Let p(x)=a_0+a_1 x+ldots+a_n x^n. If p(-2)=-1 | vedclass

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(C) We use the method of finite differences to determine the degree of the polynomial $p(x)$.Let the values of $p(x)$ for $x = -2, -1, 0, 1, 2, 3$ be

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$3$

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(C) We use the method of finite differences to determine the degree of the polynomial $p(x)$.Let the values of $p(x)$ for $x = -2, -1, 0, 1, 2, 3$ be $y_i$:$x = -2, y = -15$$x = -1, y = 1$$x = 0, y = 7$$x = 1, y = 9$$x = 2, y = 13$$x = 3, y = 25$First differences $(\Delta y)$: $1 - (-15) = 16, 7 - 1 = 6, 9 - 7 = 2, 13 - 9 = 4, 25 - 13 = 12$Second differences $(\Delta^2 y)$: $6 - 16 = -10, 2 - 6 = -4, 4 - 2 = 2, 12 - 4 = 8$Third differences $(\Delta^3 y)$: $-4 - (-10) = 6, 2 - (-4) = 6, 8 - 2 = 6$Since the third differences are constant $(6)$,the polynomial $p(x)$ is of degree $n = 3$.Thus,the smallest possible value of $n$ is $3$.

Let $p(x)=a_0+a_1 x+\ldots+a_n x^n$. If $p(-2)=-15, p(-1)=1, p(0)=7, p(1)=9, p(2)=13$ and $p(3)=25$,then the smallest possible value of $n$ is

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