Question Video: Identifying a Polynomial Function Mathematics

Which of the following is not a polynomial function? [A] 𝑓(π‘₯) = 1/(π‘₯ + 2) [B] 𝑓(π‘₯) = 2 [C] 𝑓(π‘₯) = π‘₯⁴ βˆ’ 2π‘₯Β³ + 2 [D] 𝑓(π‘₯) = π‘₯Β² βˆ’ 4 [E] 𝑓(π‘₯) = √(2π‘₯) + 3.

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Video Transcript

Which of the following is not a polynomial function? Is it (A) 𝑓 of π‘₯ equals one over π‘₯ plus two? Is it (B) 𝑓 of π‘₯ equals two? (C) 𝑓 of π‘₯ is π‘₯ to the fourth power minus two π‘₯ cubed plus two. (D) 𝑓 of π‘₯ equals π‘₯ squared minus four. Or (E) 𝑓 of π‘₯ is equal to the square root of two π‘₯ plus three.

Let’s begin by reminding ourselves what we mean when we talk about a polynomial function. A polynomial function is one that’s made up as a sum of monomial terms. So, what’s a monomial term? Well, a monomial term is a single term made up of a constant and a variable, and that variable will have a nonnegative integer exponent. So, for instance, three π‘₯𝑦 squared is a monomial. It’s the product of a constant three and some variables π‘₯ and 𝑦, and their exponents are one and two, respectively. Four π‘₯ to the power of negative seven, however, is not a monomial, and that’s because the exponent of the variable is negative seven.

So, let’s begin by identifying which of our functions are polynomials, are a sum of monomials. Well, let’s begin with (B) 𝑓 of π‘₯ equals two. This is the same as two π‘₯ to the zeroth power. Zero is nonnegative and it’s an integer. And so, we have a term made up of a constant a, variable, and a nonnegative integer exponent. So, in fact, (B) is a polynomial function.

So, what about option (C)? Well, we have π‘₯ to the fourth power. That’s a variable raised to a nonnegative integer exponent. We have negative two π‘₯ cubed, a constant times π‘₯ raised to another nonnegative integer exponent, and we just saw that two itself is indeed a monomial term. So, (C) is the sum of three monomials and it must be a polynomial function. In a similar way, (D) is also a polynomial function. It’s the sum of two monomials, π‘₯ squared and negative four.

So, what about option (E)? Well, we know that three is a constant. It’s three π‘₯ to the zeroth power, which is a monomial. But what about the square root of two times π‘₯? Well, the square root of two is in fact a constant, so we’re multiplying a constant by a variable, π‘₯ to the first power. This means root two π‘₯ is a monomial. And so, 𝑓 of π‘₯ is the sum of two monomials; it’s a polynomial.

And so, the answer must be option (A), but let’s double-check. We could rewrite 𝑓 of π‘₯ as π‘₯ plus two to the power of negative one. Now, we can’t distribute these parentheses. And so, in fact, we have a function which is a sum of a pair of monomials, but that itself is raised to a negative integer exponent. And so, 𝑓 of π‘₯ equals one over π‘₯ plus two cannot be a polynomial. And so, the answer is (A). (A) is not a polynomial function.

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