Super basic polynomial and exponent definition help

[Greg Bernhardt]

Please bare with me. Most of you know I actually don't have a great math background. In any case I'm going way back and filling in some very basic math that I have long forgot. I have some questions about terms in a polynomial.

Here is an example

$$3x^5+7x^3-5$$

1. From my book 3 and 7 are clearly coefficients, but I am unsure if 5 is also one? My understanding is that coefficients are in front of a variable.

2. Would the last term be 5 or -5? Both? Since it could be plus -5 for the same result? If you were to list that term how would you do it?

3. Even though there is no exponent listed on the 5, is $5^1$ implied and identical? Certainly it would not be $5^0$ because that always equals 1 right?

Thanks!

 

[kuruman]

The polynomial you have is of the general form ax⁵ + bx³ + cx⁰, where a = 3, b =7 and c = -5.

 

[Greg Bernhardt]

The polynomial you have is of the general form ax⁵ + bx³ + cx⁰, where a = 3, b =7 and c = -5.

ok so there is an implied variable behind the 5. I guess I get c = -5 but I am unsure why the exponent is 0? From what I read any term with an exponent of 0 is 1.

Furthermore why is subtraction actually a thing. This is example it's confusing and should just be plus -5. That makes it so clear instead of having to do a mental check.

 

[jedishrfu]

Is this homework?

 

[fresh_42]

Please bare with me. Most of you know I actually don't have a great math background. In any case I'm going way back and filling in some very basic math that I have long forgot. I have some questions about terms in a polynomial.

Here is an example

$$3x^5+7x^3-5$$

1. From my book 3 and 7 are clearly coefficients, but I am unsure if 5 is also one? My understanding is that coefficients are in front of a variable.

Basically yes, the "co" means "at" so a variable is needed, but it's more convenient to write $3x^5+7x^3-5 \triangleq (3,0,7,0,0,-5)$ and call all of them by the same term. If you like then it could be considered as $- 5= -5 \cdot 1 = -5\cdot x^0$ and then there is the variable again.

2. Would the last term be 5 or -5? Both? Since it could be + -5 for the same result? If you were to list that term how would you do it?

$-5$, see $(3,0,7,0,0,-5)$ above.

3. Even though there is no exponent listed on the 5, is $5^1$ implied and identical? Certainly it would not be $5^0$ because that always equals 1 right?

Yes, but $x^0$ is the solution here. It is called absolute term or absolute coefficient, and cannot be decoupled from the polynomial, neither mathematically nor by name. And it is really needed, as it guarantees the embedding of the numbers in the ring of polynomials $\mathbb{Q} \hookrightarrow \mathbb{Q}[x]$. Numbers are also polynomials $p(x)=1\,.$

 

[jedishrfu]

The -5 is just a convenience just like we use contractions in everyday English. You just implicitly think + -5

also the polynomial has some missing terms for $x^4, x^2$ and $x^1$ with zero coefficients.

 

[Greg Bernhardt]

Basically yes, the "co" means "at" so a variable is needed, but it's more convenient to write $3x^5+7x^3-5 \triangleq (3,0,7,0,0,-5)$ and call all of them by the same term. If you like then it could be considered as $- 5= -5 \cdot 1 = -5\cdot x^0$ and then there is the variable again.
$-5$, see $(3,0,7,0,0,-5)$ above.
Yes, but $x^0$ is the solution here. It is called absolute term or absolute coefficient, and cannot be decoupled from the polynomial, neither mathematically nor by name. And it is really needed, as it guarantees the embedding of the numbers in the ring of polynomials $\mathbb{Q} \hookrightarrow \mathbb{Q}[x]$. Numbers are also polynomials $p(x)=1\,.$

Fresh I'm at a 1st grade level here!
But I get some of it and it seems that the key to my questions really is that one needs to understand the form which @kuruman posted in #2. Perhaps have jumped ahead somewhere.

 

[jedishrfu]

@Greg Bernhardt you missed the is this homework joke in post #4

 

[kuruman]

From what I read any term with an exponent of 0 is 1.

That is absolutely correct. The constant in a polynomial can be viewed as the coefficient of the zeroth power. This allows writing the most general polynomial as
$$P=\sum_{n=0}^{N}a_nx^n$$
The coefficients $a_n$ can be any number, positive negative or zero. The highest exponent, N, denotes the order of the polynomial.

On edit: Corrected in light of comments below.

 

[Greg Bernhardt]

If you like then it could be considered as $- 5= -5 \cdot 1 = -5\cdot x^0$ and then there is the variable again.

This now makes sense, thanks!

Just to be clear, one has to learn the form and be able to recognize to apply the form when seeing that polynomial? Because inherently it doesn't seem obvious that I should know to add an $x^0$ after the -5.

 

[Greg Bernhardt]

@Greg Bernhardt you missed the is this homework joke in post #4

I didn't miss it

 

[jedishrfu]

Aha doing your kids schoolwork. There ought to be a law against that... also Khan Academy should have some good videos on polynomials.

 

[fresh_42]

Aha doing your kids schoolwork. There ought to be a law against that... also Khan Academy should have some good videos on polynomials.

Why to be satisfied with second best?

Just thought today: Maybe we don't solve your problems for you, but we definitely improve your understanding!

 

[StoneTemplePython]

That is absolutely correct. The constant in a polynomial can be viewed as the coefficient of the zeroth power. This allows writing the most general polynomial as
$$f(x)=\sum_{n=0}^{\infty}a_nx^n$$
The coefficients $a_n$ can be any number, positive negative or zero.

ummm, someone may correct me, but I don't think this is right.

My understanding is that all single variable polynomials are nilpotent with respect to the derivatives operator -- or perhaps more to the point: all polynomials are of finite degree. What you've written is a power series or perhaps generating function, which isn't a polynomial per se.

 

[member 587159]

ummm, someone may correct me, but I don't think this is right.

My understanding is that all single variable polynomials are nilpotent with respect to the derivatives operator -- or perhaps more to the point: all polynomials are of finite degree. What you've written is a power series or perhaps generating function, which isn't a polynomial per se.

The point is that such a polynomial has only a finite amount of coefficients that are non zero, so that we really can write it that way. That we write this as an infinite sum, is just notation:

In my algebra course, we rigorously defined polynomials as sequences of elements. For example, the sequence

$(1,2,3,0,0,0, \dots)$ denotes the polynomial $1 + 2x + 3x^2$. And the latter is just the way we represent the sequence of polynomials. Also, addition and multiplication are defined on those sequences.

From this, you can see that you are actually confusing polynomials and polynomial functions, but this is ok, because we can identify the two things with each other, but they are not formally the same things.

 

[fresh_42]

ummm, someone may correct me, but I don't think this is right.

You are right. There should have been the additional information almost all $a_n=0$, for otherwise it's a series and not a polynomial.