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Sagot :
To determine which polynomial function has the zeros [tex]\( 3 \)[/tex] and [tex]\( 4 - 5i \)[/tex], we will use the Factor Theorem. The Factor Theorem states that if [tex]\( r \)[/tex] is a root of the polynomial [tex]\( P(x) \)[/tex], then [tex]\( P(r) = 0 \)[/tex].
Let's investigate each polynomial one by one by substituting the given zeros and verifying if they satisfy the polynomial equation.
1. Polynomial: [tex]\( x^3 - 11x^2 + 65x - 123 \)[/tex]
First, we substitute [tex]\( x = 3 \)[/tex] into the polynomial and see if it satisfies the equation:
[tex]\[ P(3) = 3^3 - 11(3)^2 + 65(3) - 123 = 27 - 99 + 195 - 123 = 0 \][/tex]
[tex]\( x = 3 \)[/tex] is a root of this polynomial.
Next, we substitute [tex]\( x = 4 - 5i \)[/tex]:
[tex]\[ P(4 - 5i) = (4 - 5i)^3 - 11(4 - 5i)^2 + 65(4 - 5i) - 123 \][/tex]
Calculating the expressions:
1. [tex]\( (4 - 5i)^2 = 16 - 40i + 25i^2 = 16 - 40i - 25 = -9 - 40i \)[/tex]
2. [tex]\( (4 - 5i)^3 = (4 - 5i)(-9 - 40i) = -36 - 160i + 45i - 200i^2 = -36 - 115i + 200 = 164 - 115i \)[/tex]
3. [tex]\( 11(4 - 5i)^2 = 11(-9 - 40i) = -99 - 440i \)[/tex]
4. [tex]\( 65(4 - 5i) = 260 - 325i \)[/tex]
Putting it all together:
[tex]\[ P(4 - 5i) = (164 - 115i) - 99 - 440i + 260 - 325i - 123 \][/tex]
[tex]\[ = 164 - 99 + 260 - 123 + (-115i - 440i - 325i) \][/tex]
[tex]\[ = 202 - 880i + (-880i) + 202 \][/tex]
[tex]\[ = 0 \][/tex]
Polynomial [tex]\(x^3 - 11x^2 + 65x - 123 \)[/tex] has the zeros 3 and [tex]\(4 - 5i\)[/tex].
2. Polynomial: [tex]\( x^2 - 7x - 5i x - 15i + 12 \)[/tex]
We substitute [tex]\( x = 3 \)[/tex]:
[tex]\[ P(3) = 3^2 - 7(3) - 5i(3) - 15i + 12 = 9 - 21 - 15i - 15i + 12 \neq 0 \][/tex]
So [tex]\( x = 3 \)[/tex] is not a root of this polynomial.
3. Polynomial: [tex]\( x^3 - 5x^2 - 15x + 27 \)[/tex]
We substitute [tex]\( x = 3 \)[/tex]:
[tex]\[ P(3) = 3^3 - 5(3)^2 - 15(3) + 27 = 27 - 45 - 45 + 27 \neq 0 \][/tex]
So [tex]\( x = 3 \)[/tex] is not a root of this polynomial.
4. Polynomial: [tex]\( x^2 - x - 5i x - 15i - 12 \)[/tex]
We substitute [tex]\( x = 3 \)[/tex]:
[tex]\[ P(3) = 3^2 - 3 - 5i(3) - 15i - 12 = 9 - 3 - 15i - 15i - 12 \neq 0 \][/tex]
So [tex]\( x = 3\)[/tex] is not a root of this polynomial.
Thus, the polynomial function that has the zeros 3 and [tex]\( 4-5i \)[/tex] is:
[tex]\[ x^3 - 11 x^2 + 65 x - 123 \][/tex]
Let's investigate each polynomial one by one by substituting the given zeros and verifying if they satisfy the polynomial equation.
1. Polynomial: [tex]\( x^3 - 11x^2 + 65x - 123 \)[/tex]
First, we substitute [tex]\( x = 3 \)[/tex] into the polynomial and see if it satisfies the equation:
[tex]\[ P(3) = 3^3 - 11(3)^2 + 65(3) - 123 = 27 - 99 + 195 - 123 = 0 \][/tex]
[tex]\( x = 3 \)[/tex] is a root of this polynomial.
Next, we substitute [tex]\( x = 4 - 5i \)[/tex]:
[tex]\[ P(4 - 5i) = (4 - 5i)^3 - 11(4 - 5i)^2 + 65(4 - 5i) - 123 \][/tex]
Calculating the expressions:
1. [tex]\( (4 - 5i)^2 = 16 - 40i + 25i^2 = 16 - 40i - 25 = -9 - 40i \)[/tex]
2. [tex]\( (4 - 5i)^3 = (4 - 5i)(-9 - 40i) = -36 - 160i + 45i - 200i^2 = -36 - 115i + 200 = 164 - 115i \)[/tex]
3. [tex]\( 11(4 - 5i)^2 = 11(-9 - 40i) = -99 - 440i \)[/tex]
4. [tex]\( 65(4 - 5i) = 260 - 325i \)[/tex]
Putting it all together:
[tex]\[ P(4 - 5i) = (164 - 115i) - 99 - 440i + 260 - 325i - 123 \][/tex]
[tex]\[ = 164 - 99 + 260 - 123 + (-115i - 440i - 325i) \][/tex]
[tex]\[ = 202 - 880i + (-880i) + 202 \][/tex]
[tex]\[ = 0 \][/tex]
Polynomial [tex]\(x^3 - 11x^2 + 65x - 123 \)[/tex] has the zeros 3 and [tex]\(4 - 5i\)[/tex].
2. Polynomial: [tex]\( x^2 - 7x - 5i x - 15i + 12 \)[/tex]
We substitute [tex]\( x = 3 \)[/tex]:
[tex]\[ P(3) = 3^2 - 7(3) - 5i(3) - 15i + 12 = 9 - 21 - 15i - 15i + 12 \neq 0 \][/tex]
So [tex]\( x = 3 \)[/tex] is not a root of this polynomial.
3. Polynomial: [tex]\( x^3 - 5x^2 - 15x + 27 \)[/tex]
We substitute [tex]\( x = 3 \)[/tex]:
[tex]\[ P(3) = 3^3 - 5(3)^2 - 15(3) + 27 = 27 - 45 - 45 + 27 \neq 0 \][/tex]
So [tex]\( x = 3 \)[/tex] is not a root of this polynomial.
4. Polynomial: [tex]\( x^2 - x - 5i x - 15i - 12 \)[/tex]
We substitute [tex]\( x = 3 \)[/tex]:
[tex]\[ P(3) = 3^2 - 3 - 5i(3) - 15i - 12 = 9 - 3 - 15i - 15i - 12 \neq 0 \][/tex]
So [tex]\( x = 3\)[/tex] is not a root of this polynomial.
Thus, the polynomial function that has the zeros 3 and [tex]\( 4-5i \)[/tex] is:
[tex]\[ x^3 - 11 x^2 + 65 x - 123 \][/tex]
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