Let ƒ(x) = ax 3 + bx 2 + cx, x ∈ [0, 1]. Since ƒ is a polynomial function, ƒ is differentiable on the whole real line and in particular on [0, 1].

Also, ƒ(0) = 0 and ƒ(1) = a + b + c = 0.

Thus, all the conditions in the hypothesis of the Rolle’s

theorem are satisfied by the Rolle’s theorem there exists at

least one α ∈ (0, 1) such that ƒ ′ ( α ) = 0

But ƒ ′ (x) = 3ax 2 + 2bx + c and ƒ(1) = a + b + c = 0

Hence, 3ax 2 + 2bx + c = 0 has a root in [0, 1].

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