factor the difference of two squares 16x^4-81



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embizze's profile pic

Posted on

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:


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:



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




baxthum8's profile pic

Posted on

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 )`


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