Answer :
To use synthetic division and the Remainder Theorem to evaluate [tex]\( P(-1) \)[/tex], follow these steps:
1. Write down the coefficients of the polynomial [tex]\( P(x) \)[/tex]:
Let's assume the polynomial [tex]\( P(x) \)[/tex] is given by:
[tex]\[ P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \][/tex]
Write down the coefficients [tex]\( [a_n, a_{n-1}, \ldots, a_1, a_0] \)[/tex].
2. Set up the synthetic division table:
Write down the value of [tex]\( c \)[/tex] which is [tex]\(-1\)[/tex] in this case, and the coefficients in a row:
[tex]\[ \begin{array}{c|cccc} -1 & a_n & a_{n-1} & \cdots & a_0 \\ \end{array} \][/tex]
3. Perform synthetic division:
- Bring down the leading coefficient [tex]\( a_n \)[/tex] to the bottom row.
- Multiply [tex]\( a_n \)[/tex] by [tex]\( c \)[/tex] (which is [tex]\(-1\)[/tex]) and place the result under the next coefficient.
- Add the value obtained to the next coefficient and place the result in the bottom row.
- Repeat this process for all coefficients.
Let's go through a concrete example with specific coefficients to clarify:
Suppose [tex]\( P(x) = 2x^3 - 3x^2 + 4x - 5 \)[/tex].
The coefficients are [tex]\( [2, -3, 4, -5] \)[/tex].
Set up the synthetic division table:
[tex]\[ \begin{array}{c|cccc} -1 & 2 & -3 & 4 & -5 \\ \hline & & & & \\ \end{array} \][/tex]
1. Bring down the leading coefficient [tex]\( 2 \)[/tex]:
[tex]\[ \begin{array}{c|cccc} -1 & 2 & -3 & 4 & -5 \\ \hline & 2 & & & \\ \end{array} \][/tex]
2. Multiply [tex]\( 2 \)[/tex] by [tex]\(-1\)[/tex] and add to the next coefficient [tex]\( -3 \)[/tex]:
[tex]\[ \begin{array}{c|cccc} -1 & 2 & -3 & 4 & -5 \\ \hline & 2 & -5 & & \\ \end{array} \][/tex]
3. Multiply [tex]\( -5 \)[/tex] by [tex]\(-1\)[/tex] and add to the next coefficient [tex]\( 4 \)[/tex]:
[tex]\[ \begin{array}{c|cccc} -1 & 2 & -3 & 4 & -5 \\ \hline & 2 & -5 & 9 & \\ \end{array} \][/tex]
4. Multiply [tex]\( 9 \)[/tex] by [tex]\(-1\)[/tex] and add to the next coefficient [tex]\( -5 \)[/tex]:
[tex]\[ \begin{array}{c|cccc} -1 & 2 & -3 & 4 & -5 \\ \hline & 2 & -5 & 9 & -14 \\ \end{array} \][/tex]
The value obtained in the last column of the bottom row is the remainder, which is the value of [tex]\( P(-1) \)[/tex].
Thus, [tex]\( P(-1) = -14 \)[/tex].
So, the evaluated value of [tex]\( P(-1) \)[/tex] is [tex]\(\boxed{-14}\)[/tex].
1. Write down the coefficients of the polynomial [tex]\( P(x) \)[/tex]:
Let's assume the polynomial [tex]\( P(x) \)[/tex] is given by:
[tex]\[ P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \][/tex]
Write down the coefficients [tex]\( [a_n, a_{n-1}, \ldots, a_1, a_0] \)[/tex].
2. Set up the synthetic division table:
Write down the value of [tex]\( c \)[/tex] which is [tex]\(-1\)[/tex] in this case, and the coefficients in a row:
[tex]\[ \begin{array}{c|cccc} -1 & a_n & a_{n-1} & \cdots & a_0 \\ \end{array} \][/tex]
3. Perform synthetic division:
- Bring down the leading coefficient [tex]\( a_n \)[/tex] to the bottom row.
- Multiply [tex]\( a_n \)[/tex] by [tex]\( c \)[/tex] (which is [tex]\(-1\)[/tex]) and place the result under the next coefficient.
- Add the value obtained to the next coefficient and place the result in the bottom row.
- Repeat this process for all coefficients.
Let's go through a concrete example with specific coefficients to clarify:
Suppose [tex]\( P(x) = 2x^3 - 3x^2 + 4x - 5 \)[/tex].
The coefficients are [tex]\( [2, -3, 4, -5] \)[/tex].
Set up the synthetic division table:
[tex]\[ \begin{array}{c|cccc} -1 & 2 & -3 & 4 & -5 \\ \hline & & & & \\ \end{array} \][/tex]
1. Bring down the leading coefficient [tex]\( 2 \)[/tex]:
[tex]\[ \begin{array}{c|cccc} -1 & 2 & -3 & 4 & -5 \\ \hline & 2 & & & \\ \end{array} \][/tex]
2. Multiply [tex]\( 2 \)[/tex] by [tex]\(-1\)[/tex] and add to the next coefficient [tex]\( -3 \)[/tex]:
[tex]\[ \begin{array}{c|cccc} -1 & 2 & -3 & 4 & -5 \\ \hline & 2 & -5 & & \\ \end{array} \][/tex]
3. Multiply [tex]\( -5 \)[/tex] by [tex]\(-1\)[/tex] and add to the next coefficient [tex]\( 4 \)[/tex]:
[tex]\[ \begin{array}{c|cccc} -1 & 2 & -3 & 4 & -5 \\ \hline & 2 & -5 & 9 & \\ \end{array} \][/tex]
4. Multiply [tex]\( 9 \)[/tex] by [tex]\(-1\)[/tex] and add to the next coefficient [tex]\( -5 \)[/tex]:
[tex]\[ \begin{array}{c|cccc} -1 & 2 & -3 & 4 & -5 \\ \hline & 2 & -5 & 9 & -14 \\ \end{array} \][/tex]
The value obtained in the last column of the bottom row is the remainder, which is the value of [tex]\( P(-1) \)[/tex].
Thus, [tex]\( P(-1) = -14 \)[/tex].
So, the evaluated value of [tex]\( P(-1) \)[/tex] is [tex]\(\boxed{-14}\)[/tex].
