Question

Use the following multiplication problem to show how multiplying polynomials is a closed system.

(x2−1)(−2x+5)(𝑥2−1)(−2𝑥+5)

KEEP IT SIMPLE
(2 points)

Answer 1

To demonstrate that multiplying polynomials is a closed system, we will multiply the given polynomials step by step. Let's consider the polynomials:

1. \( (x^2 - 1) \)
2. \( (-2x + 5) \)

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:

1. Multiply each term of \(-2x + 5\) with \(-2x^3\):

\[
-2x^3 \cdot -2x = 4x^4
\]
\[
-2x^3 \cdot 5 = -10x^3
\]

2. Multiply each term of \(-2x + 5\) with \(5x^2\):

\[
5x^2 \cdot -2x = -10x^3
\]
\[
5x^2 \cdot 5 = 25x^2
\]

3. Multiply each term of \(-2x + 5\) with \(2x\):

\[
2x \cdot -2x = -4x^2
\]
\[
2x \cdot 5 = 10x
\]

4. 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.