Answer :
To evaluate [tex]\( P(c) \)[/tex] for the polynomial [tex]\( P(x) = x^3 - 4x^2 + 9x - 15 \)[/tex] using synthetic division and the Remainder Theorem with [tex]\( c = 2 \)[/tex], follow these steps:
### Step-by-Step Solution
Step 1: Write down the coefficients of [tex]\( P(x) \)[/tex].
The polynomial [tex]\( P(x) = x^3 - 4x^2 + 9x - 15 \)[/tex] has coefficients:
[tex]\[ 1, -4, 9, -15 \][/tex]
Step 2: Set up the synthetic division.
Since [tex]\( c = 2 \)[/tex], we use this value as the divisor in synthetic division.
The synthetic division setup looks like this:
[tex]\[ \begin{array}{r|rrrr} 2 & 1 & -4 & 9 & -15 \\ \hline & & & & \\ \end{array} \][/tex]
Step 3: Perform synthetic division.
1. Bring down the leading coefficient (1).
[tex]\[ \begin{array}{r|rrrr} 2 & 1 & -4 & 9 & -15 \\ \hline & 1 & & & \\ \end{array} \][/tex]
2. Multiply 2 by the number just written below the line (1). Write the result (2) under the next coefficient (-4).
[tex]\[ \begin{array}{r|rrrr} 2 & 1 & -4 & 9 & -15 \\ \hline & 1 & 2 & & \\ \end{array} \][/tex]
3. Add the column: [tex]\(-4 + 2 = -2\)[/tex].
[tex]\[ \begin{array}{r|rrrr} 2 & 1 & -4 & 9 & -15 \\ \hline & 1 & -2 & & \\ \end{array} \][/tex]
4. Multiply 2 by the number just written below the line (-2). Write the result (-4) under the next coefficient (9).
[tex]\[ \begin{array}{r|rrrr} 2 & 1 & -4 & 9 & -15 \\ \hline & 1 & -2 & -4 & \\ \end{array} \][/tex]
5. Add the column: [tex]\(9 + (-4) = 5\)[/tex].
[tex]\[ \begin{array}{r|rrrr} 2 & 1 & -4 & 9 & -15 \\ \hline & 1 & -2 & 5 & \\ \end{array} \][/tex]
6. Multiply 2 by the number just written below the line (5). Write the result (10) under the next coefficient (-15).
[tex]\[ \begin{array}{r|rrrr} 2 & 1 & -4 & 9 & -15 \\ \hline & 1 & -2 & 5 & 10 \\ \end{array} \][/tex]
7. Add the column: [tex]\(-15 + 10 = -5\)[/tex].
The quotient and the remainder are:
[tex]\[ \begin{array}{r|rrrr} 2 & 1 & -4 & 9 & -15 \\ \hline & 1 & -2 & 5 & -5 \\ \end{array} \][/tex]
The quotient and remainder are:
- Quotient: [tex]\( 1x^2 - 2x + 5 \)[/tex]
- Remainder: [tex]\( -5 \)[/tex]
By the Remainder Theorem, the value of [tex]\( P(c) \)[/tex] is equal to the remainder when [tex]\( P(x) \)[/tex] is divided by [tex]\( x - c \)[/tex].
Thus, we have:
[tex]\[ \boxed{1 \x^2 - 2x + 5} \][/tex]
[tex]\[ \boxed{-5} \][/tex]
So, [tex]\[ P(2) = \boxed{-5} \][/tex]
### Step-by-Step Solution
Step 1: Write down the coefficients of [tex]\( P(x) \)[/tex].
The polynomial [tex]\( P(x) = x^3 - 4x^2 + 9x - 15 \)[/tex] has coefficients:
[tex]\[ 1, -4, 9, -15 \][/tex]
Step 2: Set up the synthetic division.
Since [tex]\( c = 2 \)[/tex], we use this value as the divisor in synthetic division.
The synthetic division setup looks like this:
[tex]\[ \begin{array}{r|rrrr} 2 & 1 & -4 & 9 & -15 \\ \hline & & & & \\ \end{array} \][/tex]
Step 3: Perform synthetic division.
1. Bring down the leading coefficient (1).
[tex]\[ \begin{array}{r|rrrr} 2 & 1 & -4 & 9 & -15 \\ \hline & 1 & & & \\ \end{array} \][/tex]
2. Multiply 2 by the number just written below the line (1). Write the result (2) under the next coefficient (-4).
[tex]\[ \begin{array}{r|rrrr} 2 & 1 & -4 & 9 & -15 \\ \hline & 1 & 2 & & \\ \end{array} \][/tex]
3. Add the column: [tex]\(-4 + 2 = -2\)[/tex].
[tex]\[ \begin{array}{r|rrrr} 2 & 1 & -4 & 9 & -15 \\ \hline & 1 & -2 & & \\ \end{array} \][/tex]
4. Multiply 2 by the number just written below the line (-2). Write the result (-4) under the next coefficient (9).
[tex]\[ \begin{array}{r|rrrr} 2 & 1 & -4 & 9 & -15 \\ \hline & 1 & -2 & -4 & \\ \end{array} \][/tex]
5. Add the column: [tex]\(9 + (-4) = 5\)[/tex].
[tex]\[ \begin{array}{r|rrrr} 2 & 1 & -4 & 9 & -15 \\ \hline & 1 & -2 & 5 & \\ \end{array} \][/tex]
6. Multiply 2 by the number just written below the line (5). Write the result (10) under the next coefficient (-15).
[tex]\[ \begin{array}{r|rrrr} 2 & 1 & -4 & 9 & -15 \\ \hline & 1 & -2 & 5 & 10 \\ \end{array} \][/tex]
7. Add the column: [tex]\(-15 + 10 = -5\)[/tex].
The quotient and the remainder are:
[tex]\[ \begin{array}{r|rrrr} 2 & 1 & -4 & 9 & -15 \\ \hline & 1 & -2 & 5 & -5 \\ \end{array} \][/tex]
The quotient and remainder are:
- Quotient: [tex]\( 1x^2 - 2x + 5 \)[/tex]
- Remainder: [tex]\( -5 \)[/tex]
By the Remainder Theorem, the value of [tex]\( P(c) \)[/tex] is equal to the remainder when [tex]\( P(x) \)[/tex] is divided by [tex]\( x - c \)[/tex].
Thus, we have:
[tex]\[ \boxed{1 \x^2 - 2x + 5} \][/tex]
[tex]\[ \boxed{-5} \][/tex]
So, [tex]\[ P(2) = \boxed{-5} \][/tex]
