To demonstrate that polynomials form a closed system under multiplication, we need to show that when you multiply two polynomials, the result is also a polynomial.
Let's evaluate each option:
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\((\sqrt{x})(x+1)\)
- \(\sqrt{x}\) is not a polynomial (since a polynomial must have non-negative integer exponents). Therefore, their product \((\sqrt{x})(x+1)\) is not a polynomial.
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\((x^2+1)(x-\frac{1}{2})\)
- Both \(x^2 + 1\) and \(x - \frac{1}{2}\) are polynomials. The product of two polynomials is also a polynomial. Therefore, \((x^2 + 1)(x - \frac{1}{2})\) is a polynomial.
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\(\left(\frac{x^2}{2}\right)\left(\frac{1}{x}\right)\)
- While \(\frac{x^2}{2}\) is a polynomial, \(\frac{1}{x}\) is not a polynomial (it has a negative exponent). Therefore, their product \(\left(\frac{x^2}{2}\right)\left(\frac{1}{x}\right) = \frac{x}{2}\) is not a polynomial since \(\frac{x}{2}\) can be considered a polynomial of degree 1, but it is important to note that \(\frac{1}{x}\) disqualifies it from being categorized as a polynomial under strict definitions (which require integer exponents).
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\(x^2 + 2x + 1\)
- This expression is a polynomial itself, but it does not demonstrate closure under multiplication as it is just a single polynomial and does not involve multiplying two separate polynomials.
The only response that demonstrates polynomial multiplication resulting in another polynomial is:
\((x^2+1)(x-\frac{1}{2})\) because both factors are polynomials, and their product will also be a polynomial.
1 answer