# Finding a Polynomial Function

• February 26th 2010, 11:40 AM
TaylorM0192
Finding a Polynomial Function
Suppose you're given the following zeroes of a polynomial of degree 5, and leading coefficient 1:

$x=2+\sqrt{3}, x=1-i, x=1$. Thus, you have four factors as follows (by the complex conjugate theorem): $(x-1)(x-(2+\sqrt{3}))(x-(1-i))(x-(1+i))$ which more elagantly written is $(x-1)((x-2)-\sqrt{3})((x-1)+i)((x-1)-i).$.

Knowing that the polynomial is degree 5 with a leading coefficient of 1, how do we determine the remaiing zero?

I have some ideas, but I'm not exactly sure. Do we expand out the factors we have and simply make up a factor that will give you a leading co-efficient of 1 and then expand to create the 5th degree polynomial in expanded form? Do we say that there is an additional zero of $2-\sqrt{3}, )$?

Hmm...
• February 26th 2010, 01:26 PM
Soroban
Hello, TaylorM0192!

I think you miscounted the zeroes . . .

Quote:

Given these zeroes of a polynomial of degree 5, and leading coefficient 1:

. . $\begin{Bmatrix}2+\sqrt{3} \\ 1-i \\ 1\end{Bmatrix}$

Find the polynomial.

We already know all five zeros of the polynomial: . $\begin{Bmatrix}2 + \sqrt{3} \\ 2-\sqrt{3} \\ 1 - i \\ 1 + i \\ 1 \end{Bmatrix}$

Hence, the polynomial is: . $\bigg[x -(2 + \sqrt{3})\bigg]\,\bigg[x - (2 - \sqrt{3})\bigg]\,\bigg[x - (1-i)\bigg]\,\bigg[x - (1+i)\bigg]\,\bigg[x - 1\bigg]$

Just multiply it out . . .

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Here's a tip for the multiplication. . .

Consider: . $\bigg[x - (1-i)\bigg]\,\bigg[x-(1+i)\bigg]$

. . . . . . $=\;\bigg[(x-1) + i\bigg]\,\bigg[(x-1) - i\bigg]$

. . . . . . $=\qquad(x-1)^2 - (i^2)$

. . . . . . $=\quad\;\; x^2 - 2x + 1 + 1$

. . . . . . $=\qquad\; x^2 - 2x + 2$