Polynomials in one variable

A polynomial in one variable $x$ is any expression that can be written in the general form

where: $n$ is a non-negative whole number (so $n = 0, 1, 2, 3, \dots$ — never negative, never a fraction), each coefficient $a_n, a_{n-1}, \dots, a_0$ is a real number, and $a_n \neq 0$ so that $a_n x^n$ is genuinely the highest term present. Four pieces of vocabulary do almost all the work in this chapter. A term is a piece of the polynomial separated from the rest by a $+$ or $-$ sign — in $3x^2 - 5x + 7$ the terms are $3x^2$, $-5x$, and $7$. The coefficient of a power of $x$ is the real number multiplying it, so the coefficient of $x^2$ above is $3$; a bare $x$ (with no visible number in front) always carries coefficient $1$. The constant term is the term with no $x$ attached at all — here, $7$. The degree is the highest power of $x$ that appears with a non-zero coefficient; the degree of $3x^2 - 5x + 7$ is $2$. The one exception is the zero polynomial, the expression $0$ itself, whose degree is left undefined because there is no highest power to point to. Polynomials sort neatly by degree — constant (degree $0$, e.g. $p(x)=5$), linear (degree $1$, e.g. $2x+3$), quadratic (degree $2$, e.g. $x^2-4x+4$), cubic (degree $3$, e.g. $x^3-2x^2+x-1$) — and separately by how many terms they have: a monomial has one term ($7x^3$), a binomial has two ($x+1$), a trinomial has three ($x^2+2x+1$). These two ways of sorting are independent: a binomial can have any degree at all, so $x^{100}+1$ is still just a binomial, of degree $100$.

What breaks the rule matters just as much as what satisfies it. Expressions such as $\sqrt{x}+2$ (the power of $x$ is $\tfrac12$, a fraction), $\dfrac{3}{x}+1$ (this is $3x^{-1}+1$, a negative power), or $x^2+\dfrac{1}{x-1}$ are not polynomials, because somewhere in them a power of $x$ is not a non-negative whole number.

Question: For $p(x) = 4x^3 - 3x^2 + x - 7$, list the terms and coefficients, give the constant term and the degree, name the polynomial by degree and by term count, and then decide whether $q(x) = 2x^{-1} + x^2$ is a polynomial.

Step 1. Separate $p(x)$ at its $+$ and $-$ signs: the terms are $4x^3$, $-3x^2$, $x$, and $-7$ — four terms in total.

Step 2. Read off the coefficients directly: the coefficient of $x^3$ is $4$, of $x^2$ is $-3$, and of $x$ is $1$ (a bare $x$ always means $1\cdot x$). The constant term, the one with no $x$ attached, is $-7$.

Step 3. The exponents present are $3, 2, 1, 0$ — all non-negative whole numbers, so $p(x)$ is a polynomial. The highest of these is $3$, so its degree is $3$, making it a cubic polynomial; since it has four terms (more than three), it has no monomial/binomial/trinomial name of its own.

Step 4. Now test $q(x) = 2x^{-1} + x^2$ using the same rule: rewriting shows the first term carries exponent $-1$ on $x$, which is negative. That single failing exponent is enough to disqualify the whole expression — $q(x)$ is not a polynomial, regardless of the perfectly well-behaved $x^2$ sitting next to it.

Final answer: $p(x)$ is a cubic polynomial with four terms, constant term $-7$; $q(x)$ is not a polynomial.

• Reading the number of terms as the degree — a binomial like $x^{100}+1$ still has degree $100$, not $2$.
• Missing that $\dfrac{1}{x}$ and $\sqrt{x}$ are disguised negative and fractional exponents; any such term anywhere in the expression disqualifies it as a polynomial.
• Forgetting the invisible coefficient $1$ on a bare $x$, or dropping a minus sign when reading off a coefficient — the coefficient of $x^2$ in $5-x^2$ is $-1$, not $1$.
• Assuming a lone constant like $7$ "isn't a polynomial" — every real constant is a polynomial of degree $0$, except $0$ itself, whose degree is undefined.

Ask the tutor: If an expression mixes a square root of a number with a fraction involving the variable, ask the AI tutor to work through it term by term with you until spotting the disqualifying exponent becomes instant.