SOLUTION: factor the polynomial completely, given that the binomial followin it is a factor of the polynomial t^2+1=13/6t

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Question 34470: factor the polynomial completely, given that the binomial followin it is a factor of the polynomial

t^2+1=13/6t
Found 3 solutions by mukhopadhyay, benni1013, Prithwis:
Answer by mukhopadhyay(490)   (Show Source):
You can put this solution on YOUR website!
t^2+1=(13/6)t
=> 6(t^2+1) = 13t
=> 6t^2-13t+6 = 0
=> 6t^2-9t-4t+6 = 0
=> (2t-3)(3t-2) = 0
=> t = 3/2 or t = 2/3

Answer by benni1013(165)   (Show Source):
You can put this solution on YOUR website!
mukhoadhyah one major problem you did.
t^2+1=13/6t==>6t(t^2+1)=13/6t(6t)==>6t^3+6t=13t==>6t^3-7t=0==>t(6t^2-7)=0
6t^2-7=0==>6t^2=0==>7/6=t^2==>t=plus or minus the square root of 7/6.

Answer by Prithwis(166)   (Show Source):
You can put this solution on YOUR website!
t^2+1=(13/6)t
=> 6(t^2+1) = 13t
=> 6t^2-13t+6 = 0
=> 6t^2-9t-4t+6 = 0
=> (2t-3)(3t-2) = 0
=> t = 3/2 or t = 2/3