To demonstrate that multiplying polynomials is a closed system, we will multiply the polynomials \( (x^2 - 1) \) and \( (-2x + 5) \) and show that the result is also a polynomial.
Let's carry out the multiplication step by step:
-
Distribute each term in the first polynomial by each term in the second polynomial:
-
Multiply \( x^2 \) by both terms in \( (-2x + 5) \): \[ x^2 \cdot (-2x) = -2x^3 \] \[ x^2 \cdot 5 = 5x^2 \]
-
Multiply \(-1\) by both terms in \( (-2x + 5) \): \[ -1 \cdot (-2x) = 2x \] \[ -1 \cdot 5 = -5 \]
-
-
Combine all the results together: \[ -2x^3 + 5x^2 + 2x - 5 \]
-
Reorganize it in standard form: \[ -2x^3 + 5x^2 + 2x - 5 \]
Now, we check if the result \( -2x^3 + 5x^2 + 2x - 5 \) is a polynomial.
Polynomial Definition
A polynomial in one variable \( x \) is defined as a sum of terms of the form \( ax^n \), where \( a \) is a coefficient (which can be any real number) and \( n \) is a non-negative integer.
Analyzing the Result
- The result \( -2x^3 + 5x^2 + 2x - 5 \) has:
- The term \( -2x^3 \): where \( a = -2 \) and \( n = 3 \) (a non-negative integer).
- The term \( 5x^2 \): where \( a = 5 \) and \( n = 2 \) (a non-negative integer).
- The term \( 2x \): where \( a = 2 \) and \( n = 1 \) (a non-negative integer).
- The term \( -5 \): where \( a = -5 \) and \( n = 0 \) (a non-negative integer).
Since all terms follow the polynomial form with non-negative integer exponents, the result is indeed a polynomial.
Conclusion
The product of the two polynomials \( (x^2 - 1) \) and \( (-2x + 5) \) is \( -2x^3 + 5x^2 + 2x - 5 \), which is also a polynomial. This demonstrates that the set of polynomials is closed under multiplication since the product of any two polynomials yields another polynomial.
1 answer