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Use the following multiplication problem to show how multiplying polynomials is a closed system. (x2−1)(−2x+5)Asked by lol
Use the following multiplication problem to show how multiplying polynomials is a closed system (x^2 - 1) (-2x + 5)
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Answered by
GPT-4o mini
To demonstrate how multiplying polynomials forms a closed system, let's multiply the given polynomials: \((x^2 - 1)\) and \((-2x + 5)\). A closed system means that when we carry out the operations (in this case, multiplication), we remain within the set of polynomials.
### Step 1: Distribute the terms
We will distribute each term in the first polynomial by each term in the second polynomial:
\[
(x^2 - 1)(-2x + 5) = x^2 \cdot -2x + x^2 \cdot 5 - 1 \cdot -2x - 1 \cdot 5
\]
### Step 2: Calculate each part
1. \(x^2 \cdot -2x = -2x^3\)
2. \(x^2 \cdot 5 = 5x^2\)
3. \(-1 \cdot -2x = 2x\)
4. \(-1 \cdot 5 = -5\)
### Step 3: Combine like terms
Now, we will combine these results:
\[
-2x^3 + 5x^2 + 2x - 5
\]
### Step 4: Result as a polynomial
The result \(-2x^3 + 5x^2 + 2x - 5\) is a polynomial of degree 3. Observing this result, we see that it is still in the form of a polynomial. Thus, the multiplication of the two polynomials has resulted in another polynomial.
### Conclusion
This demonstrated that the multiplication of the polynomials \((x^2 - 1)\) and \((-2x + 5)\) produces another polynomial, indicating that the set of polynomials is closed under multiplication. In other words, multiplying polynomials together does not produce any results outside the set of polynomials.
### Step 1: Distribute the terms
We will distribute each term in the first polynomial by each term in the second polynomial:
\[
(x^2 - 1)(-2x + 5) = x^2 \cdot -2x + x^2 \cdot 5 - 1 \cdot -2x - 1 \cdot 5
\]
### Step 2: Calculate each part
1. \(x^2 \cdot -2x = -2x^3\)
2. \(x^2 \cdot 5 = 5x^2\)
3. \(-1 \cdot -2x = 2x\)
4. \(-1 \cdot 5 = -5\)
### Step 3: Combine like terms
Now, we will combine these results:
\[
-2x^3 + 5x^2 + 2x - 5
\]
### Step 4: Result as a polynomial
The result \(-2x^3 + 5x^2 + 2x - 5\) is a polynomial of degree 3. Observing this result, we see that it is still in the form of a polynomial. Thus, the multiplication of the two polynomials has resulted in another polynomial.
### Conclusion
This demonstrated that the multiplication of the polynomials \((x^2 - 1)\) and \((-2x + 5)\) produces another polynomial, indicating that the set of polynomials is closed under multiplication. In other words, multiplying polynomials together does not produce any results outside the set of polynomials.
Answered by
GPT-4o mini
Certainly! When we multiply the polynomials \((x^2 - 1)\) and \((-2x + 5)\), we find that the result is \(-2x^3 + 5x^2 + 2x - 5\). This expression is still a polynomial, which means we have not moved outside the realm of polynomials.
This illustrates the concept of a closed system: when we multiply two polynomials, the product remains a polynomial. Therefore, the set of polynomials is closed under multiplication, as the operation produces results that stay within the same set.
This illustrates the concept of a closed system: when we multiply two polynomials, the product remains a polynomial. Therefore, the set of polynomials is closed under multiplication, as the operation produces results that stay within the same set.
Answered by
GPT-4o mini
Sure! When we take the polynomials \((x^2 - 1)\) and \((-2x + 5)\) and multiply them together, the outcome is \(-2x^3 + 5x^2 + 2x - 5\). Since this result is itself a polynomial, it shows that multiplying polynomials does not lead us outside of their group.
In essence, this reinforces the idea that the multiplication of polynomials forms a closed system: no matter how many polynomials you multiply together, you will always end up with another polynomial, confirming that this structure is self-contained.
In essence, this reinforces the idea that the multiplication of polynomials forms a closed system: no matter how many polynomials you multiply together, you will always end up with another polynomial, confirming that this structure is self-contained.