Question

Define the polynomial \(P\) with the least possible degree which satisfies the following conditions:

(a) All coefficients of \(P\) are integers,
(b) All roots of \(P\) are integers,
(c) \(P(0)=-1\),
(d) \(P(3)=128\).

Answer 1

Let's analyze the given conditions:
(a) All coefficients of \(P\) are integers.
(b) All roots of \(P\) are integers. Let the integer roots be \(r_1, r_2, \ldots, r_n\). Then, we can express \(P(x)\) as:
\(P(x)=a\left(x-r_1\right)\left(x-r_2\right) \cdots\left(x-r_n\right)\), where \(a\) is an integer constant.
(c) \(P(0)=-1\). When we plug in \(x=0\), we get:
\[
-1=P(0)=a\left(-r_1\right)\left(-r_2\right) \cdots\left(-r_n\right)
\]
(d) \(P(3)=128\). When we plug in \(x=3\), we get:
\[
128=P(3)=a\left(3-r_1\right)\left(3-r_2\right) \cdots\left(3-r_n\right)
\]
Now, let's find the least degree polynomial that satisfies these conditions. Since \(P(0)=\) -1 , we know that the product of the roots must be -1 , which means there must be an odd number of roots. The simplest case would be a polynomial with one integer root.
Let's assume the polynomial has a degree of 1 . In this case, it would be a linear polynomial of the form:
\[
P(x)=a\left(x-r_1\right)
\]

From condition (c), we know that:
\[
-1=P(0)=a\left(-r_1\right)
\]
We also know that \(P(3)=128\), so:
\[
128=P(3)=a\left(3-r_1\right)
\]
Now, we need to find the integers \(a\) and \(r_1\) that satisfy these equations. From the first equation, we have:
\[
a=\frac{-1}{-r_1}=\frac{1}{r_1}
\]
However, since \(a\) must be an integer, this implies that \(r_1=1\), so we have:
\[
P(x)=a(x-1)
\]
Now, we plug in \(x=3\) into the polynomial and set it equal to 128 :
\[
128=a(3-1) \Rightarrow 128=2 a \Rightarrow a=64
\]
Thus, we find the polynomial with the least possible degree that satisfies the conditions:
\[
P(x)=64(x-1)
\]