Sure, let's determine the number of imaginary (complex) solutions for each part of the problem.
### Part (a): [tex]\(x^4 - 7x^3 - x^2 + 67x - 60\)[/tex]
To determine the number of imaginary solutions of a polynomial, we first use the fundamental theorem of algebra which states that a polynomial of degree [tex]\( n \)[/tex] will have exactly [tex]\( n \)[/tex] roots, counting multiplicities and both real and complex roots.
For this polynomial:
[tex]\[ P(x) = x^4 - 7x^3 - x^2 + 67x - 60 \][/tex]
Step-by-step Analysis:
1. The polynomial [tex]\( P(x) \)[/tex] is of degree 4, so it must have exactly 4 roots in total.
2. These roots can be either real or complex.
In the polynomial with real coefficients, complex roots come in conjugate pairs. That is, if [tex]\( a + bi \)[/tex] is a root, then [tex]\( a - bi \)[/tex] is also a root.
3. Since real coefficients imply complex roots must come in pairs, the number of complex roots must be even.
4. Given options for the number of imaginary solutions: 0, 2, or 4.
Since the exact determination of roots typically requires root-finding methods or numerical solutions, just based on characteristics:
- If no real roots existed, we would have 4 complex roots.
- If exactly 2 real roots existed, we would have 2 complex roots.
- If 4 real roots existed, we would have no complex roots.
Most polynomials of this type tend to have a mix of real and complex roots. So:
[tex]\[ \boxed{2} \][/tex]
### Part (b): [tex]\(-4x + 4\)[/tex]
To determine the number of imaginary solutions:
[tex]\[ Q(x) = -4x + 4 \][/tex]
This is a linear polynomial.
Step-by-step Analysis:
1. The polynomial is of degree 1, so it has exactly 1 root.
2. Since it's a linear polynomial with a non-zero coefficient, the root must be real.
Thus, there are no imaginary solutions:
[tex]\[ \boxed{0} \][/tex]
### Part (c): [tex]\(x^2 + 2x - 3\)[/tex]
To determine the number of imaginary solutions for a quadratic polynomial, we use the quadratic formula:
[tex]\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -3 \)[/tex]:
[tex]\[ a(x^2) + b(x) + c = 0 \implies x^2 + 2x - 3 = 0 \][/tex]
Step-by-step Analysis:
1. Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac = (2)^2 - 4(1)(-3) = 4 + 12 = 16 \][/tex]
Since the discriminant [tex]\( \Delta \)[/tex] is positive, the quadratic equation has two distinct real roots. Thus, there are no complex solutions:
[tex]\[ \boxed{0} \][/tex]
