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.
Age
16 to 18
Challenge level
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:
\[ p^2 - 2pq + q^2 + 3p - 4q + 3 = 0 \]
What is $p-q$?
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,
$$ \eqalign { &p^2-pq+2p-pq+q^2-2q+2p-2q+4=p+1 \\ &p^2 -2pq+q^2+4p-4q+4=p+1 \\ &p^2-2pq+q^2+3p-4q+3=0 } $$.
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 $q=x^2$.