {9}^(2)`

This implies



`{x}^(4) - 8 1 = ({x}^(2) -9 ) ( {x }^(2) {}^{} + {9}^{} )`




`= (x - 3)(x + 3)(x - 3i)(x + 3i)`