How would you show that the polynomial has at least two real roots?
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p(-8) > 0 p(0) < 0 p(2) > 0 Since any polynomial is continuous, it follows from the intermediate value theorem.
Originally Posted by Black p(-8) > 0 p(0) < 0 p(2) > 0 Since any polynomial is continuous, it follows from the intermediate value theorem. How does that prove there's at least 2 roots??
By the IVT, since p(0) < 0 < p(-8) and p(0) < 0 < p(2), there exist real numbers c in [-8,0] and d in [0,2] such that p(c)= p(d) = 0.
Originally Posted by CrazyCat87 How would you show that the polynomial has at least two real roots? Descartes rule of signs show that this has one positive root, it also shows that this has one negative root. Hence it has at least two real roots. CB
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