Use synthetic division and the Remainder Theorem to find each function value. Check your answer by evaluating the function at the given​ x-value.

​f(x)equals=x cubed plus 9 x squared plus 8x3+9x2+8

**a.**​ f(1)

**b.**

​ f(minus−​1)

**c.** f left parenthesis one half right parenthesisf12

**d.**​ f(10)

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Part 1

**a.**

​ f(1)equals=enter your response here**** 

​(Type an integer or a simplified​ fraction.)

Part 2

**b.**

​ f(minus−​1)equals=enter your response here 

​(Type an integer or a simplified​ fraction.)

Part 3

**c.** 

f left parenthesis one half right parenthesisf12equals=enter your response here**** 

​(Type an integer or a simplified​ fraction.) 

Part 4

**d.**

​ f(10)equals=enter your response here 

​(Type an integer or a simplified​ fraction.)


We are given the function \( f(x) = x^3 + 9x^2 + 8x \), and we need to use **synthetic division** and the **Remainder Theorem** to evaluate it at several points.

### The Remainder Theorem:
The Remainder Theorem states that if you divide a polynomial \( f(x) \) by \( x - a \), the remainder of the division will be \( f(a) \). So, by performing synthetic division, the remainder gives us the value of \( f(a) \).

### Part a: \( f(1) \)

We want to find \( f(1) \) using synthetic division.

1. Set up the synthetic division with \( x = 1 \) and the coefficients of the polynomial \( f(x) = x^3 + 9x^2 + 8x + 0 \) (since there’s no constant term, it’s 0): 

   Coefficients: \( 1, 9, 8, 0 \)

   Synthetic division setup:
   \[
   \begin{array}{r|rrrr}
   1 & 1 & 9 & 8 & 0 \\
     &   & 1 & 10 & 18 \\
   \hline
     & 1 & 10 & 18 & 18 \\
   \end{array}
   \]

2. The remainder is 18, so \( f(1) = 18 \).

### Part b: \( f(-1) \)

Now, we will find \( f(-1) \).

1. Set up the synthetic division with \( x = -1 \):

   Coefficients: \( 1, 9, 8, 0 \)

   Synthetic division setup:
   \[
   \begin{array}{r|rrrr}
   -1 & 1 & 9 & 8 & 0 \\
      &   & -1 & -8 & 0 \\
   \hline
      & 1 & 8 & 0 & 0 \\
   \end{array}
   \]

2. The remainder is 0, so \( f(-1) = 0 \).

### Part c: \( f\left(\frac{1}{2}\right) \)

Now, let's find \( f\left(\frac{1}{2}\right) \).

1. Set up the synthetic division with \( x = \frac{1}{2} \):

   Coefficients: \( 1, 9, 8, 0 \)

   Synthetic division setup:
   \[
   \begin{array}{r|rrrr}
   \frac{1}{2} & 1 & 9 & 8 & 0 \\
               &   & \frac{1}{2} & \frac{19}{2} & \frac{35}{2} \\
   \hline
               & 1 & \frac{19}{2} & \frac{35}{2} & \frac{35}{2} \\
   \end{array}
   \]

2. The remainder is \( \frac{35}{2} \), so \( f\left(\frac{1}{2}\right) = \frac{35}{2} \).

### Part d: \( f(10) \)

Finally, let's find \( f(10) \).

1. Set up the synthetic division with \( x = 10 \):

   Coefficients: \( 1, 9, 8, 0 \)

   Synthetic division setup:
   \[
   \begin{array}{r|rrrr}
   10 & 1 & 9 & 8 & 0 \\
      &   & 10 & 190 & 1980 \\
   \hline
      & 1 & 19 & 198 & 1980 \\
   \end{array}
   \]

2. The remainder is 1980, so \( f(10) = 1980 \).

### Final Answers:
- **a.** \( f(1) = 18 \)
- **b.** \( f(-1) = 0 \)
- **c.** \( f\left(\frac{1}{2}\right) = \frac{35}{2} \)
- **d.** \( f(10) = 1980 \)

Would you like more details or clarification on any step?

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Here are 5 related questions:
1. What is the general form of the Remainder Theorem and how does it work?
2. How would you apply the Remainder Theorem to polynomials of degree higher than 3?
3. Can synthetic division be used to find roots of polynomials?
4. How do you check the result of synthetic division manually by substitution?
5. How does the Remainder Theorem connect with the Factor Theorem?

**Tip:** Always

Use synthetic division and the Remainder Theorem to find each function value. Check your answer by evaluating the function at the given x-value. f(x) = x^3 + 9x^2 + 8x. a. f(1) b. f(-1) c. f(1/2) d. f(10)

Learn how to use synthetic division and the Remainder Theorem to find the function values of the polynomial f(x) = x^3 + 9x^2 + 8x at specific x-values like 1, -1, 1/2, and 10. This solution covers key concepts such as polynomial division and remainder calculations, suitable for high school students.

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