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.
\(\small{ \begin{array}{lrrrrrrrrrrrrrrrrr} & {\color{red}d_0 = 20} && 4 && 0 && 20 && 76 && 180 && \cdots \\ \text{1. Difference } && {\color{red}d_1 = -16} && -4 && 20 && 56 && 104 && \cdots \\ \text{2. Difference } &&& {\color{red}d_2 = 12} && 24 && 36 && 48 && \cdots \\ \text{2. Difference } &&&& {\color{red}d_3 = 12} && 12 && 12 && \cdots \\ \end{array} }\)
\(\begin{array}{rcl} y &=& \binom{x}{0}\cdot {\color{red}d_0 } + \binom{x}{1}\cdot {\color{red}d_1 } + \binom{x}{2}\cdot {\color{red}d_2 } + \binom{x}{3}\cdot {\color{red}d_3 } \\ \end{array}\)
\(\begin{array}{|rcl|} \hline y &=& \binom{x}{0}\cdot {\color{red} 20 } + \binom{x}{1}\cdot {\color{red} (-16) } + \binom{x}{2}\cdot {\color{red} 12 } + \binom{x}{3}\cdot {\color{red} 12 } \\ &=& 20 - 16x + \frac{x}{2}\cdot \frac{x-1}{1} \cdot 12 + \frac{x}{3}\cdot \frac{x-1}{2}\cdot \frac{x-2}{1} \cdot 12 \\ &=& 20 - 16x + 6x(x-1) + 2x(x-1)(x-2) \\ &=& 20 - 16x + 6x^2-6x + 2x \left(x^2-3x+2 \right) \\ &=& 20 - 16x + 6x^2-6x + 2x^3-6x^2+4x \\ &=& 20 - 16x -6x + 2x^3 +4x \\ \mathbf{y} & \mathbf{=} & \mathbf{20 -18x +2x^3} \\ \hline \end{array}\)