[/quote]
,
be an acute scalene triangle with circumcenter 
.
be the feet of the altitudes from 
to 
respectively. Let 
be the midpoint of 
.
be the circle with diameter 
.
be the intersection of 
and 
.
be the orthocenter of 
be the intersection of 
and 
be the lines through 
tangent to 
respectively. Let 
be the intersection of 
and 
.
be the intersection of 
and 
.
be the line through 
and let 
be the reflection of 
is tangent to 
.
and 
on 
and 
are on the same side of 
be midpoint of segment 
be midpoint of segment 
be intersection of 
with 
and 
,
,
terms is divisible by 
be the set of vertices of 
be the set of triangles with vertices all in 
of 
be the set of triangles of this type.
be the set of triangles of this type.
,
the area of ![\[\sum_{T \in \mathcal A} S_T \geq \sum_{T \in \mathcal B} S_T.\]](http://latex.artofproblemsolving.com/5/e/d/5ed261e1e812466ea3e78f3b681f6bbdb7ebccd1.png)
and orthocenter 
intersect 
respectively. Lines through 
intersects 
respectively. Point 
is a parallelogram. Prove that lines 
and 
.
![$f(x)\in\mathbb Z[x]$](http://latex.artofproblemsolving.com/2/9/5/2953d049559ec5d4020e6777d965bdfff9ebde0f.png)
have positive integer leading coefficient. Show that there exists infinte positive integer 
such that the number of digit that doesn'r equal to 
is no more than 
is parallel with line 
.
has at least one pair of positive integer solution 
for each positive integer 
.
Prove that
Let 
Prove that
Let 
Prove that
Let 
Prove that
we have 
.
.
,
,
and 
.
and 
are both sides of 
Y by coolnick, bobcats4thewin, Adventure10, Mango247
Hi everybody,
The multivariable factor theorem for polynomials as stated in the intermediate algebra book is this: "If f(a,b) is a polynomial and there is a polynomial h(b) in b such that f(h(b),b) = 0 for all b, then we can write f(a,b) = (a-h(b))g(a,b), where g(a,b) is a polynomial."
I couldn't understand the proof provided in the book (namely the part that said, "treat b [a variable] as a constant"), and I would greatly appreciate it if somebody could explain it to me. The proof is included below:
"Taking a closer look at our solution to Problem 9.18, our key step was finding a polynomial h(b) such that f(h(b),b) = 0 for all b. Specifically, we found that if h(b) = b, then f(h(b),b b) = f(b, b) = 0 for all b. This observation guided us to guess that a - h(b) (that is, a - b) is a factor of f(a,b). Viewing f(a, b) as a one-variable polynomial in a and applying the Factor Theorem made it particularly clear that we could write f(a, b) as the product of a - b and another factor. We then found that this other factor is also a polynomial."
I am really confused about how to intuitively (and rigorously if possible without getting to linear algebra or some college mathematics) justify how we can treat b as a constant (and thus viewing the polynomial as a "one-variable polynomial in a"), but in our final statement of the theorem say that both a and b are variables.
I also tried an extensive internet search for multivariable factor theorem for polynomials and nothing accessible came up. Instead, I got a bunch of super complicated mathematics (and some dealing with how get an efficient way for computer programs, etc.).
If anybody is too lazy to explain the solution, would it be possible for someone to provide me a link to a webpage that explains multivariable factor theorem for polynomials?
Thanks so much!
The multivariable factor theorem for polynomials as stated in the intermediate algebra book is this: "If f(a,b) is a polynomial and there is a polynomial h(b) in b such that f(h(b),b) = 0 for all b, then we can write f(a,b) = (a-h(b))g(a,b), where g(a,b) is a polynomial."
I couldn't understand the proof provided in the book (namely the part that said, "treat b [a variable] as a constant"), and I would greatly appreciate it if somebody could explain it to me. The proof is included below:
"Taking a closer look at our solution to Problem 9.18, our key step was finding a polynomial h(b) such that f(h(b),b) = 0 for all b. Specifically, we found that if h(b) = b, then f(h(b),b b) = f(b, b) = 0 for all b. This observation guided us to guess that a - h(b) (that is, a - b) is a factor of f(a,b). Viewing f(a, b) as a one-variable polynomial in a and applying the Factor Theorem made it particularly clear that we could write f(a, b) as the product of a - b and another factor. We then found that this other factor is also a polynomial."
I am really confused about how to intuitively (and rigorously if possible without getting to linear algebra or some college mathematics) justify how we can treat b as a constant (and thus viewing the polynomial as a "one-variable polynomial in a"), but in our final statement of the theorem say that both a and b are variables.
I also tried an extensive internet search for multivariable factor theorem for polynomials and nothing accessible came up. Instead, I got a bunch of super complicated mathematics (and some dealing with how get an efficient way for computer programs, etc.).
If anybody is too lazy to explain the solution, would it be possible for someone to provide me a link to a webpage that explains multivariable factor theorem for polynomials?
Thanks so much!
This post has been edited 1 time. Last edited by theax, May 8, 2011, 8:34 AM
as a polynomial in 
.
.
was. This would be a polynomial in 
,
is a factor of 
as 
,
.
![$f\in R[x,y]$](http://latex.artofproblemsolving.com/a/9/f/a9ff68d47cd679badd94e9362945d3d978d47123.png)
is a polynomial in two variables, where 
is, say, an integral domain, we can define ![$\hat{f} \in (R[y])[x]$](http://latex.artofproblemsolving.com/5/1/5/515e752823f416201cd4a16b6df9e74a7abc9631.png)
by 
(its coefficients are polynomials in 
)![$R[y]$](http://latex.artofproblemsolving.com/7/1/0/7106c740ca56f85c9b2c1624561da7b434c05465.png)
of polynomials in ![$h(y) \in R[y]$](http://latex.artofproblemsolving.com/9/5/a/95a25f631197f023fb258076938aa736c25a2826.png)
such that 
,
is divisible by 
,
,![$\hat{g}\in (R[y])[x]$](http://latex.artofproblemsolving.com/d/6/2/d62aeaca8b07c9deacd83f72ea90e87436c17aee.png)
having as coefficients polynomials in ![$g\in R[x,y]$](http://latex.artofproblemsolving.com/4/2/2/4225203ac4bc5185facc2decd3481b0a6a4e286f.png)
,
.
.
and 
.
and that we could implement the same method for all values of 
