PreCalc/Trig
posted by Jaclyn on .
Please check the first and help with the secondthank you
Find all possible rational roots of f(x) = 2x^4  5x^3 + 8x^2 + 4x+7
1.I took the constant which is 7 and the leading coefficient which is 2 and factored them
7 factored would be 7,2
2 factored would be 2,1
used 7 as numerator 2 as denominator came up with 1/2,1/7/2, 7 Would that be correct?
2.I don't get this one find number of possible positive and negative real roots of f(x) = x^4x^3+2x^2 + x5
I think there are 3 sign changes so there are 3 positive but I get really confused on doing the negative real zeros would I rewrite for negative f(x) = x^4 (x^3) 2x^2 (x5) so there would be 2negative real zeros
I don't get this one

No, the numbers you state would be potential rational roots.
You still have to check to see if they actually work
e.g.
f(1/2) = 2(1/2)^4  5(1/2)^3 + 8(1/2)^2 + 4(1/2) + 7 = 10.5 ≠ 0
f(1) = 2  5 + 8 + 4 + 7 ≠ 0
etc.
none of them work. A quick look at "Wolfram" shows that there are no real roots at all.
http://www.wolframalpha.com/input/?i=2x%5E4++5x%5E3+%2B+8x%5E2+%2B+4x%2B7+%3D0 
For negative zeros, substitute x for x, but do it carefully, using the exact same coefficients as originally:
(x)^4  (x)^3 + 2(x)^2 + (x)  5
or
x^4 + x^3 + 2x^2  x  5
There's only one change of sign there, so at most one negative root.