To demonstrate how multiplying polynomials is a closed system, we'll multiply the two polynomials \( (x^2 - 1)(-2x + 5) \) and show that the result is also a polynomial.

Question

Let's perform the multiplication step-by-step:

1. **Distribute each term in the first polynomial by each term in the second polynomial.**

We can apply the distributive property (also known as the FOIL method for binomials):

\[
(x^2 - 1)(-2x + 5) = x^2 \cdot (-2x) + x^2 \cdot 5 - 1 \cdot (-2x) - 1 \cdot 5
\]

2. **Calculate each term:**

- First term: \( x^2 \cdot (-2x) = -2x^3 \)
- Second term: \( x^2 \cdot 5 = 5x^2 \)
- Third term: \(-1 \cdot (-2x) = 2x\)
- Fourth term: \(-1 \cdot 5 = -5\)

3. **Combine all these results:**

\[
-2x^3 + 5x^2 + 2x - 5
\]

4. **Resulting polynomial:**

The resulting expression \( -2x^3 + 5x^2 + 2x - 5 \) is a polynomial of degree 3.

### Conclusion

The multiplication of the two polynomials \( (x^2 - 1) \) and \( (-2x + 5) \) resulted in another polynomial \( -2x^3 + 5x^2 + 2x - 5 \). This indicates that the set of polynomials is closed under multiplication—i.e., the product of two polynomials is always a polynomial.
make this way smaller

Answer 1

To demonstrate that multiplying polynomials is a closed system, we'll multiply the polynomials \( (x^2 - 1)(-2x + 5) \):

1. Distribute:
   Using the distributive property, we have: \[ (x^2 - 1)(-2x + 5) = x^2 \cdot (-2x) + x^2 \cdot 5 - 1 \cdot (-2x) - 1 \cdot 5 \]

2. Calculate each term:

• \( x^2 \cdot (-2x) = -2x^3 \)
• \( x^2 \cdot 5 = 5x^2 \)
• \( -1 \cdot (-2x) = 2x \)
• \( -1 \cdot 5 = -5 \)

3. Combine results:
   \[ -2x^3 + 5x^2 + 2x - 5 \]

4. Conclusion:
   The result \( -2x^3 + 5x^2 + 2x - 5 \) is a polynomial, demonstrating that the set of polynomials is closed under multiplication.