# Polynomial Relations

Given any two polynomials in a single variable it is always
possible to eliminate the variable and obtain a formula showing the
relationship between the two polynomials. Try this one.

## Problem

Given any two polynomials in a single variable it is always possible to eliminate the variable and obtain a formula showing the relationship between the two polynomials.

Let $p(x) = x^2 + 2x$ and $q(x) = x^2 + x + 1$. Then, using a method which does not depend on knowing the answer, show that the relationship between the polynomials is:

\[ p^2 - 2pq + q^2 + 3p - 4q + 3 = 0 \]

## Getting Started

What is $p-q$?

## Student Solutions

Good solutions to this problem were received from Tyrone of
Cyfarthfa High School in Merthyr Tydfil, and Koopa of Boston
College in the USA.

Tyrone solved the problem by relating both polynomials to
$(x+1)^2$ :

$$ \eqalign { p(x)=x^2 + 2x \Rightarrow &p
&=&(x+1)^2 - 1 \\ &p+1 &=& (x+1)^2 \\ }
$$

(1)

$$ \eqalign { q(x)=x^2 + x + 1 \Rightarrow &q
&=&(x+1)^2 - x \\ &q+x &=& (x+1)^2 \\ }
$$

(2)

$ \Rightarrow p+1=q+x $ (combining
eqns (1) and (2)).

But $x=(p+1)^{1/2}-1$ (from eqn (1)). So

$$ \eqalign { \Rightarrow p+1&=&q+ ((p+1)^{1/2}-1) \\

&=&q+ (p+1)^{1/2}-1 }$$

$$ \eqalign { p-q+2&=&(p+1)^{1/2} \\

(p-q+2)^2&=&p+1}$$.

Squaring the bracket,

$$ \eqalign { &p^2-pq+2p-pq+q^2-2q+2p-2q+4=p+1 \\ &p^2 -2pq+q^2+4p-4q+4=p+1 \\ &p^2-2pq+q^2+3p-4q+3=0 } $$.

## Teachers' Resources

Why do this problem?

It gives practice in manipulation of polynomials.

Possible approach

An easy lesson starter!

Key question

What is $p(x)-q(x)$?

Possible
extension

Learners can make up their own probems by writing down two
polynomials in $x$ and then eliminating $x$ between the
expressions. They might be asked to make up such a problem and
exchange problems with their partner. Then they can compare and
check results in pairs.

Possible support

Try a simpler example such as: find the formula relating $p$ and $q$ where $p=x+3$ and $q=x^2$.