# indeterminate forms

• May 18th 2009, 08:08 PM
VonNemo19
indeterminate forms
Could someone justify the following statement please. I'm having trouble visualizing what has been said here:

If by direct substitution of a given value c into a rational function is made and the following result occurs

$r(c)=\frac{p(c)}{q(c)}=\frac{0}{0}$

then it can be concluded that $(x-c)$ is a factor of both $p(x)$ and $q(x)$.

Anything to shed light here would be appreciated. Thanks.
• May 18th 2009, 08:12 PM
particlejohn
Quote:

Originally Posted by VonNemo19
Could someone justify the following statement please. I'm having trouble visualizing what has been said here:

If by direct substitution of a given value c into a rational function is made and the following result occurs

$r(c)=\frac{p(c)}{q(c)}=\frac{0}{0}$

then it can be concluded that $(x-c)$ is a factor of both $p(x)$ and $q(x)$.

Anything to shed light here would be appreciated. Thanks.

if x = c is a zero of a function, then x-c is a factor.
• May 18th 2009, 08:18 PM
TheEmptySet
Quote:

Originally Posted by VonNemo19
Could someone justify the following statement please. I'm having trouble visualizing what has been said here:

If by direct substitution of a given value c into a rational function is made and the following result occurs

$r(c)=\frac{p(c)}{q(c)}=\frac{0}{0}$

then it can be concluded that $(x-c)$ is a factor of both $p(x)$ and $q(x)$.

Anything to shed light here would be appreciated. Thanks.

I am assuming $p(x), q(x)$ are polynomials otherwise this does not make sense i.e

$\frac{\sin(\pi)}{\tan(\pi)}=\frac{0}{0}$ but neither have a factor of $x-\pi$

The above is using the fact that

If f is a polynomial and $f(c)=0$ then $(x-c)$ divides f(x)

example

$f(x)=x^2-3x+2$ note that

$f(1)=1^2-3(1)+2=0$ so by the above $(x-1)$ divides $x^2-3x+2$ i.e this factors into $f(x)=(x-1)(x-2)$
• May 18th 2009, 08:20 PM
VonNemo19
I can kind of see that, but if you could explain why, I'd really appreciate it.
• May 18th 2009, 08:24 PM
VonNemo19
Good stuff Empty dude. And yes of course they are polynomials. I kind of paraphrased what it said in the book. I'm no math writer.