# Find one & all polynomial of degree 4

• Mar 23rd 2011, 11:09 AM
chipwager
Find one & all polynomial of degree 4
If someone can advise me how to approach the following problem it would be greatly appreciated.

Q: Find one & all polynomials of degree 4 such that p(-1)=1, p(0)=0, p(1)=2, p(2)=1

I tried...

p(x)=ax^4+bx^3+cx^2+dx+e
p(-1)=a-b+c-d+e=1
p(0)=e=0
p(1)=a+b+c+d+e=2
p(2)=16a+8b+4c+2d+e=1

Is it correct to place it into a matrix and reduce to echelon form? But then I'm not sure what to do with it.

[1 1 1 1 2] [1 1 1 1 2]
[1 -1 1 -1 1] ~ [0 -2 0 -2 -1]
[16 8 4 2 1] [0 0 1 1/2 -9/4]
• Mar 23rd 2011, 11:39 AM
Ackbeet
Sure, you can go to row echelon form, and then do back substitution. I would, incidentally, leave out the last column there, as you already know that e = 0. You can save yourself some computations. If the system is consistent (and it better be, because having a polynomial of degree 4 that goes through those four points is definitely possible - just think about Lagrange interpolating polynomials to see that), then I would say you should end up with a 1-parameter family of solutions. What do you get?
• Mar 23rd 2011, 11:59 AM
chipwager
I did as you suggested. I went to echelon form then back substituted to row canonical form.

[1 1 1 1 2]
[1 -1 1 -1 1]
[16 8 4 2 1]

[1 1 1 1 2]
[0 -2 0 -2 -1]
[0 0 1 1/2 -9/4]

[1 0 0 -1/2 15/4]
[0 1 0 1 1/2]
[0 0 1 1/2 -9/4]

I'm not sure what I can do with this matrix though. Am I on the right track? I represented a, b, c, d and the solution in the above matrices given that e=0.
• Mar 23rd 2011, 12:29 PM
Ackbeet
You might want to check your accuracy here. It looks like your last column might have some errors in it.

What you do now is back substitution by letting one variable, say, d, be an arbitrary parameter. I'll do the first one, and you do the rest. Let the parameter t = d. Then, by the third row, we know that c + d/2 = 9/4, or c = 9/4-t/2. And continue in this manner, solving for b and a. What do you get?
• Mar 23rd 2011, 12:41 PM
chipwager
I see where you're going with this. I get a = (t/2-3/4), b = (1/2 - t), c= (9/4 - t/2) then solve for d given that e=0. Can I assume that the value for "a" is to the degree of 4 and "b"=3, etc?
• Mar 23rd 2011, 12:44 PM
Ackbeet
Not quite. You're not going to be able to eliminate t, because you have more unknowns than equations. However, you can write the parameters a through d as linear functions of t. Then you plug those into the original polynomial, along with the fact that e = 0, and you've found all the polynomials that satisfy your conditions.
• Mar 23rd 2011, 12:45 PM
chipwager
thank you so much for walking me through this! much appreciated.
• Mar 23rd 2011, 01:07 PM
Ackbeet