Question 198878
# 1


{{{x^4+2x^3-3x-6}}} Start with the given expression



{{{(x^4+2x^3)+(-3x-6)}}} Group like terms



{{{x^3(x+2)-3(x+2)}}} Factor out the GCF {{{x^3}}} out of the first group. Factor out the GCF {{{-3}}} out of the second group



{{{(x^3-3)(x+2)}}} Since we have the common term {{{x+2}}}, we can combine like terms



So {{{x^4+2x^3-3x-6}}} factors to {{{(x^3-3)(x+2)}}}




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# 2



{{{x^8-1}}} Start with the given expression.



{{{(x^4)^2-1}}} Rewrite {{{x^8}}} as {{{(x^4)^2}}}.



{{{(x^4)^2-(1)^2}}} Rewrite {{{1}}} as {{{(1)^2}}}.



Notice how we have a difference of squares {{{A^2-B^2}}} where in this case {{{A=x^4}}} and {{{B=1}}}.



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^4)^2-(1)^2=(x^4-1)(x^4+1)}}} Plug in {{{A=x^4}}} and {{{B=1}}}.



So this shows us that {{{x^8-1}}} factors to {{{(x^4-1)(x^4+1)}}}.



In other words {{{x^8-1=(x^4-1)(x^4+1)}}}.



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Now let's factor {{{x^4-1}}}



{{{x^4-1}}} Start with the given expression.



{{{(x^2)^2-1}}} Rewrite {{{x^4}}} as {{{(x^2)^2}}}.



{{{(x^2)^2-(1)^2}}} Rewrite {{{1}}} as {{{(1)^2}}}.



Notice how we have a difference of squares {{{A^2-B^2}}} where in this case {{{A=x^2}}} and {{{B=1}}}.



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-(1)^2=(x^2-1)(x^2+1)}}} Plug in {{{A=x^2}}} and {{{B=1}}}.



So this shows us that {{{x^4-1}}} factors to {{{(x^2-1)(x^2+1)}}}.



In other words {{{x^4-1=(x^2-1)(x^2+1)}}}.



So this means that {{{(x^4-1)(x^4+1)}}} factors to {{{(x^2-1)(x^2+1)(x^4+1)}}}



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Now let's factor {{{x^2-1}}}


{{{x^2-1}}} Start with the given expression.



{{{(x)^2-1}}} Rewrite {{{x^2}}} as {{{(x)^2}}}.



{{{(x)^2-(1)^2}}} Rewrite {{{1}}} as {{{(1)^2}}}.



Notice how we have a difference of squares {{{A^2-B^2}}} where in this case {{{A=x}}} and {{{B=1}}}.



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-(1)^2=(x-1)(x+1)}}} Plug in {{{A=x}}} and {{{B=1}}}.



So this shows us that {{{x^2-1}}} factors to {{{(x-1)(x+1)}}}.



In other words {{{x^2-1=(x-1)(x+1)}}}.



So this means that {{{(x^2-1)(x^2+1)(x^4+1)}}} factors to {{{(x-1)(x+1)(x^2+1)(x^4+1)}}}



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Answer:


So {{{x^8-1}}} completely factors to {{{(x-1)(x+1)(x^2+1)(x^4+1)}}}