Math question

[imported_K3N]

does anyone know why only six zeros will appear in this polynomial instead of 7?

x^7 + 7·x^6 - 19·x^5 - 125·x^4 + 151·x^3 + 273·x^2 + 171·x + 405

 

[oznerol]

Christ, is college starting back up again? It must be.

 

[Epic Fail]

Originally posted by: K3N
does anyone know why only six zeros will appear in this polynomial instead of 7?

x^7 + 7·x^6 - 19·x^5 - 125·x^4 + 151·x^3 + 273·x^2 + 171·x + 405

I see only one zero.

 

[RESmonkey]

because it's a 7 degree polynomial

 

[Toastedlightly]

factor out some zeros and go nuts. Or use mathematica.

 

[Tim]

Orange

 

[Leros]

Depends what x is

 

[Anubis]

Originally posted by: Toastedlightly
factor out some zeros and go nuts. Or use mathematica.

maple > crapmatica

 

[darkxshade]

The answer to your question OP is yes.

 

[imported_K3N]

Originally posted by: RESmonkey
because it's a 7 degree polynomial

So if it was an 8 degree polynomial, only 7 zeros would show up?

 

[jman19]

Originally posted by: Anubis

Originally posted by: Toastedlightly
factor out some zeros and go nuts. Or use mathematica.

maple > crapmatica

:thumbsup:

 

[akubi]

:beer:

cuz x=3 is repeated
http://xrjunque.nom.es/precis/rootfinder.aspx

 

[blueshoe]

Repeated zero perhaps?

x^2=0 has a double root at x=0 whereas x^2 - 1 = 0 has a roots at 1 and -1

 

[dullard]

First, I will assume your polynomial is equal to zero. Otherwise you just have a pretty looking string of numbers, symbols, and the letter x.

Factor it, and you'll know.

z^7 + 7·x^6 - 19·x^5 - 125·x^4 + 151·x^3 + 273·x^2 + 171·x + 405 = 0

Gives:

(x+5)·(x^2 + 8x +9)·(x^2+1)·(x-3)·(x-3)=0

Thus,

x+5=0 gives one root (x=-5)
x^2 + 8x +9=0 gives two roots (x=-1.354 and x=-6.646)
x^2+1=0 gives two roots (x=i and x=-i)
x-3 = 0 gives one root (x=3)
and
x-3 = 0 gives one root (x=3)

Thus, you have seven roots, but x=3 is a repeated root.

 

[dullard]

Dang, akubi beat me to the punch.

 

[DrPizza]

Originally posted by: jman19

Originally posted by: Anubis

Originally posted by: Toastedlightly
factor out some zeros and go nuts. Or use mathematica.

maple > crapmatica

:thumbsup:

mathematica > maple, but is harder to learn to use.


As to the OP's question: the degree of the polynomial is equal to the number of roots, if you count repeated roots and complex roots. It's called the fundamental theorem of algebra.

You have a 7th degree polynomial, therefore there are 7 zeros.

 

[George P Burdell]

It's the first week of school... can't you at least do your first HW assignment yourself?

:roll:

 

[DrPizza]

edit: I see someone already explained the repeated root of 3 to you. If you're looking at a graph, when you see a repeated root, the graph will have sort of a parabolic shape in the region near that root if the root appears an even number of times. If the root appears an odd number of times, then graphically, it will pass through that value similarly to the way y=x^3 passes through the origin. Derivatives from calculus explain this: because a factor is repeated, upon applying the product rule for derivatives (and chain rule) you will obtain a critical point at that x-value. If the factor appears an even number of times, then the first derivative will contain that factor an odd number of times (and vice versa). Thus, as the first derivative describes the slope of your polynomial, at your root, the derivative will take on a value of zero; i.e. as the curve passes through that zero, the slope at that point is zero. However, if the derivative contains the factor an odd number of times, (the original polynomial contained that factor an even number of times), then the sign of the first derivative will be different on opposite sides of the zero (when taking values very close to the zero). If the 1st derivative = slope is negative to the left of the zero, then the slope will be positive to the right of the zero. i.e. the function will be decreasing until the slope is zero at your zero, then the function will be increasing. The argument is similar for roots that repeat an odd number of times: with an even power on that factor in the derivative, the 1st derivative = slope will have the same sign on both sides of the root - the function will momentarily have a slope of zero at the zero, but then will continue going whichever way it was going prior to the zero, i.e. downhill, hit the zero, then back to downhill.

 

[alexvorn2]

math - a great problem.

 

[imported_K3N]

Dullard and Mr. Pizza thanks alot for the great answers

 

[chuckywang]

Originally posted by: K3N
does anyone know why only six zeros will appear in this polynomial instead of 7?

x^7 + 7·x^6 - 19·x^5 - 125·x^4 + 151·x^3 + 273·x^2 + 171·x + 405

Has to be a double zero.

 

[flxnimprtmscl]

The answer is 42. Jesus you guys are stupid.