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

.(2 points)
x2+
x+

11 answers

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:

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

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

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

  1. 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: \[ r^3 - 2r^2 + 3r \]

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

  3. Now, combine all the terms: \[ r^3 - 2r^2 + 3r - r^2 + 2r - 3 \]

  4. 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: \[ r^3 - 3r^2 + 5r - 3 \]

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

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

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 #

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.

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 #

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:

  1. The expression given is \(−3(5x + 4y)\), where \(−3\) is a constant and \(5x + 4y\) is a polynomial (a linear polynomial).
  2. The result of the multiplication is \(−15x − 12y\).
  3. 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.

it was option #2

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!