Factor `16x^4-81` :

Recognize that `16x^4=(4x^2)^2` and `81=(3^2)^2` so we have a difference of two squares.

`a^2-b^2=(a+b)(a-b)` so we can factor:

`16x^4-81=(4x^2+9)(4x^2-9)`

Now `4x^2+9` is the sum of two squares which does not factor in the real numbers; but `4x^2-9` is again the difference of two squares and will factor:

`16x^4-81=(4x^2+9)(4x^2-9)`

`=(4x^2+9)(2x+3)(2x-3)`

Since a polynomial is fully factored when written as the product of linear factors and irreducible quadratic factors, this is fully factored.

------------------------------------------------------------------

`16x^4-81=(4x^2+9)(2x+3)(2x-3)`

------------------------------------------------------------------

You can use the difference of squares rule to factor binomials that can be written in the form `a^2-b^2.`

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

`a =sqrt(a^2)` , `b =sqrt(b^2)`

To factor `16x^4 - 81`

we will find `sqrt(16x^4) = 4x^2`

Next, `sqrt(81) = 9`

Therefore: `16x^4 - 81`

**factored is:** `( 4x^2 + 9 ) ( 4x^2 - 9 )`

`<br>`