Question 183091


{{{81x^4-16}}} Start with the given expression.



{{{(9x^2)^2-16}}} Rewrite {{{81x^4}}} as {{{(9x^2)^2}}}.



{{{(9x^2)^2-(4)^2}}} Rewrite {{{16}}} as {{{(4)^2}}}.



Notice how we have a difference of squares {{{A^2-B^2}}} where in this case {{{A=9x^2}}} and {{{B=4}}}.



So let's use the difference of squares formula {{{A^2-B^2=(A+B)(A-B)}}} to factor the expression:



{{{A^2-B^2=(A+B)(A-B)}}} Start with the difference of squares formula.



{{{(9x^2)^2-(4)^2=(9x^2+4)(9x^2-4)}}} Plug in {{{A=9x^2}}} and {{{B=4}}}.



So this shows us that {{{81x^4-16}}} factors to {{{(9x^2+4)(9x^2-4)}}}.



In other words {{{81x^4-16=(9x^2+4)(9x^2-4)}}}.




Now let's factor {{{9x^2-4}}} further


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{{{9x^2-4}}} Start with the given expression.



{{{(3x)^2-4}}} Rewrite {{{9x^2}}} as {{{(3x)^2}}}.



{{{(3x)^2-(2)^2}}} Rewrite {{{4}}} as {{{(2)^2}}}.



Notice how we have a difference of squares {{{A^2-B^2}}} where in this case {{{A=3x}}} and {{{B=2}}}.



So let's use the difference of squares formula {{{A^2-B^2=(A+B)(A-B)}}} to factor the expression:



{{{A^2-B^2=(A+B)(A-B)}}} Start with the difference of squares formula.



{{{(3x)^2-(2)^2=(3x+2)(3x-2)}}} Plug in {{{A=3x}}} and {{{B=2}}}.



So this shows us that {{{9x^2-4}}} factors to {{{(3x+2)(3x-2)}}}.



In other words {{{9x^2-4=(3x+2)(3x-2)}}}.


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Answer:



So {{{81x^4-16}}} completely factors to {{{(9x^2+4)(3x+2)(3x-2)}}}