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Difference between revisions of "Multivariate factor theorem"

(Created page with "[b] The Multivariable Factor Theorem [b] 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...")
 
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[b] The Multivariable Factor Theorem [b] 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 [b]all[/b] <math>x,</math> then we can write  
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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 '''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>
  
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[b] Proof:[/b]
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''' Proof:'''
  
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Assume that <math>f(x, h(x)) = 0</math> for all <math>x</math>. We'll treat <math>x</math> [i]as a constant,[/i] so that <math>h(x)</math> is constant with respect to <math>y.</math>
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Assume that <math>f(x, h(x)) = 0</math> for all <math>x</math>. We'll treat <math>x</math> ''as a constant,'' so that <math>h(x)</math> is constant with respect to <math>y.</math>
  
 
If we divide <math>f(x,y)</math> by <math>y-h(x)</math> using polynomial long division, so that we have <cmath>\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.</cmath>
 
If we divide <math>f(x,y)</math> by <math>y-h(x)</math> using polynomial long division, so that we have <cmath>\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.</cmath>

Latest revision as of 11:06, 15 March 2024

The Multivariable Factor Theorem states that If $f(x,y)$ is a polynomial and there is a polynomial $h(x)$ such that $f(x,h(x))=0$ for all $x,$ then we can write \[f(x,y) = (y-h(x))g(x,y),\] for some polynomial $g(x,y).$

Proof:

Assume that $f(x, h(x)) = 0$ for all $x$. We'll treat $x$ as a constant, so that $h(x)$ is constant with respect to $y.$

If we divide $f(x,y)$ by $y-h(x)$ using polynomial long division, so that we have \[\underbrace{f(x,y)}_{\textrm{dividend}} = (\underbrace{y - h(x)}_{ \textrm{divisor}}) (\underbrace{q(x,y)}_{\textrm{quotient}}) + \underbrace{r(x)}_{ \textrm{remainder}} \text{ for all } x \text{ and } y.\]

Since we're treating $x$ as a constant, $y-h(x)$ is a monic, linear polynomial in $y.$ So, either $r(x)$ is the zero polynomial, in which case it has no terms with $y,$ or it has lower degree in $y$ than $y-h(x).$ This means that $r(x)$ will itself be a polynomial in $x.$

Now, if we set $y=h(x)$ in our equation, it becomes \[0 = 0(q(x, h(x))) + r(x) \text{ for all } x.\] It follows that $r(x) = 0.$

So $r(x) = 0$ for any $x,$ and so $r(x)$ is the zero polynomial!

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