Answer :
To use synthetic substitution to evaluate [tex]\( f(3) \)[/tex] for the polynomial [tex]\( f(x) = x^3 - x^2 + 3x - 4 \)[/tex], follow these detailed steps:
1. Write down the coefficients of the polynomial: For [tex]\( f(x) = x^3 - x^2 + 3x - 4 \)[/tex], the coefficients are [tex]\( [1, -1, 3, -4] \)[/tex].
2. Set up the synthetic substitution table: We will use these coefficients and the value [tex]\( x = 3 \)[/tex] for the substitution.
3. Initialize with the leading coefficient: Place the first coefficient, [tex]\( 1 \)[/tex], in the synthetic row.
[tex]\[ \begin{array}{c|cccc} 3 & 1 & -1 & 3 & -4 \\ \hline & 1 & & & \\ \end{array} \][/tex]
4. Multiply and add sequentially:
- First step: Multiply the leading coefficient by [tex]\( x = 3 \)[/tex] and add it to the next coefficient:
[tex]\[ 1 \cdot 3 + (-1) = 3 - 1 = 2 \][/tex]
Update the synthetic row:
[tex]\[ \begin{array}{c|cccc} 3 & 1 & -1 & 3 & -4 \\ \hline & 1 & 2 & & \\ \end{array} \][/tex]
- Second step: Multiply the value from the previous step by [tex]\( x = 3 \)[/tex] and add to the next coefficient:
[tex]\[ 2 \cdot 3 + 3 = 6 + 3 = 9 \][/tex]
Update the synthetic row:
[tex]\[ \begin{array}{c|cccc} 3 & 1 & -1 & 3 & -4 \\ \hline & 1 & 2 & 9 & \\ \end{array} \][/tex]
- Third step: Multiply the value from the previous step by [tex]\( x = 3 \)[/tex] and add to the next coefficient:
[tex]\[ 9 \cdot 3 + (-4) = 27 - 4 = 23 \][/tex]
Update the synthetic row:
[tex]\[ \begin{array}{c|cccc} 3 & 1 & -1 & 3 & -4 \\ \hline & 1 & 2 & 9 & 23 \\ \end{array} \][/tex]
5. Final synthetic row: The completed synthetic row is [tex]\( [1, 2, 9, 23] \)[/tex].
6. Result: The final value in the synthetic row is [tex]\( 23 \)[/tex], which is the value of [tex]\( f(3) \)[/tex].
Therefore, using synthetic substitution, [tex]\( f(3) \)[/tex] evaluates to [tex]\( 23 \)[/tex].
1. Write down the coefficients of the polynomial: For [tex]\( f(x) = x^3 - x^2 + 3x - 4 \)[/tex], the coefficients are [tex]\( [1, -1, 3, -4] \)[/tex].
2. Set up the synthetic substitution table: We will use these coefficients and the value [tex]\( x = 3 \)[/tex] for the substitution.
3. Initialize with the leading coefficient: Place the first coefficient, [tex]\( 1 \)[/tex], in the synthetic row.
[tex]\[ \begin{array}{c|cccc} 3 & 1 & -1 & 3 & -4 \\ \hline & 1 & & & \\ \end{array} \][/tex]
4. Multiply and add sequentially:
- First step: Multiply the leading coefficient by [tex]\( x = 3 \)[/tex] and add it to the next coefficient:
[tex]\[ 1 \cdot 3 + (-1) = 3 - 1 = 2 \][/tex]
Update the synthetic row:
[tex]\[ \begin{array}{c|cccc} 3 & 1 & -1 & 3 & -4 \\ \hline & 1 & 2 & & \\ \end{array} \][/tex]
- Second step: Multiply the value from the previous step by [tex]\( x = 3 \)[/tex] and add to the next coefficient:
[tex]\[ 2 \cdot 3 + 3 = 6 + 3 = 9 \][/tex]
Update the synthetic row:
[tex]\[ \begin{array}{c|cccc} 3 & 1 & -1 & 3 & -4 \\ \hline & 1 & 2 & 9 & \\ \end{array} \][/tex]
- Third step: Multiply the value from the previous step by [tex]\( x = 3 \)[/tex] and add to the next coefficient:
[tex]\[ 9 \cdot 3 + (-4) = 27 - 4 = 23 \][/tex]
Update the synthetic row:
[tex]\[ \begin{array}{c|cccc} 3 & 1 & -1 & 3 & -4 \\ \hline & 1 & 2 & 9 & 23 \\ \end{array} \][/tex]
5. Final synthetic row: The completed synthetic row is [tex]\( [1, 2, 9, 23] \)[/tex].
6. Result: The final value in the synthetic row is [tex]\( 23 \)[/tex], which is the value of [tex]\( f(3) \)[/tex].
Therefore, using synthetic substitution, [tex]\( f(3) \)[/tex] evaluates to [tex]\( 23 \)[/tex].