Posted by Lucy on Sunday, August 31, 2008 at 11:47am.
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, Sunday, August 31, 2008 at 2:25pm
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, Sunday, August 31, 2008 at 2:33pm
Oopsleft the "x" off of 12
Should read: f(x)=6+x^4+2x^25x^312x
Sorry 
PreCalculuscheck answers  David Q, Sunday, August 31, 2008 at 2:57pm
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.)