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Sagot :
Sure, we will determine whether the numbers [tex]\(-1\)[/tex] and [tex]\(1\)[/tex] are zeros of the polynomial function [tex]\( h(x) = x^4 + 2x^3 + 3x^2 - 4x - 2 \)[/tex] using synthetic division.
### Checking if [tex]\(-1\)[/tex] is a zero:
1. Write down the coefficients of the polynomial: [tex]\( [1, 2, 3, -4, -2] \)[/tex].
2. Perform synthetic division with [tex]\(-1\)[/tex]:
- Start with the leading coefficient:
[tex]\[ 1 \][/tex]
- Multiply by [tex]\(-1\)[/tex] and add to the next coefficient:
[tex]\[ 2 + (-1) \times 1 = 1 \][/tex]
- Repeat the process:
[tex]\[ 3 + (-1) \times 1 = 2 \][/tex]
[tex]\[ -4 + (-1) \times 2 = -6 \][/tex]
[tex]\[ -2 + (-1) \times (-6) = 4 \][/tex]
We end up with the synthetic division row: [tex]\( 1, 1, 2, -6, 4 \)[/tex].
The last number, [tex]\(4\)[/tex], is the remainder. Since the remainder is not [tex]\(0\)[/tex], [tex]\(-1\)[/tex] is not a zero of the polynomial.
### Checking if [tex]\(1\)[/tex] is a zero:
1. Write down the coefficients of the polynomial: [tex]\( [1, 2, 3, -4, -2] \)[/tex].
2. Perform synthetic division with [tex]\(1\)[/tex]:
- Start with the leading coefficient:
[tex]\[ 1 \][/tex]
- Multiply by [tex]\(1\)[/tex] and add to the next coefficient:
[tex]\[ 2 + 1 \times 1 = 3 \][/tex]
- Repeat the process:
[tex]\[ 3 + 1 \times 3 = 6 \][/tex]
[tex]\[ -4 + 1 \times 6 = 2 \][/tex]
[tex]\[ -2 + 1 \times 2 = 0 \][/tex]
We end up with the synthetic division row: [tex]\( 1, 3, 6, 2, 0 \)[/tex].
The last number, [tex]\(0\)[/tex], is the remainder. Since the remainder is [tex]\(0\)[/tex], [tex]\(1\)[/tex] is a zero of the polynomial.
So, to summarize:
- Is [tex]\(-1\)[/tex] a zero of [tex]\( h(x) = x^4 + 2x^3 + 3x^2 - 4x - 2 \)[/tex]? No
- Is [tex]\(1\)[/tex] a zero of [tex]\( h(x) = x^4 + 2x^3 + 3x^2 - 4x - 2 \)[/tex]? Yes
### Checking if [tex]\(-1\)[/tex] is a zero:
1. Write down the coefficients of the polynomial: [tex]\( [1, 2, 3, -4, -2] \)[/tex].
2. Perform synthetic division with [tex]\(-1\)[/tex]:
- Start with the leading coefficient:
[tex]\[ 1 \][/tex]
- Multiply by [tex]\(-1\)[/tex] and add to the next coefficient:
[tex]\[ 2 + (-1) \times 1 = 1 \][/tex]
- Repeat the process:
[tex]\[ 3 + (-1) \times 1 = 2 \][/tex]
[tex]\[ -4 + (-1) \times 2 = -6 \][/tex]
[tex]\[ -2 + (-1) \times (-6) = 4 \][/tex]
We end up with the synthetic division row: [tex]\( 1, 1, 2, -6, 4 \)[/tex].
The last number, [tex]\(4\)[/tex], is the remainder. Since the remainder is not [tex]\(0\)[/tex], [tex]\(-1\)[/tex] is not a zero of the polynomial.
### Checking if [tex]\(1\)[/tex] is a zero:
1. Write down the coefficients of the polynomial: [tex]\( [1, 2, 3, -4, -2] \)[/tex].
2. Perform synthetic division with [tex]\(1\)[/tex]:
- Start with the leading coefficient:
[tex]\[ 1 \][/tex]
- Multiply by [tex]\(1\)[/tex] and add to the next coefficient:
[tex]\[ 2 + 1 \times 1 = 3 \][/tex]
- Repeat the process:
[tex]\[ 3 + 1 \times 3 = 6 \][/tex]
[tex]\[ -4 + 1 \times 6 = 2 \][/tex]
[tex]\[ -2 + 1 \times 2 = 0 \][/tex]
We end up with the synthetic division row: [tex]\( 1, 3, 6, 2, 0 \)[/tex].
The last number, [tex]\(0\)[/tex], is the remainder. Since the remainder is [tex]\(0\)[/tex], [tex]\(1\)[/tex] is a zero of the polynomial.
So, to summarize:
- Is [tex]\(-1\)[/tex] a zero of [tex]\( h(x) = x^4 + 2x^3 + 3x^2 - 4x - 2 \)[/tex]? No
- Is [tex]\(1\)[/tex] a zero of [tex]\( h(x) = x^4 + 2x^3 + 3x^2 - 4x - 2 \)[/tex]? Yes
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