Which expression is equivalent to -4x - 6?

To find which expression is equivalent to \(-4x - 6\), we can simplify each option.

**Option a:** \(-2(2x + 3)\)
\[
-2(2x + 3) = -2 \cdot 2x - 2 \cdot 3 = -4x - 6
\]
This matches \(-4x - 6\).

**Option b:** \(3(-x + 2)\)
\[
3(-x + 2) = 3 \cdot -x + 3 \cdot 2 = -3x + 6
\]
This does not match \(-4x - 6\).

**Option c:** \(8x - 6 - 4x\)
\[
8x - 6 - 4x = (8x - 4x) - 6 = 4x - 6
\]
This does not match \(-4x - 6\).

**Option d:** \(-4x + 3(-3x - 2)\)
\[
3(-3x - 2) = 3 \cdot -3x + 3 \cdot -2 = -9x - 6
\]
Now combining it with \(-4x\):
\[
-4x - 9x - 6 = -13x - 6
\]
This does not match \(-4x - 6\).

The only expression equivalent to \(-4x - 6\) is:

**a. -2(2x + 3)**

According to the Distributive Property which expression is equivalent to 5x^2+10x+15?

a. 5(x^2+2x+3)
b. 5(x^2+10x+3)
c. 5(x^2+10x+15)
d. x(5x+10+15)

To find the expression equivalent to \(5x^2 + 10x + 15\) using the Distributive Property, we can factor out the common factor, which is \(5\).

Starting with the expression:
\[
5x^2 + 10x + 15
\]

Factoring out \(5\):
\[
5(x^2 + 2x + 3)
\]

Now, let's evaluate each option:

**Option a:** \(5(x^2 + 2x + 3)\)
- This matches our factored expression.

**Option b:** \(5(x^2 + 10x + 3)\)
- This does not match the original expression because the term \(10x\) is incorrect.

**Option c:** \(5(x^2 + 10x + 15)\)
- This does not match the original expression because the \(10x\) and \(15\) are inside the parentheses incorrectly.

**Option d:** \(x(5x + 10 + 15)\)
- Simplifying this gives:
\[
x(5x + 25) = 5x^2 + 25x
\]
- This does not match the original expression.

The only expression equivalent to \(5x^2 + 10x + 15\) is:

**a. \(5(x^2 + 2x + 3)\)**

Simplify the expression

3x^2 - 2x^2 + 4x - 6

To simplify the expression \(3x^2 - 2x^2 + 4x - 6\), combine like terms.

1. Combine the \(x^2\) terms:
\[
3x^2 - 2x^2 = (3 - 2)x^2 = 1x^2 = x^2
\]

2. The \(4x\) term remains the same since there are no other \(x\) terms to combine:
\[
4x
\]

3. The constant term \(-6\) also remains the same.

Putting it all together:
\[
x^2 + 4x - 6
\]

Thus, the simplified expression is:
\[
\boxed{x^2 + 4x - 6}
\]

Add the polynomial expressions

3y - 4y^2 + 7y + 2 and 5y^2 + 6 - 8y.

To add the polynomial expressions \(3y - 4y^2 + 7y + 2\) and \(5y^2 + 6 - 8y\), we first need to combine like terms.

### Step 1: Write down the expressions
\[
(3y - 4y^2 + 7y + 2) + (5y^2 + 6 - 8y)
\]

### Step 2: Combine like terms
Let's rearrange the terms to group them by their degrees:

- **\(y^2\) terms:** \(-4y^2 + 5y^2\)
- **\(y\) terms:** \(3y + 7y - 8y\)
- **Constant terms:** \(2 + 6\)

### Step 3: Simplify each group
1. **For \(y^2\) terms:**
\[
-4y^2 + 5y^2 = (5 - 4)y^2 = 1y^2 = y^2
\]

2. **For \(y\) terms:**
\[
3y + 7y - 8y = (3 + 7 - 8)y = (10 - 8)y = 2y
\]

3. **For constant terms:**
\[
2 + 6 = 8
\]

### Step 4: Combine all simplified terms
Now, combine the results:
\[
y^2 + 2y + 8
\]

Thus, the result of adding the two polynomial expressions is:
\[
\boxed{y^2 + 2y + 8}
\]

Subtract.

(7x^2 + 11x + 9) - (3x^2 + 5x + 4)

To subtract the polynomial expressions \( (7x^2 + 11x + 9) - (3x^2 + 5x + 4) \), we can follow these steps:

### Step 1: Distribute the negative sign
When subtracting, we will distribute the negative sign across the second polynomial:
\[
7x^2 + 11x + 9 - 3x^2 - 5x - 4
\]

### Step 2: Rearrange the expression
Now, we can rewrite the expression while grouping like terms:
\[
(7x^2 - 3x^2) + (11x - 5x) + (9 - 4)
\]

### Step 3: Combine like terms
Now we combine the like terms:

1. **For \(x^2\) terms:**
\[
7x^2 - 3x^2 = 4x^2
\]

2. **For \(x\) terms:**
\[
11x - 5x = 6x
\]

3. **For constant terms:**
\[
9 - 4 = 5
\]

### Step 4: Combine all simplified terms
Putting everything together gives us:
\[
4x^2 + 6x + 5
\]

Thus, the result of the subtraction is:
\[
\boxed{4x^2 + 6x + 5}
\]

Multiply the polynomials

(2x - 2)(3x + 5)

To multiply the polynomials \((2x - 2)(3x + 5)\), we can use the distributive property (also known as the FOIL method in the case of binomials). Here's how it works step-by-step:

### Step 1: Distribute each term in the first polynomial to each term in the second polynomial.

\[
(2x - 2)(3x + 5) = 2x \cdot 3x + 2x \cdot 5 - 2 \cdot 3x - 2 \cdot 5
\]

### Step 2: Perform the multiplications.

1. **First Term:**
\[
2x \cdot 3x = 6x^2
\]

2. **Outer Term:**
\[
2x \cdot 5 = 10x
\]

3. **Inner Term:**
\[
-2 \cdot 3x = -6x
\]

4. **Last Term:**
\[
-2 \cdot 5 = -10
\]

### Step 3: Combine all resulting terms.
Now we can combine these results:
\[
6x^2 + 10x - 6x - 10
\]

### Step 4: Combine like terms.
Combine the \(x\) terms:
\[
10x - 6x = 4x
\]

So, we have:
\[
6x^2 + 4x - 10
\]

### Final Result
Thus, the result of multiplying the polynomials is:
\[
\boxed{6x^2 + 4x - 10}
\]

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

a. x^2 + 3x/x -1
b. (2x^2 + 3x) x (x - 4)
c. 1/x - 3 x 5

To determine which response shows that polynomials form a closed system under multiplication, we need to check if the results of the multiplication in each option yield a polynomial.

### Analyzing Each Option

**Option a:** \( \frac{x^2 + 3x}{x - 1} \)

This is a **rational expression** (a ratio of two polynomials), not a polynomial. Therefore, it does not demonstrate closure under multiplication.

---

**Option b:** \( (2x^2 + 3x) \cdot (x - 4) \)

To check closure under multiplication, we will multiply these two polynomials:
\[
(2x^2 + 3x)(x - 4)
\]

Using the distributive property:
\[
= 2x^2 \cdot x + 2x^2 \cdot (-4) + 3x \cdot x + 3x \cdot (-4)
= 2x^3 - 8x^2 + 3x^2 - 12x
= 2x^3 - 5x^2 - 12x
\]

The result, \(2x^3 - 5x^2 - 12x\), is a polynomial.

---

**Option c:** \( \frac{1}{x - 3} \cdot 5 \)

This expression involves a constant multiplied by a rational expression. The result will also be a rational expression, not a polynomial:
\[
= \frac{5}{x - 3}
\]

---

### Conclusion

From the above analysis, the only option that shows that polynomials form a closed system under multiplication is:

**b. \( (2x^2 + 3x) \cdot (x - 4) \)**

Thus, the correct answer is **b**.

Divide

20x^2 + 15x + 10 by -5x

To divide the polynomial \(20x^2 + 15x + 10\) by \(-5x\), we can divide each term of the polynomial separately by \(-5x\).

### Step 1: Divide each term

1. **First term:**
\[
\frac{20x^2}{-5x} = -4x
\]

2. **Second term:**
\[
\frac{15x}{-5x} = -3
\]

3. **Third term:**
\[
\frac{10}{-5x} = -\frac{2}{x}
\]

### Step 2: Combine the results
Putting all the results together, we have:
\[
-4x - 3 - \frac{2}{x}
\]

### Conclusion
Thus, the result of dividing \(20x^2 + 15x + 10\) by \(-5x\) is:
\[
\boxed{-4x - 3 - \frac{2}{x}}
\]

The design for a rectangular solid shipping container has a volume of 30 x^3 + 5x^2 - 15x where x represents inches. The box has a height of 5x. Find the container’s cross-sectional area using polynomial division.

To find the container's cross-sectional area using polynomial division, we need to divide the volume polynomial \(30x^3 + 5x^2 - 15x\) by the height polynomial \(5x\). The cross-sectional area can be thought of as the volume divided by the height.

### Step 1: Set up the polynomial division

We are dividing:
\[
30x^3 + 5x^2 - 15x \quad \text{ by } \quad 5x
\]

### Step 2: Perform the division

1. **Divide the leading term:**
Divide the leading term of the numerator by the leading term of the denominator:
\[
\frac{30x^3}{5x} = 6x^2
\]

2. **Multiply and subtract:**
Now, multiply \(6x^2\) by \(5x\) and subtract from the original polynomial:
\[
6x^2 \cdot 5x = 30x^3
\]
\[
(30x^3 + 5x^2 - 15x) - 30x^3 = 5x^2 - 15x
\]

3. **Repeat the process:**
Now, divide the leading term \(5x^2\) by \(5x\):
\[
\frac{5x^2}{5x} = x
\]
Multiply \(x\) by \(5x\) and subtract:
\[
x \cdot 5x = 5x^2
\]
\[
(5x^2 - 15x) - 5x^2 = -15x
\]

4. **Continue dividing:**
Next, divide \(-15x\) by \(5x\):
\[
\frac{-15x}{5x} = -3
\]
Multiply \(-3\) by \(5x\) and subtract:
\[
-3 \cdot 5x = -15x
\]
\[
(-15x) - (-15x) = 0
\]

The remainder is 0, which means the division is exact.

### Step 3: Result
Putting everything together, we get:
\[
\text{Cross-sectional area} = 6x^2 + x - 3
\]

Thus, the container’s cross-sectional area is:
\[
\boxed{6x^2 + x - 3}
\]

Using the Commutative Property, prove that these algebraic expressions are equivalent. In 3–4 sentences explain your result.

Expression 1: z * 4x
Expression 2: 4xz

To prove that the expressions \( z \cdot 4x \) and \( 4xz \) are equivalent using the Commutative Property of multiplication, we can rearrange the factors in either expression. The Commutative Property states that for any real numbers \( a \) and \( b \), \( a \cdot b = b \cdot a \).

Starting with Expression 1:
\[
z \cdot 4x = (z \cdot 4) \cdot x \quad \text{(using the associative property)}
\]
Now rearranging, we can switch the order:
\[
= 4 \cdot z \cdot x = 4xz
\]

Thus, we see that both expressions can be manipulated to show they yield the same product, confirming their equivalence. Since multiplication is commutative, the two expressions \( z \cdot 4x \) and \( 4xz \) represent the same quantity, demonstrating that they are indeed equivalent.

Using the Commutative Property, prove that these algebraic expressions are equivalent. In 3–4 sentences explain your result.

Expression 1: z * 4x
Expression 2: 4xz

(original and like a 9th grader wrote it)

To show that the expressions \( z \cdot 4x \) and \( 4xz \) are equivalent using the Commutative Property of multiplication, we can rearrange the factors in either expression. The Commutative Property says that when you multiply two numbers, it doesn't matter in what order you multiply them; you will get the same result.

For Expression 1, if we take \( z \cdot 4x \), we can rewrite it as \( 4 \cdot z \cdot x \) since multiplication is associative. Now, if we rearrange the factors, we get \( 4xz \), which is exactly Expression 2. This shows that both expressions represent the same value, proving they are equivalent.

Use the Associative Property to demonstrate that Expression 1 is equivalent to Expression 2.

Expression 1: 25r - 15
Expression 2: 20r + (5r - 3) - 12

To demonstrate that the expressions \( 25r - 15 \) (Expression 1) and \( 20r + (5r - 3) - 12 \) (Expression 2) are equivalent using the Associative Property, we can simplify Expression 2 and see if it matches Expression 1.

### Step 1: Simplify Expression 2

Starting with Expression 2:
\[
20r + (5r - 3) - 12
\]

### Step 2: Use the Associative Property

According to the Associative Property, we can group terms in any order. So, we first simplify inside the parentheses:

1. Expand the parentheses:
\[
20r + 5r - 3 - 12
\]

2. Now, group \(20r\) and \(5r\) together:
\[
(20r + 5r) + (-3 - 12)
\]

3. Combine the terms:
\[
25r + (-3 - 12) = 25r - 15
\]

### Step 3: Compare the Simplified Expression to Expression 1

Now we see that:
\[
25r - 15 \quad \text{(which we started with)}
\]

This matches Expression 1 exactly.

### Conclusion

By using the Associative Property to regroup the terms in Expression 2 and simplify, we showed that both expressions are indeed equivalent. Thus, \( 25r - 15 \) is equal to \( 20r + (5r - 3) - 12 \).

Use the Distributive Property to prove that these algebraic expressions are equivalent. Explain your answer in one sentence for each expression and show your work.

Expression 1: (x - 7)(x + 4)
Expression 2: x^2 - 3x - 28

To prove that the expressions \( (x - 7)(x + 4) \) (Expression 1) and \( x^2 - 3x - 28 \) (Expression 2) are equivalent using the Distributive Property, we will expand Expression 1 and simplify it to see if it matches Expression 2.

### Step 1: Expand Expression 1 using the Distributive Property

Start with Expression 1:
\[
(x - 7)(x + 4)
\]

#### Step 1.1: Apply the Distributive Property
Using the distributive property, we can multiply each term in the first binomial by each term in the second binomial:
\[
= x(x) + x(4) - 7(x) - 7(4)
\]

#### Step 1.2: Simplify the expression
Now calculating each term:
\[
= x^2 + 4x - 7x - 28
\]

Combine the like terms \(4x\) and \(-7x\):
\[
= x^2 - 3x - 28
\]

### Step 2: Conclusion
We have expanded Expression 1 to get \( x^2 - 3x - 28 \), which is exactly Expression 2. Thus, using the Distributive Property, we have proven that \( (x - 7)(x + 4) \) is equivalent to \( x^2 - 3x - 28 \).

Use the Distributive Property to prove that these algebraic expressions are equivalent. Explain your answer in one sentence for each expression and show your work.

Expression 1: (x - 7)(x + 4)
Expression 2: x^2 - 3x - 28

(original and like a 9th grader wrote it)

up-vote

To show that the expressions \( (x - 7)(x + 4) \) (Expression 1) and \( x^2 - 3x - 28 \) (Expression 2) are equivalent using the Distributive Property, we need to expand Expression 1 and see if it results in Expression 2.

### Step 1: Expand Expression 1

We start with Expression 1:
\[
(x - 7)(x + 4)
\]

### Step 2: Use the Distributive Property

We can use the Distributive Property by multiplying each term in the first group by each term in the second group:

1. **First, multiply \(x\) by both terms in \( (x + 4) \)**:
\[
x \cdot x + x \cdot 4 = x^2 + 4x
\]

2. **Next, multiply \(-7\) by both terms in \( (x + 4) \)**:
\[
-7 \cdot x - 7 \cdot 4 = -7x - 28
\]

### Step 3: Combine all the terms

Now, we put everything together:
\[
x^2 + 4x - 7x - 28
\]

### Step 4: Combine Like Terms

Now combine the \(4x\) and \(-7x\):
\[
x^2 + (4x - 7x) - 28 = x^2 - 3x - 28
\]

### Conclusion

Now we see that after expanding Expression 1, we get \( x^2 - 3x - 28 \), which is exactly Expression 2. Thus, we have proven that \( (x - 7)(x + 4) \) is equivalent to \( x^2 - 3x - 28 \) using the Distributive Property.

Use the Distributive Property to prove that these algebraic expressions are equivalent. Explain your answer in one sentence for each expression.

Expression 1: (x - 7)(x + 4)
Expression 2: x^2 - 3x - 28

(original and like a 9th grader wrote it)

To prove that the expressions \( (x - 7)(x + 4) \) (Expression 1) and \( x^2 - 3x - 28 \) (Expression 2) are equivalent using the Distributive Property, we will expand Expression 1 step by step.

### Step 1: Expand Expression 1

Starting with Expression 1:
\[
(x - 7)(x + 4)
\]

### Step 2: Use the Distributive Property

Now, we apply the Distributive Property, which means we will multiply each term in the first parentheses by each term in the second parentheses:

1. **First multiply \(x\) by both terms in \((x + 4)\)**:
\[
x \cdot x + x \cdot 4 = x^2 + 4x
\]

2. **Next, multiply \(-7\) by both terms in \((x + 4)\)**:
\[
-7 \cdot x - 7 \cdot 4 = -7x - 28
\]

### Step 3: Combine All Terms

Now, we combine all the terms from the two parts:
\[
x^2 + 4x - 7x - 28
\]

### Step 4: Combine Like Terms

Next, we combine the \(4x\) and \(-7x\):
\[
x^2 + (4x - 7x) - 28 = x^2 - 3x - 28
\]

### Conclusion

Now we see that after expanding Expression 1, we end up with \( x^2 - 3x - 28 \), which is exactly Expression 2. This shows that \( (x - 7)(x + 4) \) is equivalent to \( x^2 - 3x - 28 \) using the Distributive Property.

Use the Distributive Property to prove that these algebraic expressions are equivalent. Explain your answer in 3-4 sentences for each expression.

Expression 1: (x - 7)(x + 4)
Expression 2: x^2 - 3x - 28

(original and like a 9th grader wrote it)

To prove that the expressions \( (x - 7)(x + 4) \) (Expression 1) and \( x^2 - 3x - 28 \) (Expression 2) are equivalent, we can use the Distributive Property step by step.

### Expression 1: \( (x - 7)(x + 4) \)

We start with Expression 1, which is \( (x - 7)(x + 4) \). Using the Distributive Property, we multiply each term in the first parentheses by each term in the second parentheses. First, we take \( x \) from the first parentheses and distribute it to both terms in the second parentheses, giving us \( x \cdot x + x \cdot 4 = x^2 + 4x \). Then, we take \(-7\) from the first parentheses and distribute it to both terms in the second parentheses, resulting in \(-7 \cdot x - 7 \cdot 4 = -7x - 28\).

Now we combine all the terms together: \( x^2 + 4x - 7x - 28 \). After combining the like terms \( 4x \) and \(-7x\), we get \( x^2 - 3x - 28 \).

### Expression 2: \( x^2 - 3x - 28 \)

Now let’s look at Expression 2, which is already written as \( x^2 - 3x - 28 \). This expression clearly shows the terms with \( x^2 \), the linear term \(-3x\), and the constant term \(-28\). We can see that this expression is the expanded result from Expression 1 after applying the Distributive Property.

Since we have shown that expanding Expression 1 leads us precisely to the form in Expression 2, we have proven that \( (x - 7)(x + 4) \) is equivalent to \( x^2 - 3x - 28 \). This confirms that both expressions represent the same polynomial.

In 3–4 sentences explain why the following operation is a closed system under subtraction.

(8x^2 + 14x + 10) - (3x^2 + 9x + 8)

The operation of subtraction involving polynomials is a closed system because subtracting one polynomial from another always results in another polynomial. In the example \( (8x^2 + 14x + 10) - (3x^2 + 9x + 8) \), we rearrange the terms to perform the subtraction, which results in the new polynomial \( (8x^2 - 3x^2) + (14x - 9x) + (10 - 8) \). This gives us \( 5x^2 + 5x + 2 \), which is still a polynomial. Thus, since both the input expressions and the result are polynomials, we can conclude that the operation is closed under subtraction.

In 3–4 sentences explain why the following operation is a closed system under subtraction.

(8x^2 + 14x + 10) - (3x^2 + 9x + 8)

(original and like a 9th grader wrote it)

The operation of subtraction with polynomials is a closed system because when we subtract one polynomial from another, we always end up with another polynomial. In the case of \( (8x^2 + 14x + 10) - (3x^2 + 9x + 8) \), if we do the subtraction, we rearrange it to \( 8x^2 - 3x^2 + 14x - 9x + 10 - 8 \). This simplifies to \( 5x^2 + 5x + 2 \), which is also a polynomial. Since both the original expressions and the result are polynomials, it shows that the system is closed under subtraction.

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

(x^2 - 2)(-3x + 7)

To show that multiplying polynomials is a closed system, we will multiply the polynomials \( (x^2 - 2) \) and \( (-3x + 7) \) and demonstrate that the result is also a polynomial.

### Step 1: Set Up the Multiplication

We start with the expression \( (x^2 - 2)(-3x + 7) \).

### Step 2: Use the Distributive Property

We can apply the Distributive Property (also known as the FOIL method for binomials) to multiply each term in the first polynomial by each term in the second polynomial:
1. **First, multiply \( x^2 \) by both terms in \((-3x + 7)\)**:
- \( x^2 \cdot (-3x) = -3x^3 \)
- \( x^2 \cdot 7 = 7x^2 \)

2. **Next, multiply \(-2\) by both terms in \((-3x + 7)\)**:
- \(-2 \cdot (-3x) = 6x\)
- \(-2 \cdot 7 = -14\)

### Step 3: Combine All Terms

Now, we combine all of the results from above:
\[
-3x^3 + 7x^2 + 6x - 14
\]

### Step 4: Final Expression

The final expression, \( -3x^3 + 7x^2 + 6x - 14 \), is a polynomial since it consists of terms with non-negative integer exponents and combines constants.

### Conclusion

Since the product of the two polynomials \( (x^2 - 2) \) and \( (-3x + 7) \) resulted in another polynomial, we have shown that multiplying polynomials is a closed system. This means that multiplying any two polynomials will always yield another polynomial.