PreCalculuscheck answers
posted by Lucy .
Find the number of possible negative real zeros for f(x)=6+x^4+2x^25x^312.
Answer: 0
2) Approximate the real zeros of f(x)=2x^43x^22 to the nearest tenth.
Answer: no real roots

PreCalculuscheck answers 
David Q
The first function takes the value zero at approximately X= 0.9 and again at around 4.63, with a minimum value between them somewhere around 3.46; also has a gradient of zero at X=0.
Are you sure the equation is correctly written down, given that it's apparently got two constants in it but no linear term? 
PreCalculuscheck answers 
Lucy
Oopsleft the "x" off of 12
Should read: f(x)=6+x^4+2x^25x^312x
Sorry 
PreCalculuscheck answers 
David Q
I suspected as much :) Your first function appears to have two zeros, but both are for positive X (approx. 0.5 and 5.0). The second one appears to have a couple of zeros somewhere around 1.4 and +1.4: would you like to check that out? (That's just from a quick sketch of the graph between X = 2 and X = +2.)
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