Contradiction in polynomials

[levin irmak]

p(x)=a₁x¹+a₂x²+......a_nxⁿ is a polynomial

 

if the polynomial is divisible by x+c where c is real number we get;

 

p(x)= q(x).(x+c)

 

x+c=0

 

x=-c if plug that in to the division we get;

 

q(x).(x+c)/(x+c)

 

p(x)= 0/0

 

although this is a clear contradiction, why is it still possible to use it in mathematical calculations and give us right answer???

 

sincerely yours;

[swansont]

although this is a clear contradiction

 

Illegal operations can't give you a valid contradiction — you divided by zero.

[DrRocket]

p(x)=a₁x¹+a₂x²+......a_nxⁿ is a polynomial

 

if the polynomial is divisible by x+c where c is real number we get;

 

p(x)= q(x).(x+c)

 

x+c=0

 

x=-c if plug that in to the division we get;

 

q(x).(x+c)/(x+c)

 

p(x)= 0/0

 

although this is a clear contradiction, why is it still possible to use it in mathematical calculations and give us right answer???

 

sincerely yours;

 

You can divide by the abstract polynomial x+c in the polynomial ring over the real numbers, and you are not dividing by the zero polynomial, so the operation is perfectly correct. Later you can evaluate the resulting function at x=-c if you like, and if you do it properly you will be dividing by zero.

Edited October 9, 2011 by DrRocket

[mathematic]

p(x)=a₁x¹+a₂x²+......a_nxⁿ is a polynomial

 

if the polynomial is divisible by x+c where c is real number we get;

 

p(x)= q(x).(x+c)

 

x+c=0

 

x=-c if plug that in to the division we get;

 

q(x).(x+c)/(x+c)

 

p(x)= 0/0

 

although this is a clear contradiction, why is it still possible to use it in mathematical calculations and give us right answer???

 

sincerely yours;

Once you divide by (x+c) you will get q(x), not p(x).

Edited October 9, 2011 by mathematic

[levin irmak]

Once you divide by (x+c) you will get q(x), not p(x).

 

p(x)= q(x). x+c/x+c

 

it is still p(x) but written by its factors

 

 

 

 

You can divide by the abstract polynomial x+c in the polynomial ring over the real numbers, and you are not dividing by the zero polynomial, so the operation is perfectly correct. Later you can evaluate the resulting function at x=-c if you like, and if you do it properly you will be dividing by zero.

 

 

this made it clear, still it turn out to be the same question , is 1/(1/0) =o or undefined ???