Question Video: Identifying a Polynomial Function Mathematics

Which of the following is a polynomial function? [A] 𝑓(π‘₯) = π‘₯Β² + 2π‘₯ + 4 [B] 𝑓(π‘₯) = 2π‘₯⁻² [C] 𝑓(π‘₯) = π‘₯⁻² + 2π‘₯ + 4 [D] 𝑓(π‘₯) = 1/π‘₯ [E] 𝑓(π‘₯) = √(π‘₯) + 4

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Video Transcript

Which of the following is a polynomial function? Is it (A) 𝑓 of π‘₯ equals π‘₯ squared plus two π‘₯ plus four, (B) 𝑓 of π‘₯ equals two π‘₯ to the power of negative two, (C) 𝑓 of π‘₯ equals π‘₯ to the power of negative two plus two π‘₯ plus four, (D) 𝑓 of π‘₯ equals one over π‘₯, or (E) 𝑓 of π‘₯ is equal to the square root of π‘₯ plus four?

In order to answer this question, we’re going to need to remind ourselves what we mean when we talk about a polynomial function. A polynomial function is a function that’s made up as a sum of monomial functions. In this case, we call each term a monomial term. But what’s a monomial function? A monomial is an expression which is made up as the product of some constant and a variable, and any exponents of those variables must be nonnegative integers.

So, let’s look closely at all of the functions given. We want to identify which of these is a sum of terms, which are products of constants and variables, and where any exponents are nonnegative integers. In fact, it might make sense to disregard any options where the exponents are negative. For instance, consider option (B). This is two π‘₯ to the power of negative two. Since this exponent is negative, we can disregard option (B). Similarly, option (C) has two monomial terms. Those are two π‘₯ and four. But we have π‘₯ to the power of negative two. Once again, since this is a negative integer, we can disregard option (C). So, we’re left with three remaining options.

Now (D) is equal to one over π‘₯. And we might recall that one over π‘₯ can be represented as a single power of π‘₯. It’s π‘₯ to the power of negative one. So, since π‘₯ is raised to a negative power, we can also actually disregard option (D). Similarly, if we consider option (E), we know that the square root of π‘₯ is equivalent to π‘₯ to the power of one-half. So, we can rewrite 𝑓 of π‘₯ here as π‘₯ to the power of one-half plus four. But of course, we know that the exponents must be integers. That’s whole numbers. One-half is not a whole number, so we disregard option (E).

And so, we see we’re left with option (A). Let’s double-check that each of these terms is a monomial. π‘₯ squared is a variable raised to a nonnegative integer exponent. Two π‘₯ is two times π‘₯ to the first power, so it’s a monomial. And four is equivalent to four π‘₯ to the zeroth power, so it’s also a monomial. And so, the answer is (A). 𝑓 of π‘₯ equals π‘₯ squared plus two π‘₯ plus four is a polynomial function.

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