Question

Multiply the polynomial expression (x+2)(x−4)
.(2 points)
( )x^2+ ( )x + ( )

Answer 1

Multiply to simplify the polynomial expression (r−1)(r^2−2r+3)
.(2 points)
( )r^3 + ( )r^2+ ( )r + ( )

Answer 2

Simplify the polynomial expression (xy−2)(x^2+1)

( )x^3y + ( )x^2 + ( )xy + ( )

Answer 3

Which of the following shows that polynomials form a closed system under multiplication?

Option #1: 3(2/x+5)=6/x+15

Option #2: 8x+7=5x+3x+1+6

Option #3: (2x^2−4)(3y+6)=6x^2y+12x^2−12y−24

Option #4: 2x^1/2⋅3=6x−−√x

(1 point)
Option #

Answer 4

To multiply the polynomial expression \((x+2)(x-4)\), we can use the distributive property (also known as the FOIL method for binomials).

\[
(x + 2)(x - 4) = x(x) + x(-4) + 2(x) + 2(-4)
\]

Now, let's simplify that step by step:

1. \(x \cdot x = x^2\)
2. \(x \cdot (-4) = -4x\)
3. \(2 \cdot x = 2x\)
4. \(2 \cdot (-4) = -8\)

Now, combine all these:

\[
x^2 - 4x + 2x - 8
\]

Next, combine like terms:

\(-4x + 2x = -2x\)

So, we have:

\[
x^2 - 2x - 8
\]

Thus, the final expression is:

\[
1x^2 + (-2)x + (-8)
\]

So, filling in the blanks:

\((1)x^2 + (-2)x + (-8)\).

Answer 5

To multiply and simplify the polynomial expression \((r-1)(r^2-2r+3)\), we can use the distributive property.

We will distribute \(r - 1\) to each term in \(r^2 - 2r + 3\):

1. **Distributing \(r\)**:
\[
r \cdot (r^2) = r^3
\]
\[
r \cdot (-2r) = -2r^2
\]
\[
r \cdot (3) = 3r
\]

2. **Distributing \(-1\)**:
\[
-1 \cdot (r^2) = -r^2
\]
\[
-1 \cdot (-2r) = 2r
\]
\[
-1 \cdot (3) = -3
\]

Now, let's combine all the terms:

\[
r^3 - 2r^2 + 3r - r^2 + 2r - 3
\]

Now, combine like terms:

- The \(r^2\) terms: \(-2r^2 - r^2 = -3r^2\)
- The \(r\) terms: \(3r + 2r = 5r\)

Putting it all together, we get:

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

Thus, the final simplified expression is:

\[
(1)r^3 + (-3)r^2 + (5)r + (-3)
\]

So, filling in the blanks:

\((1)r^3 + (-3)r^2 + (5)r + (-3)\).

Answer 6

To simplify the polynomial expression \((xy - 2)(x^2 + 1)\), we will use the distributive property.

We will distribute the terms of the first polynomial \(xy - 2\) to each term in the second polynomial \(x^2 + 1\):

1. **Distributing \(xy\)**:
- \(xy \cdot x^2 = xyx^2 = x^3y\)
- \(xy \cdot 1 = xy\)

2. **Distributing \(-2\)**:
- \(-2 \cdot x^2 = -2x^2\)
- \(-2 \cdot 1 = -2\)

Now, combine all these results together:

\[
x^3y - 2x^2 + xy - 2
\]

Now, let's rearrange and group the terms:

\[
x^3y + xy - 2x^2 - 2
\]

Thus, the simplified expression is:

\[
(1)x^3y + (-2)x^2 + (1)xy + (-2)
\]

So, filling in the blanks:

\((1)x^3y + (-2)x^2 + (1)xy + (-2)\).

Answer 7

To determine which option shows that polynomials form a closed system under multiplication, we need to recall that a set is closed under an operation if applying that operation to members of the set always produces a member of the set. In this case, we want to check if multiplying two polynomials results in a polynomial.

Let's analyze each option:

1. **Option #1:** \(3\left(\frac{2}{x}+5\right) = \frac{6}{x} + 15\)
- This is not an example of polynomial multiplication, as it involves a rational expression (\(\frac{2}{x}\)), which is not a polynomial.

2. **Option #2:** \(8x + 7 = 5x + 3x + 1 + 6\)
- This is an expression showing addition and equating two polynomial expressions. It does not demonstrate multiplication.

3. **Option #3:** \((2x^2 - 4)(3y + 6) = 6x^2y + 12x^2 - 12y - 24\)
- This shows the product of two polynomials \((2x^2 - 4)\) and \((3y + 6)\), which results in another polynomial. Therefore, this option demonstrates that polynomials are closed under multiplication.

4. **Option #4:** \(2x^{1/2} \cdot 3 = 6\sqrt{x}\)
- This involves a term \(x^{1/2}\) which is a radical expression, and thus it is not considered a polynomial.

Based on this analysis, the correct option that shows that polynomials form a closed system under multiplication is:

**Option #3.**