Question

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

Answer 1

Apply difference of two squares,

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

This implies


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



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

Answer 2

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

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