Question 416748: Need to solve by factoring:
t^(5/3)-25t^(-1/3)=0
Book shows we take out t^(-1/3) and end up with t^(-1/3)[t^2-25]=0
Why do we take out the t^(-1/3)?
Thanks!!
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Need to solve by factoring:
t^(5/3)-25t^(-1/3)=0
Book shows we take out t^(-1/3) and end up with t^(-1/3)[t^2-25]=0
Why do we take out the t^(-1/3)
..
t^(5/3)-25t^(-1/3)=0
t^(-1/3)(t^2-25)
..
t^(-1/3)=1/t^(1/3)=0
t=infinity
t^2-25=0
(t+5)(t-5)=0
t+5=0
t=-5
t-5=0
t=5
Solution:
t=-5,5,or infinity
..
When factoring this type of expression, the rule is to take out the term with the smallest exponent which is t^-(1/3) The reason for doing this could be seen by realizing that the factored form when expanded should look exactly like the original expression. For any kind of factoring, part of the process is to check whether the factored form when expanded returns to the original expression. Note that:
t^(-1/3)*t^2 = t^(-1/3)*t^(6/3) = t^(5/3) (The first term)
hope this helps
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