To demonstrate that polynomials form a closed system under multiplication, we need to verify that when you multiply any two polynomials, the result is also a polynomial.
Let's evaluate each option:
A. \((x^2+1)(x-1/2)\)
- Both \(x^2 + 1\) and \(x - 1/2\) are polynomials. Multiplying them will yield another polynomial.
- Result: This represents a polynomial.
B. \((\sqrt{x})(x+1)\)
- Here, \(\sqrt{x}\) can be rewritten as \(x^{1/2}\), which is not a polynomial since polynomials must have whole number (non-negative integer) exponents.
- Result: This does not represent a polynomial.
C. \((x^2/2)(1/x)\)
- This simplifies to \(\frac{x^2}{2} \cdot \frac{1}{x} = \frac{x}{2}\), which is a polynomial. While the multiplication does yield a polynomial, one of the factors here \(1/x\) is not a polynomial.
- Result: This does not represent a polynomial because \(1/x\) is not a polynomial.
D. \(x^2 + 2x + 1\)
- This expression is clearly a polynomial.
- Result: This represents a polynomial.
From these evaluations, the correct response that illustrates that polynomials form a closed system under multiplication is:
A: \((x^2+1)(x-1/2)\)
Though D is a polynomial itself, it doesn’t reflect the multiplication of two polynomials. Therefore, A clearly demonstrates the closure property.
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