SOLUTION: Factor completely x^4+2x^3-3x-6 and this one x^8-1

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Question 198878: Factor completely x^4+2x^3-3x-6
and this one x^8-1
Found 2 solutions by jim_thompson5910, Earlsdon:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
# 1

x%5E4%2B2x%5E3-3x-6 Start with the given expression


%28x%5E4%2B2x%5E3%29%2B%28-3x-6%29 Group like terms


x%5E3%28x%2B2%29-3%28x%2B2%29 Factor out the GCF x%5E3 out of the first group. Factor out the GCF -3 out of the second group


%28x%5E3-3%29%28x%2B2%29 Since we have the common term x%2B2, we can combine like terms


So x%5E4%2B2x%5E3-3x-6 factors to %28x%5E3-3%29%28x%2B2%29





# 2


x%5E8-1 Start with the given expression.


%28x%5E4%29%5E2-1 Rewrite x%5E8 as %28x%5E4%29%5E2.


%28x%5E4%29%5E2-%281%29%5E2 Rewrite 1 as %281%29%5E2.


Notice how we have a difference of squares A%5E2-B%5E2 where in this case A=x%5E4 and B=1.


So let's use the difference of squares formula A%5E2-B%5E2=%28A-B%29%28A%2BB%29 to factor the expression:


A%5E2-B%5E2=%28A-B%29%28A%2BB%29 Start with the difference of squares formula.


%28x%5E4%29%5E2-%281%29%5E2=%28x%5E4-1%29%28x%5E4%2B1%29 Plug in A=x%5E4 and B=1.


So this shows us that x%5E8-1 factors to %28x%5E4-1%29%28x%5E4%2B1%29.


In other words x%5E8-1=%28x%5E4-1%29%28x%5E4%2B1%29.


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Now let's factor x%5E4-1


x%5E4-1 Start with the given expression.


%28x%5E2%29%5E2-1 Rewrite x%5E4 as %28x%5E2%29%5E2.


%28x%5E2%29%5E2-%281%29%5E2 Rewrite 1 as %281%29%5E2.


Notice how we have a difference of squares A%5E2-B%5E2 where in this case A=x%5E2 and B=1.


So let's use the difference of squares formula A%5E2-B%5E2=%28A-B%29%28A%2BB%29 to factor the expression:


A%5E2-B%5E2=%28A-B%29%28A%2BB%29 Start with the difference of squares formula.


%28x%5E2%29%5E2-%281%29%5E2=%28x%5E2-1%29%28x%5E2%2B1%29 Plug in A=x%5E2 and B=1.


So this shows us that x%5E4-1 factors to %28x%5E2-1%29%28x%5E2%2B1%29.


In other words x%5E4-1=%28x%5E2-1%29%28x%5E2%2B1%29.


So this means that %28x%5E4-1%29%28x%5E4%2B1%29 factors to %28x%5E2-1%29%28x%5E2%2B1%29%28x%5E4%2B1%29


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Now let's factor x%5E2-1

x%5E2-1 Start with the given expression.


%28x%29%5E2-1 Rewrite x%5E2 as %28x%29%5E2.


%28x%29%5E2-%281%29%5E2 Rewrite 1 as %281%29%5E2.


Notice how we have a difference of squares A%5E2-B%5E2 where in this case A=x and B=1.


So let's use the difference of squares formula A%5E2-B%5E2=%28A-B%29%28A%2BB%29 to factor the expression:


A%5E2-B%5E2=%28A-B%29%28A%2BB%29 Start with the difference of squares formula.


%28x%29%5E2-%281%29%5E2=%28x-1%29%28x%2B1%29 Plug in A=x and B=1.


So this shows us that x%5E2-1 factors to %28x-1%29%28x%2B1%29.


In other words x%5E2-1=%28x-1%29%28x%2B1%29.


So this means that %28x%5E2-1%29%28x%5E2%2B1%29%28x%5E4%2B1%29 factors to %28x-1%29%28x%2B1%29%28x%5E2%2B1%29%28x%5E4%2B1%29


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Answer:

So x%5E8-1 completely factors to %28x-1%29%28x%2B1%29%28x%5E2%2B1%29%28x%5E4%2B1%29

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Factor completely:
1) x%5E4%2B2x%5E3-3x-6 Factor by grouping:
%28x%5E4%2B2x%5E3%29-%283x%2B6%29 Notice the sign change in the second group. Now factor x%5E3 from the first group and 3 from the second group.
x%5E3%28x%2B2%29-3%28x%2B2%29 Finally, factor the common %28x%2B2%29
highlight%28%28x%2B2%29%28x%5E3-3%29%29
2) x%5E8-1 Rewrite as the difference of squares.
%28x%5E4%29%5E2-1%5E2 Factor.
%28x%5E4-1%29%28x%5E4%2B1%29 Factor the %28x%5E4-1%29 as the difference of squares %28%28x%5E2%29%5E2-1%5E2%29%28x%5E4%2B1%29 Factor the %28x%5E2%29%5E2-1%5E2
%28x%5E2-1%29%28x%5E2%2B1%29%28x%5E4%2B1%29 Factor the x%5E2-1 as the difference of squares.
highlight%28%28x-1%29%28x%2B1%29%28x%5E2%2B1%29%28x%5E4%2B1%29%29