Question

find the error your classmate says that polynomials are not closed under addition and gives this counterexample, which he says is not a polynomial: ((x^2 - 2x) + (-x^2 + 2x) = 0) because the sum is 0. what definition can you provide to show him correct the error?
both (0x^2 + 0x) and (0) are (\boldsymbol{\text{select choice}})
i can remind my classmate that (0x^2) and (0x) are both (\boldsymbol{\text{select choice}}) because they are a product of a number, (0), and one or more variables. (0) is also a (\boldsymbol{\text{select choice}}) because...
therefore, the solution is (\boldsymbol{\text{select choice}})

Explanation:

Step 1: Recall Polynomial Closure Under Addition

Polynomials are closed under addition, meaning the sum of two polynomials is always a polynomial. Let's analyze the given counterexample: \((x^2 - 2x)+(-x^2 + 2x)\).

Step 2: Simplify the Expression

Combine like terms:

• For the \(x^2\) terms: \(x^2+(-x^2)=x^2 - x^2 = 0\)
• For the \(x\) terms: \(-2x + 2x=0\)

So, \((x^2 - 2x)+(-x^2 + 2x)=0 + 0x=0\), and \(0\) is a polynomial (a constant polynomial, which can be written as \(0x^n+\dots+0x + 0\) for any non - negative integer \(n\)).

Step 3: Identify the Error

The classmate's error is in claiming that the result \(0\) is not a polynomial. Polynomials include constant polynomials (like \(0\), \(5\), \(-3\) etc.). Also, when we have terms like \(0x^2\) and \(0x\), these are valid polynomial terms (since a polynomial is an expression of the form \(a_nx^n+\dots+a_1x + a_0\) where \(n\) is a non - negative integer and \(a_i\) are constants, and here \(a_i = 0\) for the non - constant terms and \(a_0 = 0\) for the constant term in the case of \(0\)).

Answer:

The classmate's error is thinking that \(0\) (the result of \((x^2 - 2x)+(-x^2 + 2x)\)) is not a polynomial. In fact, \(0\) is a polynomial (a constant polynomial), and polynomials are closed under addition. Also, \(0x^2\) and \(0x\) are valid polynomial terms (since a polynomial can have coefficients of \(0\) for some powers of \(x\)).