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Sagot :
Let's find the polynomial function step by step.
We know the polynomial must have a leading coefficient of 1 and the roots [tex]\((7+i)\)[/tex], [tex]\((5-i)\)[/tex], [tex]\((7-i)\)[/tex], and [tex]\((5+i)\)[/tex], each with multiplicity 1.
To construct the polynomial, we will use the fact that if [tex]\(r\)[/tex] is a root of the polynomial, then [tex]\((x - r)\)[/tex] is a factor of the polynomial.
1. Start with the roots given:
- [tex]\(7 + i\)[/tex]
- [tex]\(7 - i\)[/tex]
- [tex]\(5 - i\)[/tex]
- [tex]\(5 + i\)[/tex]
2. Form the factors of the polynomial based on these roots:
- [tex]\((x - (7 + i))\)[/tex]
- [tex]\((x - (7 - i))\)[/tex]
- [tex]\((x - (5 - i))\)[/tex]
- [tex]\((x - (5 + i))\)[/tex]
Thus, the polynomial can be written as:
[tex]\[ f(x) = (x - (7+i))(x - (7-i))(x - (5-i))(x - (5+i)) \][/tex]
Now simplify each pair of factors:
[tex]\[ (x - (7+i))(x - (7-i)) = \left((x - 7) - i\right) \left((x - 7) + i\right) \][/tex]
Using the difference of squares formula, [tex]\(a^2 - b^2\)[/tex]:
[tex]\[ = ((x - 7)^2 - i^2) = (x - 7)^2 - (-1) = (x - 7)^2 + 1 \][/tex]
Similarly,
[tex]\[ (x - (5-i))(x - (5+i)) = \left((x - 5) + i\right) \left((x - 5) - i\right) \][/tex]
Using the difference of squares again:
[tex]\[ = ((x - 5)^2 - i^2) = (x - 5)^2 - (-1) = (x - 5)^2 + 1 \][/tex]
Putting it all together, the polynomial function is:
[tex]\[ f(x) = \left((x - 7)^2 + 1\right) \left((x - 5)^2 + 1\right) \][/tex]
Upon reviewing the options given in the question:
1. [tex]\((x+7)(x-i)(x+5)(x+i)\)[/tex]
2. [tex]\((x-7)(x-1)(x-5)(x+1)\)[/tex]
3. [tex]\((x-(7-i))(x-(5+i))(x-(7+i))(x-(5-i))\)[/tex]
4. [tex]\((x+(7-i))(x+(5+i)(x+(7+i))(x+(5-i))\)[/tex]
The correct answer matches option 3, [tex]\((x-(7-i))(x-(5+i))(x-(7+i))(x-(5-i))\)[/tex].
So the polynomial function with the given roots and a leading coefficient of 1 is:
[tex]\[ f(x) = (x - (7+i))(x - (7-i))(x - (5-i))(x - (5+i)) \][/tex]
The correct answer is:
[tex]\[ \boxed{(x-(7-i))(x-(5+i))(x-(7+i))(x-(5-i))} \][/tex]
We know the polynomial must have a leading coefficient of 1 and the roots [tex]\((7+i)\)[/tex], [tex]\((5-i)\)[/tex], [tex]\((7-i)\)[/tex], and [tex]\((5+i)\)[/tex], each with multiplicity 1.
To construct the polynomial, we will use the fact that if [tex]\(r\)[/tex] is a root of the polynomial, then [tex]\((x - r)\)[/tex] is a factor of the polynomial.
1. Start with the roots given:
- [tex]\(7 + i\)[/tex]
- [tex]\(7 - i\)[/tex]
- [tex]\(5 - i\)[/tex]
- [tex]\(5 + i\)[/tex]
2. Form the factors of the polynomial based on these roots:
- [tex]\((x - (7 + i))\)[/tex]
- [tex]\((x - (7 - i))\)[/tex]
- [tex]\((x - (5 - i))\)[/tex]
- [tex]\((x - (5 + i))\)[/tex]
Thus, the polynomial can be written as:
[tex]\[ f(x) = (x - (7+i))(x - (7-i))(x - (5-i))(x - (5+i)) \][/tex]
Now simplify each pair of factors:
[tex]\[ (x - (7+i))(x - (7-i)) = \left((x - 7) - i\right) \left((x - 7) + i\right) \][/tex]
Using the difference of squares formula, [tex]\(a^2 - b^2\)[/tex]:
[tex]\[ = ((x - 7)^2 - i^2) = (x - 7)^2 - (-1) = (x - 7)^2 + 1 \][/tex]
Similarly,
[tex]\[ (x - (5-i))(x - (5+i)) = \left((x - 5) + i\right) \left((x - 5) - i\right) \][/tex]
Using the difference of squares again:
[tex]\[ = ((x - 5)^2 - i^2) = (x - 5)^2 - (-1) = (x - 5)^2 + 1 \][/tex]
Putting it all together, the polynomial function is:
[tex]\[ f(x) = \left((x - 7)^2 + 1\right) \left((x - 5)^2 + 1\right) \][/tex]
Upon reviewing the options given in the question:
1. [tex]\((x+7)(x-i)(x+5)(x+i)\)[/tex]
2. [tex]\((x-7)(x-1)(x-5)(x+1)\)[/tex]
3. [tex]\((x-(7-i))(x-(5+i))(x-(7+i))(x-(5-i))\)[/tex]
4. [tex]\((x+(7-i))(x+(5+i)(x+(7+i))(x+(5-i))\)[/tex]
The correct answer matches option 3, [tex]\((x-(7-i))(x-(5+i))(x-(7+i))(x-(5-i))\)[/tex].
So the polynomial function with the given roots and a leading coefficient of 1 is:
[tex]\[ f(x) = (x - (7+i))(x - (7-i))(x - (5-i))(x - (5+i)) \][/tex]
The correct answer is:
[tex]\[ \boxed{(x-(7-i))(x-(5+i))(x-(7+i))(x-(5-i))} \][/tex]
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