Tatiana was asked to construct a polynomial function with the following zeros and multiplicities. Which function should she write?

Question

Zero Multiplicity
x=1
3
x=81
5(1 point)
Responses

f(x)=(x+1)3(x+81)5
f left parenthesis x right parenthesis equals left parenthesis x plus 1 right parenthesis cubed left parenthesis x plus 81 right parenthesis superscript 5 baseline

f(x)=(x−1)3(x−81)5
f left parenthesis x right parenthesis equals left parenthesis x minus 1 right parenthesis cubed left parenthesis x minus 81 right parenthesis superscript 5 baseline

f(x)=(x−1)5(x−81)3
f left parenthesis x right parenthesis equals left parenthesis x minus 1 right parenthesis superscript 5 baseline left parenthesis x minus 81 right parenthesis cubed

f(x)=(x+1)5(x+81)3

Answer 1

To construct the polynomial function based on the given zeros and their respective multiplicities, we need to follow these steps:

1. Identify the zeros and their multiplicities:

   • Zero at \( x = 1 \) with a multiplicity of \( 3 \)
   • Zero at \( x = 81 \) with a multiplicity of \( 5 \)

2. Form the factors from the zeros:

   • The factor corresponding to the zero at \( x = 1 \) would be \( (x - 1)^3 \).
   • The factor corresponding to the zero at \( x = 81 \) would be \( (x - 81)^5 \).

3. Combine the factors to create the polynomial: Thus, the polynomial function can be expressed as: \[ f(x) = (x - 1)^3 (x - 81)^5 \]

Now, let's compare this with the provided options:

• \( f(x) = (x + 1)^3 (x + 81)^5 \) (Incorrect)
• \( f(x) = (x - 1)^3 (x - 81)^5 \) (Correct)
• \( f(x) = (x - 1)^5 (x - 81)^3 \) (Incorrect)
• \( f(x) = (x + 1)^5 (x + 81)^3 \) (Incorrect)

The correct response for the function that Tatiana should write is: \[ \boxed{f(x) = (x - 1)^3 (x - 81)^5} \]

Answer 2

Which of the following is a factor of x3−5x2−2x+24
, given that x−3
is one of the linear factors?(1 point)
Responses

x+2
x plus 2

x−2
x minus 2

x+4
x plus 4

x2−8x+22

Answer 3

To determine which of the provided options is also a factor of the polynomial \( x^3 - 5x^2 - 2x + 24 \) given that \( x - 3 \) is a factor, we can use synthetic division or polynomial long division to divide the polynomial by \( x - 3 \).

1. Perform synthetic division of \( x^3 - 5x^2 - 2x + 24 \) by \( x - 3 \):

Using synthetic division, we will set \( 3 \) (the zero of \( x - 3 \)) and set up the coefficients from the polynomial \( 1, -5, -2, 24 \):

\[ \begin{array}{r|rrrr} 3 & 1 & -5 & -2 & 24 \ & & 3 & -6 & -24 \ \hline & 1 & -2 & -8 & 0 \ \end{array} \]

The resulting polynomial from the division is \( x^2 - 2x - 8 \), and the remainder is \( 0 \), confirming that \( x - 3 \) is indeed a factor.

2. Now, we factor \( x^2 - 2x - 8 \):

To factor the quadratic \( x^2 - 2x - 8 \), we look for two numbers that multiply to \(-8\) and add to \(-2\). The numbers \(-4\) and \(2\) satisfy these conditions:

\[ x^2 - 2x - 8 = (x - 4)(x + 2) \]

3. Identify the linear factors from the quadratic:

From \( (x - 4)(x + 2) \), we see that the linear factors are \( x - 4 \) and \( x + 2 \).

4. Choose the correct option:

The answer choices provided are:

• \( x + 2 \) (This is a factor)
• \( x - 2 \)
• \( x + 4 \)
• \( x^2 - 8x + 22 \)

Since \( x + 2 \) is one of the factors derived from our polynomial division, the correct response is:

\[ \boxed{x + 2} \]

Answer 4

Janet is trying to solve an equation and gets stuck at x2(2x−1)+3x(2x−1)−4(2x−1)=0
. Which of the following is a complete set of solutions to Janet’s equation?(1 point)
Responses

−4
, 1
negative 4 , 1

12
Start Fraction 1 over 2 End Fraction

12
, −4
, 1
Start Fraction 1 over 2 End Fraction , negative 4 , 1

12
, 4, −1

Answer 5

To solve the equation

\[ x^2(2x - 1) + 3x(2x - 1) - 4(2x - 1) = 0, \]

we notice that \( (2x - 1) \) is a common factor. We can factor it out from each term:

\[ (2x - 1)(x^2 + 3x - 4) = 0. \]

Now we have two factors set to zero:

1. For the first factor: \[ 2x - 1 = 0 \implies 2x = 1 \implies x = \frac{1}{2}. \]

2. For the second factor, we need to solve the quadratic equation: \[ x^2 + 3x - 4 = 0. \]

We can factor this quadratic expression. To do this, we need two numbers that multiply to \(-4\) and add to \(3\). These numbers are \(4\) and \(-1\):

\[ (x + 4)(x - 1) = 0. \]

Now we set each factor to zero:

• For \( x + 4 = 0 \): \[ x = -4. \]

• For \( x - 1 = 0 \): \[ x = 1. \]

3. Compile all solutions:

The complete set of solutions to Janet's equation is

• \( x = \frac{1}{2} \)
• \( x = -4 \)
• \( x = 1 \)

Thus, the complete set of solutions is:

\[ \boxed{\frac{1}{2}, -4, 1}. \]

Answer 6

Factor the following high-degree polynomial completely.

48x6−3x2


(1 point)
Responses

3x2(4x2+1)(2x+1)(2x−1)
3 x squared left parenthesis 4 x squared plus 1 right parenthesis left parenthesis 2 x plus 1 right parenthesis left parenthesis 2 x minus 1 right parenthesis

3x2(8x2+1)(8x2−1)
3 x squared left parenthesis 8 x squared plus 1 right parenthesis left parenthesis 8 x squared minus 1 right parenthesis

3x2(4x2+1)(4x+1)(4x−1)
3 x squared left parenthesis 4 x squared plus 1 right parenthesis left parenthesis 4 x plus 1 right parenthesis left parenthesis 4 x minus 1 right parenthesis

3x2(4x2+1)(4x2−1)

Answer 7

To factor the polynomial \( 48x^6 - 3x^2 \) completely, let’s follow these steps:

1. Factor out the greatest common factor (GCF):

   The GCF of \( 48x^6 \) and \( -3x^2 \) is \( 3x^2 \). So we can factor this out:

   \[ 48x^6 - 3x^2 = 3x^2(16x^4 - 1). \]

2. Factor the expression inside the parentheses:

   The term \( 16x^4 - 1 \) is a difference of squares and can be factored as follows:

   \[ 16x^4 - 1 = (4x^2)^2 - 1^2 = (4x^2 - 1)(4x^2 + 1). \]

3. Further factor \( 4x^2 - 1 \):

   The term \( 4x^2 - 1 \) is also a difference of squares:

   \[ 4x^2 - 1 = (2x)^2 - 1^2 = (2x - 1)(2x + 1). \]

4. Putting it all together:

   Now we combine all the factors obtained from the previous steps:

   \[ 48x^6 - 3x^2 = 3x^2(4x^2 - 1)(4x^2 + 1) = 3x^2(2x - 1)(2x + 1)(4x^2 + 1). \]

Now, we can compare the fully factored form of the polynomial with the provided options:

• \( 3x^2(4x^2 + 1)(2x + 1)(2x - 1) \)
• \( 3x^2(8x^2 + 1)(8x^2 - 1) \)
• \( 3x^2(4x^2 + 1)(4x + 1)(4x - 1) \)
• \( 3x^2(4x^2 + 1)(4x^2 - 1) \)

The correct factored form is:

\[ \boxed{3x^2(4x^2 + 1)(2x + 1)(2x - 1)}. \]

Answer 8

What are the zeros of f(x)=−4(2x+3)3(6−x)
?

(1 point)
Responses

x=−32,6,−4
x equals negative Start Fraction 3 over 2 End Fraction comma 6 comma negative 4

x=32,−6
x equals Start Fraction 3 over 2 End Fraction comma negative 6

x=−32,6
x equals negative Start Fraction 3 over 2 End Fraction comma 6

x=−32,6,0

Answer 9

To find the zeros of the function

\[ f(x) = -4(2x + 3)^3(6 - x), \]

we need to set \( f(x) = 0 \) and solve for \( x \).

1. Identify the factors that can give us zeros:

Since \( f(x) \) is a product of factors, it is equal to zero when any of the factors equal zero.

• First factor: \[ (2x + 3)^3 = 0 \] Solving this gives: \[ 2x + 3 = 0 \implies 2x = -3 \implies x = -\frac{3}{2}. \]

• Second factor: \[ (6 - x) = 0 \] Solving this gives: \[ 6 - x = 0 \implies x = 6. \]

2. List all the zeros found:
   • From \( (2x + 3)^3 = 0 \), we have a zero at \( x = -\frac{3}{2} \) with a multiplicity of \( 3 \).
   • From \( (6 - x) = 0 \), we have a zero at \( x = 6 \).

Thus, the zeros of the function \( f(x) \) are:

• \( x = -\frac{3}{2} \)
• \( x = 6 \)

3. Check the provided options:
   • \( x = -\frac{3}{2}, 6, -4 \) (this is incorrect because it includes \(-4\) which is not a zero)
   • \( x = \frac{3}{2}, -6 \) (this is incorrect)
   • \( x = -\frac{3}{2}, 6 \) (this is correct)
   • \( x = -\frac{3}{2}, 6, 0 \) (this is incorrect because it includes \(0\) which is not a zero)

The correct response is:

\[ \boxed{x = -\frac{3}{2}, 6}. \]