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Answer :
Certainly! Let's analyze each polynomial and determine the number of possible imaginary (complex) roots for each one.
1. Polynomial [tex]\( a \)[/tex]: [tex]\( x^4 - 7x^3 - x^2 + 67x - 60 \)[/tex]
- Degree: 4
- According to the Fundamental Theorem of Algebra, a polynomial of degree [tex]\( n \)[/tex] will have exactly [tex]\( n \)[/tex] roots, some of which may be complex.
- Complex roots (non-real) always come in conjugate pairs.
- The number of complex roots must be even.
2. Polynomial [tex]\( b \)[/tex]: [tex]\( -4x + 4 \)[/tex]
- Degree: 1
- A linear polynomial has exactly one real root and no complex roots (since it's of odd degree and coefficients are real).
3. Polynomial [tex]\( c \)[/tex]: [tex]\( x^2 + 2x - 3 \)[/tex]
- Degree: 2
- For a quadratic polynomial, if the discriminant ([tex]\( b^2 - 4ac \)[/tex]) is negative, the roots are complex.
- Discriminant: [tex]\( 2^2 - 4(1)(-3) = 4 + 12 = 16 \)[/tex] (positive)
- Therefore, both roots are real and there are no complex roots.
4. Polynomial [tex]\( d \)[/tex]: [tex]\( x^5 - 9x^4 + 13x^3 + 69x^2 - 194x + 120 \)[/tex]
- Degree: 5
- An odd-degree polynomial with real coefficients has at least one real root, implying it can have 0, 2, or 4 complex roots (since complex roots come in pairs).
5. Polynomial [tex]\( e \)[/tex]: [tex]\( x^3 - 3x^2 - 13x + 15 \)[/tex]
- Degree: 3
- An odd-degree polynomial with real coefficients will have at least one real root.
- For a cubic polynomial, this means it can have either 0 or 2 complex roots.
Here’s a summary of the number of possible complex roots for each polynomial:
a) 0, 2, or 4 complex roots (must be even)
b) 0 complex roots (linear polynomial)
c) 0 complex roots (quadratic with positive discriminant)
d) 0, 2, or 4 complex roots (odd degree and complex roots come in pairs)
e) 0, or 2 complex roots (cubic polynomial)
Therefore, the final answer is:
a) 0, 2, or 4,
b) 0,
c) 0,
d) 0, 2, or 4,
e) 0 or 2.
1. Polynomial [tex]\( a \)[/tex]: [tex]\( x^4 - 7x^3 - x^2 + 67x - 60 \)[/tex]
- Degree: 4
- According to the Fundamental Theorem of Algebra, a polynomial of degree [tex]\( n \)[/tex] will have exactly [tex]\( n \)[/tex] roots, some of which may be complex.
- Complex roots (non-real) always come in conjugate pairs.
- The number of complex roots must be even.
2. Polynomial [tex]\( b \)[/tex]: [tex]\( -4x + 4 \)[/tex]
- Degree: 1
- A linear polynomial has exactly one real root and no complex roots (since it's of odd degree and coefficients are real).
3. Polynomial [tex]\( c \)[/tex]: [tex]\( x^2 + 2x - 3 \)[/tex]
- Degree: 2
- For a quadratic polynomial, if the discriminant ([tex]\( b^2 - 4ac \)[/tex]) is negative, the roots are complex.
- Discriminant: [tex]\( 2^2 - 4(1)(-3) = 4 + 12 = 16 \)[/tex] (positive)
- Therefore, both roots are real and there are no complex roots.
4. Polynomial [tex]\( d \)[/tex]: [tex]\( x^5 - 9x^4 + 13x^3 + 69x^2 - 194x + 120 \)[/tex]
- Degree: 5
- An odd-degree polynomial with real coefficients has at least one real root, implying it can have 0, 2, or 4 complex roots (since complex roots come in pairs).
5. Polynomial [tex]\( e \)[/tex]: [tex]\( x^3 - 3x^2 - 13x + 15 \)[/tex]
- Degree: 3
- An odd-degree polynomial with real coefficients will have at least one real root.
- For a cubic polynomial, this means it can have either 0 or 2 complex roots.
Here’s a summary of the number of possible complex roots for each polynomial:
a) 0, 2, or 4 complex roots (must be even)
b) 0 complex roots (linear polynomial)
c) 0 complex roots (quadratic with positive discriminant)
d) 0, 2, or 4 complex roots (odd degree and complex roots come in pairs)
e) 0, or 2 complex roots (cubic polynomial)
Therefore, the final answer is:
a) 0, 2, or 4,
b) 0,
c) 0,
d) 0, 2, or 4,
e) 0 or 2.
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