Equation about polynomials that implies polynomials are zero

In summary, the conversation discusses a mathematical problem involving real polynomials P, Q, and R. The goal is to demonstrate that if P, Q, and R belong to R[X], then P²-XQ²=XR² implies that P=Q=R=0. The conversation goes on to discuss different methods of proving this statement, including using degrees and considering the fact that the square of a real polynomial is always greater than or equal to 0.
  • #1
mohlam12
154
0
Hi everyone,

I have to demonsrate that for every real polynomial, P Q and R, I have :

P²=X(Q²+R²) ==imply==> P=Q=R=0

Using degrees, we can easily demonsrate the above. However, I'm looking for another way, without using that.
 
Physics news on Phys.org
  • #2


What is X? :confused:
 
  • #3


X is the variable ! We can rewrite the problem this way :
if P(X), Q(X) and R(X) belongs to R[X], then writing P²(X) = X(Q²(X) + R²(X)) means P(X)=Q(X)=R(X)=0

Someone adviced me to start with demonstrating that constant coefficients are equal to 0 ? :s
 
  • #4


The problem still makes no sense. Surely you don't mean "for every real polynomial, P Q and R" and then say "P= Q= R= 0".

Suppose Q= R= x and P= 2x3. Does that contradict what you are trying to prove? Are you assuming that P, Q, and R are have the same degree?
 
  • #5


My bad ! Here's the exact question :
"Demonstrate that if P, Q, and R belong to R[X], therefore P² - XQ² = XR² imply that P=Q=R=0"

Big logic mistake in my first post -.- sorry
 
  • #6


I think the "degree argument" you mentioned is still the simplest and best way to go. If n is the higher degree of Q and R, then the degree of xQ2+ R2 is 2n+ 1 so the degree of P would have to be (2n+1)/2, not an integer.
 
  • #7


At the risk of pointing out the obvious, [itex]f(x)^2 \geq 0[/itex] for every real x...
 

1. What is the equation for polynomials that implies they are zero?

The equation for polynomials that implies they are zero is known as the "zero polynomial" or "null polynomial." It is written as P(x) = 0, where P(x) represents the polynomial function and 0 represents the constant zero.

2. How do you solve an equation about polynomials that implies they are zero?

To solve an equation about polynomials that implies they are zero, you can use the zero product property. This property states that if the product of two or more factors is equal to zero, then at least one of the factors must be equal to zero. Therefore, you can set each factor of the polynomial equal to zero and solve for the variable to find the roots or solutions.

3. Can an equation about polynomials imply they are zero without having any real solutions?

Yes, an equation about polynomials can imply they are zero without having any real solutions. This can occur when the polynomial has complex or imaginary roots. For example, the polynomial x^2 + 1 has no real roots, but it can still be written as P(x) = 0, where P(x) represents the polynomial function and 0 represents the constant zero.

4. Are there any special cases for equations about polynomials that imply they are zero?

One special case for equations about polynomials that imply they are zero is when the polynomial has no degree or is a constant polynomial. In this case, the equation would simply be written as c = 0, where c is a constant. This implies that the polynomial is equal to zero for all values of x.

5. How can equations about polynomials that imply they are zero be applied in real life?

Equations about polynomials that imply they are zero have many real-life applications, such as in physics, engineering, and economics. They can be used to model and solve problems related to motion, growth and decay, and optimization. For example, the equation P(x) = 0 can represent the equilibrium point in a supply and demand curve, where the quantity demanded equals the quantity supplied.

Similar threads

  • Calculus and Beyond Homework Help
Replies
24
Views
616
  • Calculus and Beyond Homework Help
Replies
13
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
577
  • Calculus and Beyond Homework Help
Replies
9
Views
960
  • Calculus and Beyond Homework Help
Replies
7
Views
606
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
12
Views
2K
  • Calculus and Beyond Homework Help
Replies
5
Views
353
  • Calculus and Beyond Homework Help
Replies
5
Views
535
  • Calculus and Beyond Homework Help
Replies
3
Views
985
Back
Top