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How to tell whether a function is a polynomial

Look at every exponent

A polynomial in $x$ is a finite sum of constant multiples of nonnegative integer powers of $x$ . Negative or fractional powers remaining after simplification are not allowed. A constant denominator is fine. If factors cancel in a rational expression, also check its original domain: cancellation does not fill a missing input value.

Problem

Which are polynomial functions? (a) $v (x) = 7 x^{6}$ ; (b) $h (x) = - 1/ x^{3}$ ; (c) $u (x) = 3 x - 9 x^{5}$ ; (d) $f (x) = x^{9} - 7 + 3 x^{- 6}$ .

Answer

Only (a) is a polynomial function.

A negative coefficient is fine; a negative or fractional exponent of x is not.

Step-by-step solution

1. Apply the definition

Every exponent of $x$ in a polynomial must be a nonnegative integer. Constants are allowed.

2. Classify the four functions

(a) $7 x^{6}$ has exponent 6, so it is a polynomial. (b) $- 1/ x^{3} = - x^{- 3}$ has a negative exponent. (c) $x = x^{1/2}$ has a fractional exponent. (d) The term $3 x^{- 6}$ has a negative exponent. Thus (b), (c) and (d) are not polynomials.

More examples: degree, leading term and constant term

These additional problems were solved on 6 October 2026.

Example 1. Classify $f (x) = 3 x + x^{2}$ and write it in standard form.

f (x) = x^{2} + 3 x .

Both exponents are nonnegative integers, so this is a polynomial of degree 2. Its leading term is $x^{2}$ , its leading coefficient is 1, and its constant term is 0. A missing written constant does not make the function non-polynomial.

Example 2. Classify $g (x) = (5 - x^{2}) /6$ . The denominator is a constant, so division merely changes the coefficients:

g (x) = - \frac{1}{6} x^{2} + \frac{5}{6} .

This is a polynomial of degree 2, with leading term $- x^{2} /6$ and constant term $5/6$ . Polynomial coefficients need not be integers.

Example 3. Classify $F (x) = (x^{3} - 3) / x^{6}$ . On its original domain $x \neq = 0$ ,

F (x) = x^{- 3} - 3 x^{- 6} .

The negative powers do not cancel. The function is not a polynomial, so no polynomial degree, leading term or constant term should be assigned to it.

Example 4. Could two straight rays meeting at $(- 4, 5)$ in an upward V-shape form the graph of a polynomial? No. The graph has a corner, so its derivative is not defined at the meeting point. Every polynomial is differentiable at every real input. A smooth appearance is necessary for polynomial graphs, but is not by itself sufficient to prove a curve is polynomial.

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