To demonstrate that multiplying polynomials is a closed system, we will multiply the given polynomials step by step. Let's consider the polynomials:
- \( (x^2 - 1) \)
- \( (-2x + 5) \)
We can start by multiplying these two expressions together.
Step 1: Multiply \( (x^2 - 1) \) and \( (-2x + 5) \)
Using the distributive property (also known as FOIL for binomials):
\[ (x^2 - 1)(-2x + 5) = x^2(-2x) + x^2(5) - 1(-2x) - 1(5) \]
Calculating each term:
- \( x^2 \cdot -2x = -2x^3 \)
- \( x^2 \cdot 5 = 5x^2 \)
- \( -1 \cdot -2x = 2x \)
- \( -1 \cdot 5 = -5 \)
Now, combine these results:
\[ -2x^3 + 5x^2 + 2x - 5 \]
Step 2: Multiply the result by itself
Now we take our result and multiply it by \( (x^2 - 1)(-2x + 5) \) again:
\[ (-2x^3 + 5x^2 + 2x - 5)(-2x + 5) \]
We can distribute this again:
- Multiply each term of \(-2x + 5\) with \(-2x^3\):
\[ -2x^3 \cdot -2x = 4x^4 \] \[ -2x^3 \cdot 5 = -10x^3 \]
- Multiply each term of \(-2x + 5\) with \(5x^2\):
\[ 5x^2 \cdot -2x = -10x^3 \] \[ 5x^2 \cdot 5 = 25x^2 \]
- Multiply each term of \(-2x + 5\) with \(2x\):
\[ 2x \cdot -2x = -4x^2 \] \[ 2x \cdot 5 = 10x \]
- Multiply each term of \(-2x + 5\) with \(-5\):
\[ -5 \cdot -2x = 10x \] \[ -5 \cdot 5 = -25 \]
Step 3: Combine all the terms
Now combine all those results together:
\[ 4x^4 + (-10x^3 - 10x^3) + (25x^2 - 4x^2) + (10x + 10x) - 25 \]
Simplify each part:
- The \(x^3\) terms: \(-10x^3 - 10x^3 = -20x^3\)
- The \(x^2\) terms: \(25x^2 - 4x^2 = 21x^2\)
- The \(x\) terms: \(10x + 10x = 20x\)
So the final result is:
\[ 4x^4 - 20x^3 + 21x^2 + 20x - 25 \]
Conclusion
The final result \(4x^4 - 20x^3 + 21x^2 + 20x - 25\) is also a polynomial. Since we started with polynomials and ended with a polynomial after performing multiplication, this demonstrates that the multiplication of polynomials is a closed system. The result is still within the set of polynomials.
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