Answer :
To find [tex]\( f(-3) \)[/tex] and [tex]\( f(4) \)[/tex] for the given function [tex]\( f(x) = x^3 + 3x^2 + 2x - 50 \)[/tex] using synthetic substitution, we use the following steps:
### For [tex]\( f(-3) \)[/tex]:
1. Write the coefficients of the polynomial: [tex]\( 1, 3, 2, -50 \)[/tex].
2. Set up the synthetic division with [tex]\(-3\)[/tex] as the divisor.
3. Bring down the first coefficient, [tex]\( 1 \)[/tex], immediately.
4. Multiply [tex]\(-3\)[/tex] by the number just written below (start with [tex]\( 1 \)[/tex]) and write the result under the next coefficient:
[tex]\[
-3 \times 1 = -3
\][/tex]
5. Add this result to the next coefficient:
[tex]\[
3 + (-3) = 0
\][/tex]
6. Repeat the multiplication with the new sum:
[tex]\[
-3 \times 0 = 0
\][/tex]
7. Add to the next coefficient:
[tex]\[
2 + 0 = 2
\][/tex]
8. Multiply and add once more:
[tex]\[
-3 \times 2 = -6
\][/tex]
[tex]\[
-50 + (-6) = -56
\][/tex]
The final result under the line is [tex]\( f(-3) = -56 \)[/tex].
### For [tex]\( f(4) \)[/tex]:
1. Use the same coefficients: [tex]\( 1, 3, 2, -50 \)[/tex].
2. Set up the synthetic division with [tex]\( 4 \)[/tex] as the divisor.
3. Bring down the first coefficient, [tex]\( 1 \)[/tex].
4. Multiply and add step-by-step:
[tex]\[
4 \times 1 = 4 \quad \Rightarrow \quad 3 + 4 = 7
\][/tex]
[tex]\[
4 \times 7 = 28 \quad \Rightarrow \quad 2 + 28 = 30
\][/tex]
[tex]\[
4 \times 30 = 120 \quad \Rightarrow \quad -50 + 120 = 70
\][/tex]
The result under the line is [tex]\( f(4) = 70 \)[/tex].
Thus, the values are [tex]\( f(-3) = -56 \)[/tex] and [tex]\( f(4) = 70 \)[/tex].
The correct answer choice is:
d. [tex]\(-56, 70\)[/tex]
### For [tex]\( f(-3) \)[/tex]:
1. Write the coefficients of the polynomial: [tex]\( 1, 3, 2, -50 \)[/tex].
2. Set up the synthetic division with [tex]\(-3\)[/tex] as the divisor.
3. Bring down the first coefficient, [tex]\( 1 \)[/tex], immediately.
4. Multiply [tex]\(-3\)[/tex] by the number just written below (start with [tex]\( 1 \)[/tex]) and write the result under the next coefficient:
[tex]\[
-3 \times 1 = -3
\][/tex]
5. Add this result to the next coefficient:
[tex]\[
3 + (-3) = 0
\][/tex]
6. Repeat the multiplication with the new sum:
[tex]\[
-3 \times 0 = 0
\][/tex]
7. Add to the next coefficient:
[tex]\[
2 + 0 = 2
\][/tex]
8. Multiply and add once more:
[tex]\[
-3 \times 2 = -6
\][/tex]
[tex]\[
-50 + (-6) = -56
\][/tex]
The final result under the line is [tex]\( f(-3) = -56 \)[/tex].
### For [tex]\( f(4) \)[/tex]:
1. Use the same coefficients: [tex]\( 1, 3, 2, -50 \)[/tex].
2. Set up the synthetic division with [tex]\( 4 \)[/tex] as the divisor.
3. Bring down the first coefficient, [tex]\( 1 \)[/tex].
4. Multiply and add step-by-step:
[tex]\[
4 \times 1 = 4 \quad \Rightarrow \quad 3 + 4 = 7
\][/tex]
[tex]\[
4 \times 7 = 28 \quad \Rightarrow \quad 2 + 28 = 30
\][/tex]
[tex]\[
4 \times 30 = 120 \quad \Rightarrow \quad -50 + 120 = 70
\][/tex]
The result under the line is [tex]\( f(4) = 70 \)[/tex].
Thus, the values are [tex]\( f(-3) = -56 \)[/tex] and [tex]\( f(4) = 70 \)[/tex].
The correct answer choice is:
d. [tex]\(-56, 70\)[/tex]
