### Video Transcript

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

Letβs begin by reminding ourselves what we mean when we talk about a polynomial function. A polynomial function is one thatβs made up as a sum of monomial terms. So, whatβs 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₯π¦ squared is a monomial. Itβs the product of a constant three and some variables π₯ and π¦, and their exponents are one and two, respectively. Four π₯ to the power of negative seven, however, is not a monomial, and thatβs because the exponent of the variable is negative seven.

So, letβs begin by identifying which of our functions are polynomials, are a sum of monomials. Well, letβs begin with (B) π of π₯ equals two. This is the same as two π₯ to the zeroth power. Zero is nonnegative and itβs 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. Thatβs 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. Itβs the sum of two monomials, π₯ squared and negative four.

So, what about option (E)? Well, we know that three is a constant. Itβs 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 weβre multiplying a constant by a variable, π₯ to the first power. This means root two π₯ is a monomial. And so, π of π₯ is the sum of two monomials; itβs a polynomial.

And so, the answer must be option (A), but letβs double-check. We could rewrite π of π₯ as π₯ plus two to the power of negative one. Now, we canβt 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, π of π₯ equals one over π₯ plus two cannot be a polynomial. And so, the answer is (A). (A) is not a polynomial function.