To show that polynomials form a closed system under multiplication, we must demonstrate that the product of any two polynomials is also a polynomial.
Let’s denote two polynomials as follows:
\[ P(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 \] \[ Q(x) = b_m x^m + b_{m-1} x^{m-1} + \ldots + b_1 x + b_0 \]
where \(P(x)\) is a polynomial of degree \(n\) and \(Q(x)\) is a polynomial of degree \(m\), and \(a_i\) and \(b_j\) are coefficients (which can be real or complex numbers).
To find the product \(R(x) = P(x) \cdot Q(x)\), we calculate:
\[ R(x) = (a_n x^n + a_{n-1} x^{n-1} + \ldots + a_0)(b_m x^m + b_{m-1} x^{m-1} + \ldots + b_0) \]
We distribute all terms of \(P(x)\) across all terms of \(Q(x)\). This results in:
\[ R(x) = \sum_{i=0}^{n} \sum_{j=0}^{m} a_i b_j x^{i+j} \]
The degree of the resulting polynomial \(R(x)\) will be at most \(n + m\) (specifically, the leading term will be \(a_n \cdot b_m \cdot x^{n+m}\), assuming \(a_n \neq 0\) and \(b_m \neq 0\)).
The coefficients \(a_i b_j\) are also well-defined (since they are products of coefficients from the original polynomials), and they sum to form the resulting polynomial.
Thus, the product \(R(x)\) remains a polynomial, proving that the set of polynomials is closed under multiplication.
In conclusion, since the product of two polynomials yields another polynomial, we have shown that the polynomials form a closed system under multiplication.
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