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### Fourth-Degree Polynomial

```
Date: 04/26/2001 at 09:08:06
From: Richard Miller
Subject: Fourth-degree polynomial how?

Help! Our teacher gave us a question on graphing polynomials as a
just-for-fun problem and said it can be done. We want to come up with
a solution.

Construct a fourth degree polynomial that contains all of the
following points: (-1,2), (1,-2), (3,7), (2,1), (0,3).

Any help would be great! Thanks (-:
```

```
Date: 04/26/2001 at 10:20:24
From: Doctor Paul
Subject: Re: Fourth-degree polynomial how?

This is not a difficult problem. You should be aware that n points
uniquely determine a polynomial of degree n-1. For n = 2, this fact
reduces to the statement that two points uniquely determine a line.

You know that the general form of a 4th-degree polynomial is:

y = ax^4 + bx^3 + cx^2 + dx + e

where a, b, c, d, and e are real numbers.

Now all you need to do is take each point (x,y) - there are five of
them - and substitute the appropriate values for x and y into the
general equation given above. This will give you a system of five
equations with five unknowns (a, b, c, d, and e are the unknowns) that
you can solve pretty easily using a TI graphing calculator.

Let's do it:

The point (-1,2) means that x = -1 and y = 2. Substituting these
values into the equation y = ax^4 + bx^3 + cx^2 + dx + e gives:

2 = a - b + c - d + e

Skipping the details, the other four equations are:

(1,-2), (3,7), (2,1), (0,3)

-2 = a + b + c + d + e
7 = 81a + 27b + 9c + 3d + e
1 = 16a + 8b + 4c + 2d + e
3 = 0a + 0b + 0c + 0d + e

Now we have the necessary five equations and five unknowns. We can
rewrite this system of equations using matrix multiplication as
follows:

Ax = b where

[ 1    -1    1    -1    1]
[                        ]
[ 1     1    1     1    1]
[                        ]
A =  [81    27    9     3    1]
[                        ]
[16     8    4     2    1]
[                        ]
[ 0     0    0     0    1]

and

[ 2]
[  ]
[-2]
[  ]
b =  [ 7]
[  ]
[ 1]
[  ]
[ 3]

Solving Ax = b for x gives x = A^(-1) * b. Use a TI Graphing
calculator to compute the inverse of A, and then multiply this inverse
by b to obtain:

[-19]
[---]
[24 ]
[   ]
[47 ]
[-- ]
[12 ]
[   ]
x =  [-53]
[---]
[24 ]
[   ]
[-71]
[---]
[12 ]
[   ]
[ 3 ]

Interpreting these results means that

a = -19/24
b =  47/12
c = -53/24
d = -71/12
e =  3

So the polynomial is:

y = (-19/24)x^4 + (47/12)x^3 - (53/24)x^2 - (71/12)x + 3

- Doctor Paul, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra
High School Linear Algebra

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