Explanation Given information: Polynomial: P ( x ) = a + b x + c x 2 + d x 3 . The provided polynomial satisfies the equation a + b 2 + c 3 + d 4 = 0 . Calculation: Integrate the polynomial P ( x ) = a + b x + c x 2 + d x 3 with reference to x , to get the following: ∫ P ( x ) d x = a x + b x 2 2 + c x 3 3 + d x 4 4 Let f x = ∫ P x d x = a x + b x 2 2 + c x 3 3 + d x 4 4 ⋯ ⋯ 1 As per the Rolle's Theorem. Assume that a function f is continuous on a , b and differentiable on a , b with the property f a = f b , then, there presence at least one point c in a , b as f ′ c = 0 . Substitute in x = 0 in the equation 1 , to get the following: f 0 = a 0 + b 0 2 2 + c 0 3 3 + d 0 4 4 = 0 Plug in x = 1 in the equation 1 , to obtain the following: f 1 = a 1 + b 1 2 2 + c 1 3 3 + d 1 4 4 = a + b 2 + c 3 + d 4 = 0 Since, a + b 2 + c 3 + d 4 = 0 It follows that f 0 = f 1 = 0 . By using Rolle's Theorem, there is a point c in 0 , 1 , such that, f ′ c = 0 . As f is the integral of the polynomial P x = a + b x + c x 2 + d x 3 , so, f ′ x = P x . This indicates that, the equation P ( x ) = 0 has a root c in 0 , 1 . Assume the n th degree polynomial be the following: P ( x ) = a 0 + a 1 x + a 2 x 2 + … + a n x n . Suppose that the polynomial P ( x ) = a 0 + a 1 x + a 2 x 2 + … + a n x n satisfies the following equation: a 0 + a 1 2 + a 2 3 + … + a n n + 1 = 0 Integrate the polynomial P ( x ) = a 0 + a 1 x + a 2 x 2 + … + a n x n with respect to x , to obtain the following: ∫ P ( x ) d x = a 0 x + a 1 x 2 2 + a 2 x 3 3 + … + a n x n + 1 n + 1

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Chapter 5, Problem 13P

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To prove: The equation Px=0 has a solution between 0 and 1 .

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To prove : The equation P x = 0 has a solution between 0 and 1 .

Solution Summary: The author explains how the polynomial P(x)=0 satisfies the equation a+b2+

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To prove: The equation Px=0 has a solution between 0 and 1 .

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