Difference between revisions of "Multivariate factor theorem"
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| − | ''' The Multivariable Factor Theorem '''states that If <math>f(x,y)</math> is a polynomial and there is a polynomial <math>h(x)</math> such that <math>f(x,h(x))=0</math> for | + | ''' The Multivariable Factor Theorem '''states that If <math>f(x,y)</math> is a polynomial and there is a polynomial <math>h(x)</math> such that <math>f(x,h(x))=0</math> for '''all''' <math>x,</math> then we can write |
<cmath> f(x,y) = (y-h(x))g(x,y),</cmath> for some polynomial <math>g(x,y).</math> | <cmath> f(x,y) = (y-h(x))g(x,y),</cmath> for some polynomial <math>g(x,y).</math> | ||
Latest revision as of 11:06, 15 March 2024
The Multivariable Factor Theorem states that If 
is a polynomial and there is a polynomial 
such that 
for all 
then we can write
![\[f(x,y) = (y-h(x))g(x,y),\]](http://latex.artofproblemsolving.com/5/f/8/5f8fa527ee7b032b7c241f300ce7a80d2de73c5a.png)
for some polynomial 
Proof:
Assume that 
for all 
. We'll treat as a constant, so that is constant with respect to 
If we divide by 
using polynomial long division, so that we have ![\[\underbrace{f(x,y)}_{\textrm{dividend}} = (\underbrace{y - h(x)}_{ \textrm{divisor}}) (\underbrace{q(x,y)}_{\textrm{quotient}}) + \underbrace{r(x)}_{ \textrm{remainder}} \text{ for all } x \text{ and } y.\]](http://latex.artofproblemsolving.com/f/1/6/f16058aaab8887e6a6e77c7a272dc2aded15248e.png)
Since we're treating as a constant, is a monic, linear polynomial in So, either 
is the zero polynomial, in which case it has no terms with 
or it has lower degree in 
than 
This means that will itself be a polynomial in 
Now, if we set 
in our equation, it becomes ![\[0 = 0(q(x, h(x))) + r(x) \text{ for all } x.\]](http://latex.artofproblemsolving.com/7/c/8/7c8fe4295a1bdda35254d92ccb6a4dd3b1673cf8.png)
It follows that 
So 
for any and so is the zero polynomial!
