+128408 Here's 3
5 4 - 1 -10 -23 -40
- 1 -5 - 9 -13 -17
-4 -4 -4 -4
Since the second differences are all the same, this is a quadratic
To find the polynomial, we have this system :
a + b + c = 5
4a + 2b + c = 4
9a + 3b + c = -1
Subtract the first equation from the second and the second equation from the third and we have
3a + b = -1
5a + b = -5 subtract the first equation from the second and we have
2a = -4
a = -2
And using 3a + b = -1 to find b, we have.......3(-2) + b = -1 → b = 5
And using a + b + c = 5 to find c we have -2 + 5 + c = 5 → c = 2
So the generating polynomial is : -2x^2 + 5x + 2
Here's 4
20 4 0 20 76 180
-16 -4 20 56 104
12 24 36 48
12 12 12
Since the third differences are the same, this is a cubic polynomial
And to find the generating polynomal, we have this system
a + b + c + d = 20
8a + 4b + 2c + d = 4
27a + 9b + 3c + d = 0
64a + 16b + 4c + d = 20
Solving this system in a similar manner to (3) we have that a = 2, b = -6, c = -12, d = 36
So the generating polynomial is 2x^3 - 6x^2 -12x + 36
+26367 Help! I don't know how to solve these 2 questions. (Algebra II)
3.
\(\small{ \begin{array}{lrrrrrrrrrrrrrrrrr} & {\color{red}d_0 = 5} && 4 && -1 && -10 && -23 && -40 && \cdots \\ \text{1. Difference } && {\color{red}d_1 = -1} && -5 && -9 && -13 && -17 && \cdots \\ \text{2. Difference } &&& {\color{red}d_2 = -4} && -4 && -4 && -4 && \cdots \\ \end{array} }\)
\(\begin{array}{rcl} y &=& \binom{x}{0}\cdot {\color{red}d_0 } + \binom{x}{1}\cdot {\color{red}d_1 } + \binom{x}{2}\cdot {\color{red}d_2 } \\ \end{array}\)
\( \begin{array}{|rcl|} \hline y &=& \binom{x}{0}\cdot {\color{red} 5 } + \binom{x}{1}\cdot {\color{red} (-1) } + \binom{x}{2}\cdot {\color{red} (-4) } \\ &=& 5 - x - \frac{x}{2}\cdot \frac{x-1}{1} \cdot 4 \\ &=& 5 - x - 2x(x-1) \\ &=& 5 - x - 2x^2+2x \\ \mathbf{y} & \mathbf{=} & \mathbf{5 + x - 2x^2} \\ \hline \end{array}\)
4.
\(\small{ \begin{array}{lrrrrrrrrrrrrrrrrr} & {\color{red}d_0 = 20} && 4 && 0 && 20 && 76 && 180 && \cdots \\ \text{1. Difference } && {\color{red}d_1 = -16} && -4 && 20 && 56 && 104 && \cdots \\ \text{2. Difference } &&& {\color{red}d_2 = 12} && 24 && 36 && 48 && \cdots \\ \text{2. Difference } &&&& {\color{red}d_3 = 12} && 12 && 12 && \cdots \\ \end{array} }\)
\(\begin{array}{rcl} y &=& \binom{x}{0}\cdot {\color{red}d_0 } + \binom{x}{1}\cdot {\color{red}d_1 } + \binom{x}{2}\cdot {\color{red}d_2 } + \binom{x}{3}\cdot {\color{red}d_3 } \\ \end{array}\)
\(\begin{array}{|rcl|} \hline y &=& \binom{x}{0}\cdot {\color{red} 20 } + \binom{x}{1}\cdot {\color{red} (-16) } + \binom{x}{2}\cdot {\color{red} 12 } + \binom{x}{3}\cdot {\color{red} 12 } \\ &=& 20 - 16x + \frac{x}{2}\cdot \frac{x-1}{1} \cdot 12 + \frac{x}{3}\cdot \frac{x-1}{2}\cdot \frac{x-2}{1} \cdot 12 \\ &=& 20 - 16x + 6x(x-1) + 2x(x-1)(x-2) \\ &=& 20 - 16x + 6x^2-6x + 2x \left(x^2-3x+2 \right) \\ &=& 20 - 16x + 6x^2-6x + 2x^3-6x^2+4x \\ &=& 20 - 16x -6x + 2x^3 +4x \\ \mathbf{y} & \mathbf{=} & \mathbf{20 -18x +2x^3} \\ \hline \end{array}\)
