Question

which algebraic expression is a trinomial?
○ $x^3 + x^2 - \sqrt{x}$
○ $2x^3 - x^2$
○ $4x^3 + x^2 - \frac{1}{x}$
○ $x^6 - x + \sqrt{6}$

Explanation:

Step1: Recall trinomial definition

A trinomial is a polynomial with three terms, and a polynomial consists of terms with non - negative integer exponents of the variable and real coefficients.

Step2: Analyze option 1

For the expression \(x^{3}+x^{2}-\sqrt{x}=x^{3}+x^{2}-x^{\frac{1}{2}}\), the exponent of \(x\) in the third term is \(\frac{1}{2}\), which is not a non - negative integer. So, it is not a polynomial, let alone a trinomial.

Step3: Analyze option 2

The expression \(2x^{3}-x^{2}\) has two terms, so it is a binomial, not a trinomial.

Step4: Analyze option 3

For the expression \(4x^{3}+x^{2}-\frac{1}{x}=4x^{3}+x^{2}-x^{- 1}\), the exponent of \(x\) in the third term is \(- 1\), which is not a non - negative integer. So, it is not a polynomial, let alone a trinomial.

Step5: Analyze option 4

The expression \(x^{6}-x+\sqrt{6}\) has three terms. The exponents of \(x\) in the terms \(x^{6}\), \(-x\) are \(6\) and \(1\) respectively (both non - negative integers), and \(\sqrt{6}\) is a constant term (which can be considered as a term with \(x^{0}\) since \(x^{0} = 1\) for \(x
eq0\)). So, it is a trinomial.

Answer:

\(x^{6}-x+\sqrt{6}\) (the fourth option: \(x^{6}-x+\sqrt{6}\))