To demonstrate that the multiplication of polynomials forms a closed system, we can follow through with the multiplication of the given expression \((x^2 - 1)(-2 + 5)\).
First, let's clarify what we're working with. The problem seems to contain a minor mistake in the second polynomial; we should assume you meant to multiply \((x^2 - 1)(-2 + 5x)\) instead of just \((-2 + 5)\) for a polynomial operation. For the sake of demonstration, I’ll proceed with the assumption of multiplying \((x^2 - 1)(5x - 2)\).
Step 1: Write Down the Polynomials
- First polynomial: \(P(x) = x^2 - 1\)
- Second polynomial: \(Q(x) = 5x - 2\)
Step 2: Distribute Each Term
We will distribute each term of the first polynomial to each term of the second polynomial:
\[ (x^2 - 1)(5x - 2) \]
Distributing \(x^2\):
\[ x^2 \cdot 5x + x^2 \cdot (-2) = 5x^3 - 2x^2 \]
Distributing \(-1\):
\[ -1 \cdot 5x + (-1) \cdot (-2) = -5x + 2 \]
Step 3: Combine Like Terms
Now we can combine all the terms from our distribution:
\[ 5x^3 - 2x^2 - 5x + 2 \]
Conclusion
The final result is:
\[ 5x^3 - 2x^2 - 5x + 2 \]
This result is still a polynomial, confirming closure under multiplication since we started with two polynomials and ended up with another polynomial.
Summary
In this example, we showed that multiplying the polynomials \((x^2 - 1)\) and \((5x - 2)\) resulted in another polynomial \(5x^3 - 2x^2 - 5x + 2\). Therefore, the set of polynomials is closed under multiplication, as multiplying any two polynomials yields another polynomial in the same set.