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# Help! I don't know how to solve these 2 questions. (Algebra II)

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Help! I don't know how to solve these 2 questions. (Algebra II)

Guest Sep 5, 2017
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#1
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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

CPhill  Sep 5, 2017
#2
+18564
0

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.

$$\small{ \begin{array}{lrrrrrrrrrrrrrrrrr} & {\color{red}d_0 = 20} && 4 && 0 && 20 && 76 && 180 && \cdots \\ \text{1. Difference } && {\color{red}d_1 = -16} && -4 && 20 && 56 && 104 && \cdots \\ \text{2. Difference } &&& {\color{red}d_2 = 12} && 24 && 36 && 48 && \cdots \\ \text{2. Difference } &&&& {\color{red}d_3 = 12} && 12 && 12 && \cdots \\ \end{array} }$$

$$\begin{array}{rcl} y &=& \binom{x}{0}\cdot {\color{red}d_0 } + \binom{x}{1}\cdot {\color{red}d_1 } + \binom{x}{2}\cdot {\color{red}d_2 } + \binom{x}{3}\cdot {\color{red}d_3 } \\ \end{array}$$

$$\begin{array}{|rcl|} \hline y &=& \binom{x}{0}\cdot {\color{red} 20 } + \binom{x}{1}\cdot {\color{red} (-16) } + \binom{x}{2}\cdot {\color{red} 12 } + \binom{x}{3}\cdot {\color{red} 12 } \\ &=& 20 - 16x + \frac{x}{2}\cdot \frac{x-1}{1} \cdot 12 + \frac{x}{3}\cdot \frac{x-1}{2}\cdot \frac{x-2}{1} \cdot 12 \\ &=& 20 - 16x + 6x(x-1) + 2x(x-1)(x-2) \\ &=& 20 - 16x + 6x^2-6x + 2x \left(x^2-3x+2 \right) \\ &=& 20 - 16x + 6x^2-6x + 2x^3-6x^2+4x \\ &=& 20 - 16x -6x + 2x^3 +4x \\ \mathbf{y} & \mathbf{=} & \mathbf{20 -18x +2x^3} \\ \hline \end{array}$$

heureka  Sep 6, 2017
edited by heureka  Sep 6, 2017
edited by heureka  Sep 6, 2017

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