Determine if a function is a polynomial

[datafiend]

I'm going through polynomials and the the problem:
\(\displaystyle g\left(x\right)= (4+x^3)/3 \) IS NOT A POLYNOMIAL FUNCTION.

I don't get it. The answer says \(\displaystyle x\ne0\), it's not a polynomial.
How did you deduce that?

Going down the rabbit hole...and it's the third week.

 

[evinda]

I'm going through polynomials and the the problem:
\(\displaystyle g\left(x\right)= (4+x^3)/3 \) IS NOT A POLYNOMIAL FUNCTION.

I don't get it. The answer says \(\displaystyle x\ne0\), it's not a polynomial.
How did you deduce that?

Going down the rabbit hole...and it's the third week.

A polynomial of degree $n$ is a function of the form
$$f(x)=a_n x^n+a_{n-1}x^{n-1}+ \dots + a_2 x^2+a_1 x+a_0$$

where each coefficient $a_k$ is a real number, $a_n \neq 0$, and $n$ is a non-negative integer.

In your case, $n=3 \in \mathbb{N}$, $a_3=\frac{1}{3} \in \mathbb{R} \setminus \{ 0 \}, a_2=a_1=0 \in \mathbb{R}, a_0=\frac{4}{3} \in \mathbb{R}$

Therefore, $g(x)=\frac{1}{3} x^3+\frac{4}{3}$ is a polynomial function of degree $3$.

 

[datafiend]

I typed in the problem wrong. It should be

\(\displaystyle g(x)=(4+x^3)/x\)

Sorry.

 

[MarkFL]

First, do you see why $x\ne0$?

Second, can this be written in the form given above by evinda?

 

[mathbalarka]

If $f(x)$ and $g(x)$ be polynomials (look at evinda's definition) then we know that $f(x) + g(x)$ and $f(x)g(x)$ are also polynomials. (why?). However, If $P(x) = (4 + x^3)/x$ is a polynomial, then $x \cdot P(x) - x^3 = 4$ is a polynomial too. Prove that a polynomial with nonzero coefficients can never be everywhere constant on $\Bbb R$, thus arrive at a contradiction.

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Rather easily, prove that a polynomial is everywhere defined on $\Bbb R$. However, $P(x) = (4+x^3)/x$ is not defined at $x = 0$ (why?). Thus conclude that $P(x)$ is not a polynomial. (EDIT : Oh, I see this was the intended answer. Silly me)

 

[datafiend]

I think intuitively that x CAN'T be a zero when it's in the denominator.

 

[evinda]

I think intuitively that x CAN'T be a zero when it's in the denominator.

Yes, right! (Yes)

When we try to divide by zero, things stop making sense..