Question 182024


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



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



{{{(x^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=x^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.



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



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



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


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Now let's factor {{{x^2-4}}} (which is a difference of squares)



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



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



{{{(x)^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=x}}} 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.



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



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



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



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So this means that {{{(x^2+4)(x^2-4)}}} factors even further to {{{(x^2+4)(x+2)(x-2)}}} 




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Answer:



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