Chapter 8.7, Problem 27WE

To determine

To show: that $h$ must be a factor of $d$ .

Explanation of Solution

Given information:

  $h$ must be a factor of $d$ .

Proof: let consider a equation,

  $\begin{array}{l}a{\left(\frac{h}{k}\right)}^3+b{\left(\frac{h}{k}\right)}^2+c\left(\frac{h}{k}\right)+d=0\\ a{h}^3+b{h}^{2}k+ch{k}^2+d{k}^3=0\text{ [multiply by }{k}^3]\\ d{k}^3=-a{h}^3-b{h}^{2}k+-ch{k}^2\text{ [subtract]}\\ \frac{d{k}^3}{h}=-a{h}^2-bhk+-c{k}^2\text{ [divide by }h]\end{array}$

Since, $\frac{d{k}^3}{h}$ is an integer, all the prime factors of $h$ must divide out with the prime factors of $d{k}^3$ but $h$ and $k$ have no common prime factor.

Therefor $h$ must be a factor of $d$ .