SOLUTION: Look at the following polynomial: Ρ(x) = x3 + x2 - 3x - 1. The only possible rational roots are 1 and -1, neither one of which are roots. But the polynomial does have roots. C

Question 1043720: Look at the following polynomial: Ρ(x) = x3 + x2 - 3x - 1. The only possible rational roots are 1 and -1, neither one of which are roots. But the polynomial does have roots. Consider Ρ(0) and Ρ(2) = 5, the polynomial has a negative output at x=0 and a positive output at x=2. So the polynomial must equal zero somewhere between 0 and 2. how you would go about finding the root?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Look at the following polynomial: Ρ(x) = x3 + x2 - 3x - 1. The only possible rational roots are 1 and -1, neither one of which are roots. But the polynomial does have roots. Consider Ρ(0) and Ρ(2) = 5, the polynomial has a negative output at x=0 and a positive output at x=2. So the polynomial must equal zero somewhere between 0 and 2. how you would go about finding the root?
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Determine the sign of f(1)
If it is negative there is a root between x = 1 and x = 2
If it is positive there is a root between x = 0 and x = 1
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Cheers,
Stan H.
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SOLUTION: Look at the following polynomial: &#929;(x) = x3 + x2 - 3x - 1. The only possible rational roots are 1 and -1, neither one of which are roots. But the polynomial does have roots. C

SOLUTION: Look at the following polynomial: Ρ(x) = x3 + x2 - 3x - 1. The only possible rational roots are 1 and -1, neither one of which are roots. But the polynomial does have roots. C