# The set of 4-degree polynomials (Linear algebra)

1. Oct 22, 2009

### nietzsche

1. The problem statement, all variables and given/known data

If P is the set of all 4-degree polynomials, and W is the subset of all 4-degree polynomials such that p(-2) = p(2), find a set S such that W = span(S).

2. Relevant equations

3. The attempt at a solution

My guess is that one set that works is x^4, x^2, and 1. My reasoning is that if x^3 is included, then you could take the coefficients of everything to be zero, and (-2)^3 =/= 2^3. Can't include x for similar reasons.

Really not sure, I feel like it is possible to construct a polynomial where p(-2) = p(2) but does include an x^3 or an x.

2. Oct 22, 2009

### Dick

Uh, 'I feel like' doesn't really help. Why don't you try and construct a polynomial such that ax^3+bx=p(x) and p(2)=p(-2)? Since you only have one restriction on the five dimensional subspace P, you really would expect W to have dimension 4.

3. Oct 22, 2009

### nietzsche

ah, it took me a while but i got one.

x^3 + 3x^2 - 4x - 12

i guess i don't really understand the question then. how could such a set be constructed?

4. Oct 22, 2009

### Dick

It took too long because you were working too hard. You already know the span of 1, x^2 and x^4 are in W. The question you asked, and suspected is true, is whether a combination of x and x^3 are in W. If p(x)=ax^3+bx, then p(2)=8a+2b., p(-2)=(-8a-2b). The condition for them to be equal is 4a+b=0, right? So right again, you should include x^3-4x in the spanning vectors. You didn't need to fuss with the even power terms. You already know they work.

5. Oct 22, 2009

### nietzsche

oh, i see. it makes perfect sense now. oh well, i guess i will just get part marks for that question then... thanks for the explanation.

6. Oct 23, 2009

### HallsofIvy

Staff Emeritus
A general fourth degree polynomial is of the form $p(x)= ax^4+ bx^3+ cx^2+ dx+ e$. Then the condition that p(2)= p(-2) is that $p(2)= 16a+ 8b+ 4c+ 2d+ e= p(-2)= 16a- 8b+ 4c- 2d+ e$. a, c, and e cancel immediately leaving $8b+ 2d= -8b- 2d[itex] or [itex]16b+ 4d= 0$. That reduces to d= -4b while a, c, and e can be anything. Taking a= 1, b= c= d= e= 0 gives $x^4$. Taking b= 1, a= c= e= 0, d= -4 gives $x^3- 4x$. Taking c= 1, a= b= d= e= 0 gives $x^2$. Taking e= 1, a= b= c= d= 0 gives $1$. A basis for this space is {$x^4$, $x^3- 4x$, $x^2$, 1}.