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Sagot :
Let's analyze each polynomial to determine the number of imaginary (complex) solutions:
### a) [tex]\(x^3 - x^2 - 36x + 36\)[/tex]
1. Degree: The polynomial is of degree 3.
2. Possible Roots: A cubic (degree 3) polynomial can have at most 3 roots, and imaginary (complex) roots come in conjugate pairs.
3. Roots Analysis: Since imaginary roots exist in pairs, this polynomial can have 0 or 2 imaginary roots. Determining the exact number would involve solving the polynomial and examining the roots.
Given that this polynomial has [tex]\(0\)[/tex] imaginary solutions.
### b) [tex]\(-2x + 2\)[/tex]
1. Degree: The polynomial is of degree 1.
2. Possible Roots: A linear polynomial has exactly one root, which is real.
3. Roots Analysis: Since it's a linear polynomial, it cannot have any imaginary roots.
This polynomial has [tex]\(0\)[/tex] imaginary solutions.
### c) [tex]\(x^5 - 43x^3 + 6x^2 + 252x - 216\)[/tex]
1. Degree: The polynomial is of degree 5.
2. Possible Roots: A quintic (degree 5) polynomial can have at most 5 roots. Imaginary roots must come in pairs, so the possible imaginary roots could be 0, 2, or 4.
3. Roots Analysis: Determining the exact number would involve solving and examining the polynomial's roots.
Given that this polynomial has [tex]\(0\)[/tex] imaginary solutions.
### d) [tex]\(x^2 - 7x + 6\)[/tex]
1. Degree: The polynomial is of degree 2.
2. Possible Roots: A quadratic (degree 2) polynomial can have at most 2 roots. If it has imaginary roots, they must occur in pairs.
3. Roots Analysis: For a quadratic polynomial to have imaginary roots, its discriminant ([tex]\(b^2 - 4ac\)[/tex]) must be less than 0.
Given this polynomial has [tex]\(0\)[/tex] imaginary solutions.
### e) [tex]\(x^4 - 3x^3 - 34x^2 + 108x - 72\)[/tex]
1. Degree: The polynomial is of degree 4.
2. Possible Roots: A quartic (degree 4) polynomial can have at most 4 roots, and imaginary roots can occur in pairs.
3. Roots Analysis: The possible numbers of imaginary roots are 0, 2, or 4.
Given that this polynomial has [tex]\(0\)[/tex] imaginary solutions.
### Summary
The number of imaginary (complex) solutions for each polynomial is:
- a) 0
- b) 0
- c) 0
- d) 0
- e) 0
Thus, the answer is: [tex]\( \mathbf{0, \, 0, \, 0, \, 0, \, 0} \)[/tex]
### a) [tex]\(x^3 - x^2 - 36x + 36\)[/tex]
1. Degree: The polynomial is of degree 3.
2. Possible Roots: A cubic (degree 3) polynomial can have at most 3 roots, and imaginary (complex) roots come in conjugate pairs.
3. Roots Analysis: Since imaginary roots exist in pairs, this polynomial can have 0 or 2 imaginary roots. Determining the exact number would involve solving the polynomial and examining the roots.
Given that this polynomial has [tex]\(0\)[/tex] imaginary solutions.
### b) [tex]\(-2x + 2\)[/tex]
1. Degree: The polynomial is of degree 1.
2. Possible Roots: A linear polynomial has exactly one root, which is real.
3. Roots Analysis: Since it's a linear polynomial, it cannot have any imaginary roots.
This polynomial has [tex]\(0\)[/tex] imaginary solutions.
### c) [tex]\(x^5 - 43x^3 + 6x^2 + 252x - 216\)[/tex]
1. Degree: The polynomial is of degree 5.
2. Possible Roots: A quintic (degree 5) polynomial can have at most 5 roots. Imaginary roots must come in pairs, so the possible imaginary roots could be 0, 2, or 4.
3. Roots Analysis: Determining the exact number would involve solving and examining the polynomial's roots.
Given that this polynomial has [tex]\(0\)[/tex] imaginary solutions.
### d) [tex]\(x^2 - 7x + 6\)[/tex]
1. Degree: The polynomial is of degree 2.
2. Possible Roots: A quadratic (degree 2) polynomial can have at most 2 roots. If it has imaginary roots, they must occur in pairs.
3. Roots Analysis: For a quadratic polynomial to have imaginary roots, its discriminant ([tex]\(b^2 - 4ac\)[/tex]) must be less than 0.
Given this polynomial has [tex]\(0\)[/tex] imaginary solutions.
### e) [tex]\(x^4 - 3x^3 - 34x^2 + 108x - 72\)[/tex]
1. Degree: The polynomial is of degree 4.
2. Possible Roots: A quartic (degree 4) polynomial can have at most 4 roots, and imaginary roots can occur in pairs.
3. Roots Analysis: The possible numbers of imaginary roots are 0, 2, or 4.
Given that this polynomial has [tex]\(0\)[/tex] imaginary solutions.
### Summary
The number of imaginary (complex) solutions for each polynomial is:
- a) 0
- b) 0
- c) 0
- d) 0
- e) 0
Thus, the answer is: [tex]\( \mathbf{0, \, 0, \, 0, \, 0, \, 0} \)[/tex]
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