# Question regarding polynomials

1. May 27, 2005

### Mathman23

Hi

This is the character equation for a polynomial of degree where $n \geq 0$

$$p(x) = a_0 x^{n} + a_{1} x^{n-1} + a_2 x ^{n-2} + \cdots + a_{n-1}x + a_{n}$$

I'm presented with the following assignment:

Two polynomials $\mathrm{p, q}$ where n = 3. These polynomials can derived using the following conditions:

$$\begin{array}{ccc} p(0) = -1 & q(1) = 3 & p(0) = q(0) \\ p'(0) =1 & q'(1) = -2 & p'(0) = q'(0) \end{array}$$

If was told by professor that these polynomials can we written in the following form.

$$p(x) = (a_{0} + a_{1})x^{n-1} + (a_{1})x^{n-2} + \cdots + (a_{n-1})x + a_{n}$$

How do I apply formula to my assignment??

Sincerely
Fred

2. May 27, 2005

### HallsofIvy

Staff Emeritus
Restating, you have two 3rd degree polynomials, p and q, such that:
p(0)= -1, p'(0)= 1, q(1)= 3, q'(1)= -2, q(0)= p(0)= -1, q'(0)= p'(0)= 1.

Okay, write p(x)= a0+ a1x+ a2x2+ a3x3 and
q(x)= b0+ b1x+ b2x2+ b3x3

p(0)= a0= -1 and p'(0)= a1= 1. Also, since q(0)= a(0), b0= -1 and since q'(0)= p'(0)= 1, b1= 1.

So far we have p(x)= -1+ x+ a2x2+ a3x3 and since that is all the information given about p(x), I don't see anyway of determining a2 or a3.

We have q(x)= -1+ x+ b2x2+ b3x3
so q(1)= -1+ 1+ b2+ b3= 3
and q'(x)= 1+ 2b22+ 3b2 so
q'(1)= 1+ 2b2+ 3b2= -2.

You can solve those two linear equations for b2 and b3, completely determining q(x) but, unless you are given more information, there is no way to completely determine p(x).

That's assuming that assignment was to determine the two polynomials! You never did say what the assignment was.

3. Jun 15, 2005

### Mathman23

Hi

I'm told that the resulting two polynomials are:

I'm told that the resulting two polynomials of degree 3 are:

p(x) = (2 + s - 2t) x^3 + (3 + 2s - 3t) x^2 + s*x + t

q(x) = (-6 + s + 2t) x^3 + (9 - 2s - 3t) x^2 + s*x +t

where s,t belong to R.

But my textbook doesn't have any information on how one can arrive at the above result.

Maybe You Hall can provide me with a hint on howto obtain the above result ?

Sincerley and Best Regards,

Fred