Let's determine the number of possible imaginary (complex) solutions for each of the given polynomials. To find out how many complex roots each polynomial has, it is important to first recall some key properties about polynomial roots:
1. Real coefficients and complex roots: If a polynomial has real coefficients, its complex roots must appear in conjugate pairs. For example, if [tex]\(a + bi\)[/tex] is a root, then [tex]\(a - bi\)[/tex] must also be a root.
Given this information, let's consider each polynomial to determine the number of complex solutions:
### Part (a) [tex]\( x^5 + 13 x^4 + 36 x^3 - 148 x^2 - 832 x - 960 \)[/tex]
This is a 5th-degree polynomial. It could have up to 5 roots in total. By determining root properties, we found that there are no imaginary (complex) solutions.
### Part (b) [tex]\( x^4 + 11 x^3 + 14 x^2 - 176 x - 480 \)[/tex]
This is a 4th-degree polynomial. Thus, it can have up to 4 roots. By viewing the determined properties, we have that there are no imaginary (complex) solutions.
### Part (c) [tex]\( x^2 + 10 x + 24 \)[/tex]
This is a 2nd-degree polynomial. As a quadratic, it can have up to 2 roots. The number of imaginary (complex) solutions found is zero.
### Part (d) [tex]\( x^3 + 15 x^2 + 74 x + 120 \)[/tex]
This is a 3rd-degree polynomial and could have up to 3 roots. Based on polynomial analysis, it shows zero imaginary (complex) solutions.
### Part (e) [tex]\( -4 x - 16 \)[/tex]
This is a 1st-degree polynomial (a linear polynomial), which can have just 1 root. It's clear that since it is linear and non-zero, there are no imaginary (complex) solutions.
### Summary of Results
Given the analysis of the given polynomials, we can conclude the number of possible imaginary (complex) solutions for each polynomial:
a) 0 (imaginary solutions)
b) 0 (imaginary solutions)
c) 0 (imaginary solutions)
d) 0 (imaginary solutions)
e) 0 (imaginary solutions)
The answer is: 0, 0, 0, 0, 0.
