Six statements about polynomials are given. Select all the statements that are true.(4 points)

Question

Responses

12x−10x5−7+3x4
has a leading coefficient of 12.
12 x minus 10 x to the 5th power minus 7 plus 3 x to the 4th power has a leading coefficient of 12.

5a2b−6ab3c+3a5
is a fifth degree polynomial.
5 A squared b minus 6 A b cubed c plus 3 A to the 5th power is a fifth degree polynomial.

12x−10x5−7+3x4
has a leading term of −10x5
12 x minus 10 x to the 5th power minus 7 plus 3 x to the 4th power has a leading term of negative 10 x to the 5th power

12x−10x5−7+3x4
is equivalent to −10x5+3x4+12x−7
12 x minus 10 x to the 5th power minus 7 plus 3 x to the 4th power is equivalent to negative 10 x to the 5th power plus 3 x to the 4th power plus 12 x minus 7

5a2b−6ab3c+3a5
contains three terms.
5 A squared b minus 6 A b cubed c plus 3 A to the 5th power contains three terms.

5a2b−6ab3c+3a5
has a leading coefficient of 5.

Answer 1

Let's analyze each of the statements one by one to determine which are true.

1. "12x−10x5−7+3x4 has a leading coefficient of 12."

   • False. The leading term of the polynomial is \(-10x^5\), so the leading coefficient is \(-10\), not \(12\).

2. "5a2b−6ab3c+3a5 is a fifth degree polynomial."

   • True. The highest degree term is \(3a^5\), which has degree \(5\). Therefore, the polynomial is indeed a fifth degree polynomial.

3. "12x−10x5−7+3x4 has a leading term of −10x5."

   • True. The leading term is \(-10x^5\) since it is the term with the highest degree in the polynomial.

4. "12x−10x5−7+3x4 is equivalent to −10x5+3x4+12x−7."

   • True. The given expression can be rearranged as \(-10x^5 + 3x^4 + 12x - 7\), which maintains the same terms and combines them appropriately.

5. "5a2b−6ab3c+3a5 contains three terms."

   • True. The polynomial consists of three distinct terms: \(5a^2b\), \(-6ab^3c\), and \(3a^5\).

6. "5a2b−6ab3c+3a5 has a leading coefficient of 5."

   • False. The term \(3a^5\) is the leading term, and its coefficient is \(3\), not \(5\). Therefore, the leading coefficient is \(3\).

Final True Statements:

• "5a2b−6ab3c+3a5 is a fifth degree polynomial."
• "12x−10x5−7+3x4 has a leading term of −10x5."
• "12x−10x5−7+3x4 is equivalent to −10x5+3x4+12x−7."
• "5a2b−6ab3c+3a5 contains three terms."

Answer 2

Caleb's work to find the difference (−4.6k2+2.5k−3.8)−(3.6k+5.1k2)

Given: (−4.6k2+2.5k−3.8)−(3.6k+5.1k2)

Step 1: −4.6k2+2.5k−3.8−3.6k+5.1k2

Step 2: −4.6k2+5.1k2+2.5k−3.6k−3.8

Step 3: 0.5k2−1.1k−3.8

Complete the statements.

(2 points)
Put responses in the correct input to answer the question. Select a response, navigate to the desired input and insert the response. Responses can be selected and inserted using the space bar, enter key, left mouse button or touchpad. Responses can also be moved by dragging with a mouse.
Caleb Response area.

The difference of (−4.6k2+2.5k−3.8)−(3.6k+5.1k2)
is

Answer 3

To complete the statement with the correct information from Caleb's work:

The difference of \((-4.6k^2+2.5k−3.8)−(3.6k+5.1k^2)\) is \(0.5k^2−1.1k−3.8\).

So, you would put the response:

0.5k²−1.1k−3.8

in the Caleb Response area.

Answer 4

Find the product of (4y+5)(4y−4)
in standard form. You can use the Algebra tiles to help you.

Answer 5

To find the product of \((4y + 5)(4y - 4)\), we can use the distributive property (also known as the FOIL method for binomials).

Here are the steps for multiplying the two binomials:

1. First: Multiply the first terms in each binomial: \[ 4y \cdot 4y = 16y^2 \]

2. Outer: Multiply the outer terms in the expression: \[ 4y \cdot (-4) = -16y \]

3. Inner: Multiply the inner terms: \[ 5 \cdot 4y = 20y \]

4. Last: Multiply the last terms in each binomial: \[ 5 \cdot (-4) = -20 \]

Now, we combine all these results: \[ 16y^2 + (-16y) + 20y - 20 \]

Combine like terms (\(-16y + 20y = 4y\)): \[ 16y^2 + 4y - 20 \]

So, the product of \((4y + 5)(4y - 4)\) in standard form is: \[ \boxed{16y^2 + 4y - 20} \]

Answer 6

Which are equivalent expressions for (4−2z2−z)(3z+5)
? Select all that apply.(3 points)
Responses

6z3−10z2+5z+12z+20−3z2
6 z cubed minus 10 z squared plus 5 z plus 12 z plus 20 minus 3 z squared

−6z3−13z2+7z+20
negative 6 z cubed minus 13 z squared plus 7 z plus 20

−6z3−7z2−2z+20
negative 6 z cubed minus 7 z squared minus 2 z plus 20

−6z3−10z2−3z2−5z+12z+20
negative 6 z cubed minus 10 z squared minus 3 z squared minus 5 z plus 12 z plus 20

−6z3−13z2−5z+12z+20

Answer 7

To determine which expressions are equivalent to \((4−2z^2−z)(3z+5)\), let's first expand the expression:

\[ (4 - 2z^2 - z)(3z + 5) \]

We will use the distributive property (also known as the FOIL method for two binomials):

1. Distributing \(4\):

   • \(4 \cdot 3z = 12z\)
   • \(4 \cdot 5 = 20\)

2. Distributing \(-2z^2\):

   • \(-2z^2 \cdot 3z = -6z^3\)
   • \(-2z^2 \cdot 5 = -10z^2\)

3. Distributing \(-z\):

   • \(-z \cdot 3z = -3z^2\)
   • \(-z \cdot 5 = -5z\)

Now we combine all these results: \[ (-6z^3) + (12z + 20) + (-10z^2) + (-3z^2) + (-5z) \]

Combine like terms:

• The \(z^3\) term: \(-6z^3\)
• The \(z^2\) terms: \(-10z^2 - 3z^2 = -13z^2\)
• The \(z\) terms: \(12z - 5z = 7z\)

So the final combined expression is: \[ -6z^3 - 13z^2 + 7z + 20 \]

Now let's compare this with the given response options:

1. \(6z^3−10z^2+5z+12z+20−3z^2\)

   • Simplifying gives: \[ 6z^3 + (-10z^2 - 3z^2) + (5z + 12z) + 20 = 6z^3 - 13z^2 + 17z + 20 \quad \text{(not equivalent)} \]

2. \(-6z^3−13z^2+7z+20\)

   • This matches exactly. \quad \text{(equivalent)}

3. \(-6z^3−7z^2−2z+20\)

   • This does not match. \quad \text{(not equivalent)}

4. \(-6z^3−10z^2−3z^2−5z+12z+20\)

   • Simplifying gives: \[ -6z^3 + (-10z^2 - 3z^2) + (12z - 5z) + 20 = -6z^3 - 13z^2 + 7z + 20 \quad \text{(equivalent)} \]

5. \(-6z^3−13z^2−5z+12z+20\)

   • Simplifying gives: \[ -6z^3 - 13z^2 + (12z - 5z) + 20 = -6z^3 - 13z^2 + 7z + 20 \quad \text{(equivalent)} \]

Final Equivalent Expressions:

• \(-6z^3−13z^2+7z+20\)
• \(-6z^3−10z^2−3z^2−5z+12z+20\)
• \(-6z^3−13z^2−5z+12z+20\)

Answer 8

Ahmed found the product of (23g2−6n)2
using the area model. His partial work is shown. What is the partial product that goes in the orange box? What is the partial product that goes in the blue box? What is the product of (23g2−6n)2
written in standard form?(3 points)

Answer 9

To find the product of \((23g^2 - 6n)^2\) using the area model, we can use the algebraic identity for squaring a binomial:

\[ (a - b)^2 = a^2 - 2ab + b^2 \]

Where \(a = 23g^2\) and \(b = 6n\).

We'll calculate each part:

Step 1: Finding the partial products

1. The Orange Box (first partial product - \(a^2\)):

   • This is the square of the first term: \[ (23g^2)^2 = 529g^4 \]

2. The Blue Box (second partial product - \(-2ab\)):

   • This is twice the product of the first and second terms: \[ -2(23g^2)(6n) = -276g^2n \]

Step 2: Finding the final product

Now, we also need to calculate the last part, which is \(b^2\):

3. The last term (\(b^2\)):
   • This is the square of the second term: \[ (6n)^2 = 36n^2 \]

Step 3: Combining all the parts into standard form

So now we can combine all the parts we calculated: \[ (23g^2 - 6n)^2 = 529g^4 - 276g^2n + 36n^2 \]

Conclusion

• Partial product for the Orange Box: \(529g^4\)
• Partial product for the Blue Box: \(-276g^2n\)
• Product in standard form: \(529g^4 - 276g^2n + 36n^2\)

These results can be summarized as follows:

• Orange Box: \(529g^4\)
• Blue Box: \(-276g^2n\)
• Final Product: \(529g^4 - 276g^2n + 36n^2\)

Answer 10

Ahmed found the product of (2/3g2−6n)2
using the area model. His partial work is shown. What is the partial product that goes in the orange box? What is the partial product that goes in the blue box? What is the product of (23g2−6n)2
written in standard form?(3 points)

Answer 11

To find the product of \(\left(\frac{2}{3}g^2 - 6n\right)^2\) using the area model, we can use the squared binomial formula:

\[ (a - b)^2 = a^2 - 2ab + b^2 \]

Where:

• \(a = \frac{2}{3}g^2\)
• \(b = 6n\)

Step 1: Calculating the partial products

1. The Orange Box (first partial product, \(a^2\)):

   • This is the square of the first term: \[ \left(\frac{2}{3}g^2\right)^2 = \frac{4}{9}g^4 \]

2. The Blue Box (second partial product, \(-2ab\)):

   • This is twice the product of the first and second terms: \[ -2\left(\frac{2}{3}g^2\right)(6n) = -2 \times \frac{2}{3} \times 6 \times g^2 \times n \]
   • This simplifies to: \[ -2 \times \frac{12}{3}g^2n = -8g^2n \]

Step 2: Finding the last term, \(b^2\)

3. Last term (\(b^2\)):
   • This is the square of the second term: \[ (6n)^2 = 36n^2 \]

Step 3: Combining the results

Now we can combine all parts: \[ \left(\frac{2}{3}g^2 - 6n\right)^2 = \frac{4}{9}g^4 - 8g^2n + 36n^2 \]

Conclusion

• Partial Product for the Orange Box: \(\frac{4}{9}g^4\)
• Partial Product for the Blue Box: \(-8g^2n\)
• Final Product (in standard form): \[ \frac{4}{9}g^4 - 8g^2n + 36n^2 \]

These results are summarized as follows:

• Orange Box: \(\frac{4}{9}g^4\)
• Blue Box: \(-8g^2n\)
• Final Product: \(\frac{4}{9}g^4 - 8g^2n + 36n^2\)