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^2-5x^3-12.

Answer: 0

2) Approximate the real zeros of f(x)=2x^4-3x^2-2 to the nearest tenth.

Answer: no real roots

- Pre-Calculus-check 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?

- Pre-Calculus-check answers -
**Lucy**, Sunday, August 31, 2008 at 2:33pm
Oops---left the "x" off of 12

Should read: f(x)=6+x^4+2x^2-5x^3-12x

Sorry

- Pre-Calculus-check 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.)

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