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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− | + | ''' 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> | ||
− | + | ''' Proof:''' | |
− | Assume that <math>f(x, h(x)) = 0</math> for all <math>x</math>. We'll treat <math>x</math> | + | 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 12:06, 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
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
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
It follows that
So for any
and so
is the zero polynomial!