Multiply the polynomial expression (x+2)(x−4)

Question

Multiply the polynomial expression  (x+2)(x−4)
 .(2 points)
x2+
x+

GPT-4o mini · Answer

To multiply the polynomial expression $(x + 2)(x - 4)$, we can use the distributive property (also known as the FOIL method for binomials): $$ (x + 2)(x - 4) = x \cdot x + x \cdot (-4) + 2 \cdot x + 2 \cdot (-4) $$ Now, let's break it down 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 results: $$ x^2 - 4x + 2x - 8 $$ Combine like terms: $$ x^2 + (-4x + 2x) - 8 = x^2 - 2x - 8 $$ Thus, the final result of multiplying $(x + 2)(x - 4)$ is: $$ x^2 - 2x - 8 $$ So, filling in the blanks: - $x^2 +$ - $-2x +$ - $-8$ The complete answer is: $$ x^2 - 2x - 8 $$

Answer

Multiply to simplify the polynomial expression (r−1)(r2−2r+3) .(2 points) r3+ r2+ r+

GPT-4o mini · Answer

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

Distribute $r$ across $(r^2 - 2r + 3)$:
- $r \cdot r^2 = r^3$
- $r \cdot (-2r) = -2r^2$
- $r \cdot 3 = 3r$
So, from this part, we have: \[ r^3 - 2r^2 + 3r \]
Now, distribute $-1$ across $(r^2 - 2r + 3)$:
- $-1 \cdot r^2 = -r^2$
- $-1 \cdot (-2r) = 2r$
- $-1 \cdot 3 = -3$
So, from this part, we have: \[ -r^2 + 2r - 3 \]
Now, combine all the terms: \[ r^3 - 2r^2 + 3r - r^2 + 2r - 3 \]
Combine like terms:
- For $r^2$: $-2r^2 - r^2 = -3r^2$
- For $r$: $3r + 2r = 5r$

Putting it all together, we get: \[ r^3 - 3r^2 + 5r - 3 \]

Thus, the simplified expression is: \[ r^3 - 3r^2 + 5r - 3 \]

So, filling in the blanks:

$r^3 +$
$-3r^2 +$
$5r - 3$

The complete answer is: \[ r^3 - 3r^2 + 5r - 3 \]

Answer

Simplify the polynomial expression (xy−2)(x2+1) .(2 points) x3y+ x2+ xy+

GPT-4o mini · Answer

To simplify the polynomial expression $(xy - 2)(x^2 + 1)$, we will use the distributive property. 1. Distribute $xy$ across $(x^2 + 1)$: - $xy \cdot x^2 = x^3y$ - $xy \cdot 1 = xy$ From this part, we have: $$ x^3y + xy $$ 2. Now, distribute $-2$ across $(x^2 + 1)$: - $-2 \cdot x^2 = -2x^2$ - $-2 \cdot 1 = -2$ From this part, we have: $$ -2x^2 - 2 $$ 3. Now, combine all the terms: $$ x^3y + xy - 2x^2 - 2 $$ As a result, the simplified expression is: $$ x^3y - 2x^2 + xy - 2 $$ Filling in the blanks: - $x^3y +$ - $-2x^2 +$ - $xy - 2$ The complete answer can be presented as: $$ x^3y - 2x^2 + xy - 2 $$

Answer

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

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

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

Option #3: (2x2−4)(3y+6)=6x2y+12x2−12y−24

Option #4: 2x12⋅3=6x−−√

(1 point)
Option #

GPT-4o mini · Answer

To determine which option demonstrates that polynomials form a closed system under multiplication, we need to find the option that shows the product of two polynomials resulting in another polynomial.

Let's evaluate each option:

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

This shows the multiplication of a polynomial $(2x + 5)$ by a constant (which is acceptable), resulting in a polynomial. Although it shows a valid operation, it is not explicitly demonstrating multiplication of two polynomials.

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

This option shows an equation and a simplification of a polynomial expression. It doesn’t involve multiplication or demonstrate the closure property.

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

This option shows the multiplication of two polynomials $(2x^2 - 4)$ and $(3y + 6)$, resulting in another polynomial. This clearly demonstrates closure under multiplication.

Option #4: $2x^{12} \cdot 3=6x^{-−sqrt}$

This option is incorrect as it improperly represents polynomial terms, especially with the radical in the result. It does not demonstrate closure since the result is not a polynomial.

The correct answer is:

Option #3.

Answer

Which of the following options correctly describes if the following polynomial forms a closed system under multiplication?

−3(5x+4y)=−15x−12y

Option #1: Yes, because the result of multiplying the polynomials is also a polynomial.

Option #2: No, because the exponent of x is not a positive integer.

Option #3: No, because multiplying the polynomials resulted in subtraction.

(1 point)
Option #

GPT-4o mini · Answer

To determine whether the given polynomial $−3(5x+4y)=−15x−12y$ demonstrates closure under multiplication, we need to evaluate the result of multiplying polynomials and whether that result is indeed a polynomial.

Evaluation:

The expression given is $−3(5x + 4y)$, where $−3$ is a constant and $5x + 4y$ is a polynomial (a linear polynomial).
The result of the multiplication is $−15x − 12y$.
Both the original polynomial and the resulting expression are polynomials because they are expressed as sums of terms where the variables have non-negative integer exponents.

Analysis of the Options:

Option #1: Yes, because the result of multiplying the polynomials is also a polynomial.

This is correct. The multiplication of the constant $-3$ and the polynomial $5x + 4y$ produces another polynomial $−15x − 12y$.

Option #2: No, because the exponent of x is not a positive integer.

This is incorrect. In the resulting polynomial $−15x − 12y$, the exponent of $x$ is $1$, which is a positive integer.

Option #3: No, because multiplying the polynomials resulted in subtraction.

This is incorrect. The result being a sum or a difference does not affect whether the polynomials form a closed system under multiplication. Polynomials can have addition and subtraction as operations and still be considered polynomials.

Conclusion:

The correct choice is:

Option #1: Yes, because the result of multiplying the polynomials is also a polynomial.

Answer

it was option #2

GPT-4o mini · Answer

I apologize for the confusion. Let's carefully examine the reasoning behind the statement that the polynomial forms a closed system under multiplication and why Option #2 might be suggested.

The definition of a polynomial requires that its exponents be non-negative integers. The expression in question was:

\[ -3(5x + 4y) = -15x - 12y \]

In this case, both $x$ and $y$ have non-negative integer exponents in the final result (the exponent of $x$ is 1, which is indeed a positive integer), and therefore, the result is still a polynomial.

Option #2 states: "No, because the exponent of x is not a positive integer."

This reasoning seems to be incorrect because $x$ has an exponent of 1 in the resulting polynomial, which is a positive integer. Therefore, the conclusion that was drawn for Option #2 is not accurate according to the definition of polynomials.

The correct conclusion should indeed align with Option #1, which states that multiplying the original polynomial indeed produces a polynomial.

If there is any additional context or reasoning you would like to provide for why Option #2 was considered the right answer, please let me know, and I’d be happy to review it further. Thank you for your understanding!