Question

1. use synthetic division and the remainder theorem to evaluate $f(5)$.

$f(x)=x^2 - 7x - 2$
show your work here
5 | 1 -7 -2
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$f(5)=$

1. use synthetic division and the remainder theorem to evaluate $f(-2)$.

$f(x)=x^4 + 3x^3 + 5x^2 - 5x + 5$
show your work here

Explanation:

Problem 14: Evaluate \( f(5) \) for \( f(x) = x^2 - 7x - 2 \) using Synthetic Division and Remainder Theorem

Step 1: Set up Synthetic Division

For \( f(x) = x^2 - 7x - 2 \), the coefficients are \( 1 \) (for \( x^2 \)), \( -7 \) (for \( x \)), and \( -2 \) (constant term). We use \( x = 5 \) for synthetic division.
The setup is:
\[

$$\begin{array}{r|rrr} 5 & 1 & -7 & -2 \\ & & & \\ \hline & & & \\ \end{array}$$

\]

Step 2: Bring down the leading coefficient

Bring down the \( 1 \) (the coefficient of \( x^2 \)):
\[

$$\begin{array}{r|rrr} 5 & 1 & -7 & -2 \\ & & & \\ \hline & 1 & & \\ \end{array}$$

\]

Step 3: Multiply and add

Multiply the number we just brought down (\( 1 \)) by \( 5 \): \( 1 \times 5 = 5 \). Add this to the next coefficient (\( -7 \)): \( -7 + 5 = -2 \).
\[

$$\begin{array}{r|rrr} 5 & 1 & -7 & -2 \\ & & 5 & \\ \hline & 1 & -2 & \\ \end{array}$$

\]

Step 4: Multiply and add again

Multiply the new number (\( -2 \)) by \( 5 \): \( -2 \times 5 = -10 \). Add this to the last coefficient (\( -2 \)): \( -2 + (-10) = -12 \).
\[

$$\begin{array}{r|rrr} 5 & 1 & -7 & -2 \\ & & 5 & -10 \\ \hline & 1 & -2 & -12 \\ \end{array}$$

\]

By the Remainder Theorem, the remainder when \( f(x) \) is divided by \( x - 5 \) is \( f(5) \). The remainder from synthetic division is \( -12 \), so \( f(5) = -12 \).

Step 1: Set up Synthetic Division

For \( f(x) = x^4 + 3x^3 + 5x^2 - 5x + 5 \), the coefficients are \( 1 \) (for \( x^4 \)), \( 3 \) (for \( x^3 \)), \( 5 \) (for \( x^2 \)), \( -5 \) (for \( x \)), and \( 5 \) (constant term). We use \( x = -2 \) for synthetic division.
The setup is:
\[

$$\begin{array}{r|rrrrr} -2 & 1 & 3 & 5 & -5 & 5 \\ & & & & & \\ \hline & & & & & \\ \end{array}$$

\]

Step 2: Bring down the leading coefficient

Bring down the \( 1 \) (the coefficient of \( x^4 \)):
\[

$$\begin{array}{r|rrrrr} -2 & 1 & 3 & 5 & -5 & 5 \\ & & & & & \\ \hline & 1 & & & & \\ \end{array}$$

\]

Step 3: Multiply and add (first time)

Multiply the number we just brought down (\( 1 \)) by \( -2 \): \( 1 \times (-2) = -2 \). Add this to the next coefficient (\( 3 \)): \( 3 + (-2) = 1 \).
\[

$$\begin{array}{r|rrrrr} -2 & 1 & 3 & 5 & -5 & 5 \\ & & -2 & & & \\ \hline & 1 & 1 & & & \\ \end{array}$$

\]

Step 4: Multiply and add (second time)

Multiply the new number (\( 1 \)) by \( -2 \): \( 1 \times (-2) = -2 \). Add this to the next coefficient (\( 5 \)): \( 5 + (-2) = 3 \).
\[

$$\begin{array}{r|rrrrr} -2 & 1 & 3 & 5 & -5 & 5 \\ & & -2 & -2 & & \\ \hline & 1 & 1 & 3 & & \\ \end{array}$$

\]

Step 5: Multiply and add (third time)

Multiply the new number (\( 3 \)) by \( -2 \): \( 3 \times (-2) = -6 \). Add this to the next coefficient (\( -5 \)): \( -5 + (-6) = -11 \).
\[

$$\begin{array}{r|rrrrr} -2 & 1 & 3 & 5 & -5 & 5 \\ & & -2 & -2 & -6 & \\ \hline & 1 & 1 & 3 & -11 & \\ \end{array}$$

\]

Step 6: Multiply and add (fourth time)

Multiply the new number (\( -11 \)) by \( -2 \): \( -11 \times (-2) = 22 \). Add this to the last coefficient (\( 5 \)): \( 5 + 22 = 27 \).
\[

$$\begin{array}{r|rrrrr} -2 & 1 & 3 & 5 & -5 & 5 \\ & & -2 & -2 & -6 & 22 \\ \hline & 1 & 1 & 3 & -11 & 27 \\ \end{array}$$

\]

By the Remainder Theorem, the remainder when \( f(x) \) is divided by \( x + 2 \) (since \( x - (-2) = x + 2 \)) is \( f(-2) \). The remainder from synthetic division is \( 27 \), so \( f(-2) = 27 \).

Answer:

For the synthetic division table:

• The first empty box (under -7) is \( 5 \)
• The second empty box (under -2) is \( -10 \)
• The bottom row boxes are \( 1 \), \( -2 \), \( -12 \)

\( f(5) = \boldsymbol{-12} \)

Problem 15: Evaluate \( f(-2) \) for \( f(x) = x^4 + 3x^3 + 5x^2 - 5x + 5 \) using Synthetic Division and Remainder Theorem