anonymous
  • anonymous
What is y^4-y^3=0 How do you factor this out?
Mathematics
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SOLVED
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chestercat
  • chestercat
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Nnesha
  • Nnesha
what's the common factor ? well its soo easy to find common factor of variables `variable with the smallest degree would be the common factor `
anonymous
  • anonymous
How would you do it though?
DanJS
  • DanJS
you are really multiplying by y^2 / y^2 this will pull a y^2 term out of each of those

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Nnesha
  • Nnesha
we should take out the common factor so first ) what's a common factor ???
DanJS
  • DanJS
left with y^2*(each divided by y^2)
anonymous
  • anonymous
@DanJS my book did it by y(y^3-1) =y(y-1)(y^2+y+1) I dont understand how they do it
Nnesha
  • Nnesha
well ^^take out y from y^4 -y^3 \[\huge\rm y(y^3-1)\] y^3-1 is same as (y^3 -1^3)
Nnesha
  • Nnesha
then you can apply the rule \[\huge\rm x^3-y^3=(x-y)(x^2+xy+y^2)\]
DanJS
  • DanJS
is the original problem y^4 - y^3
DanJS
  • DanJS
= y^3 * (y-1) = 0
DanJS
  • DanJS
y is 0 or 1
anonymous
  • anonymous
Yes
Nnesha
  • Nnesha
i guess it's y(y^3-1) ??
anonymous
  • anonymous
Shouldn't it factor out as y^3(y-1)?
anonymous
  • anonymous
Doesn't y(y^3-1) = y^4-y
Nnesha
  • Nnesha
ye....... and then set it equal to zero if you want to solve to for y but factor is y^3(y-1) that looks correct 2 me :/
mathstudent55
  • mathstudent55
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mathstudent55
  • mathstudent55
\(y^4 - y^3 = 0\) \(y^3(y - 1) = 0\) \(y^3 = 0\) or \(y - 1 = 0\) Now solve both equations for y.
Nnesha
  • Nnesha
\(\color{blue}{\text{Originally Posted by}}\) @Anzhbar my book did it by y(y^3-1) =y(y-1)(y^2+y+1) I dont understand how they do it \(\color{blue}{\text{End of Quote}}\) i guess you're looking at the wrong question .....:/
mathstudent55
  • mathstudent55
|dw:1443387331458:dw|

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