[/quote]
,
and 
are complex numbers such that
complex numbers with 
such that
Find the value of 
.
denote the set of natural numbers and let 
denote 
ordered pairs in 
.
distinct elements in the set 
for 
whose product is a perfect cube.
be an interior point of a triangle 
and 
meet the sides 
in 
respectively. Show that ![\[ \frac{AP}{PD} = \frac{AF}{FB} + \frac{AE}{EC}. \]](http://latex.artofproblemsolving.com/7/5/9/759482bace109eb619aaea692625f7854faa9fec.png)
,
and 
on the points 
,
and 
,
intersect the circumference on the point 
.
meet the segment 
at its midpoint if and only if 
.
Where 
are positive integers and 
is a non-square positive integer. If 
denotes the n-th set of solution to the equation with 
Then, prove that 
Prove that
Let 
Prove that
Let 
Prove that
-
,
,
.![$$\sqrt[3]{(a^2+1)(b^2+1)(c^2+1)} \le \left(\frac{a+b+c}{3}\right)^2 + 1.$$](http://latex.artofproblemsolving.com/2/0/7/207934bca53047858a3820f37705171ff0abfd45.png)
with 
.![\[
\frac{a^5}{b^2 + 2c^3} + \frac{2b^5}{3c + a^6} + \frac{c^7}{a + b^4} \geq 2.
\]](http://latex.artofproblemsolving.com/f/f/5/ff5043705025592aeefc55a764a8b7170e13ff4b.png)
be an integer. Prove that there is a set of 
points in the plane such that the distance between any two points is irrational and each set of three points determines a non-degenerate triangle with rational area.
.
on 
respectively. Circumcircle 
of triangle 
intersects 
.
intersects 
at 
.
cuts 
respectively. Construct parallelogram 
.
are collinear.
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 
