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April 1, 2015

April 1, 2015

Posted by **Jaclyn** on Monday, September 12, 2011 at 10:53am.

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^4-x^3+2x^2 + x-5

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 -(x-5) so there would be 2negative real zeros

I don't get this one

- Pre-Calc/Trig -
**Reiny**, Monday, September 12, 2011 at 12:40pmNo, 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

- Pre-Calc/Trig -
**Steve**, Wednesday, September 21, 2011 at 12:57pmFor 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.

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