Answer:
The first and third polynomials have roots in x = 3 and x = -i. (A, C)
Step-by-step explanation:
The quickest form to determine if [tex]x = 3[/tex] and [tex]x = -i[/tex] are roots consist in evaluating each polynomial and proving that result is zero.
A. [tex]f(x) = x^{3}-3\cdot x^{2}+x-3[/tex]
x = 3
[tex]f(3) = 3^{3}-3\cdot (3)^{2}+3-3[/tex]
[tex]f(3) = 27-27+3-3[/tex]
[tex]f(3) = 0[/tex]
x = -i
[tex]f(-i) = (-i)^{3}-3\cdot (-i)^{2}-i-3[/tex]
[tex]f(-i) = i + 3-i-3[/tex]
[tex]f(-i) = 0[/tex]
B. [tex]f(x) = x^{5}+9\cdot x^{4}+28\cdot x^{3} + 36\cdot x^{2}+27\cdot x +27[/tex]
x = 3
[tex]f(3) = 3^{5}+9\cdot (3)^{4}+28\cdot (3)^{3}+36\cdot (3)^{2}+27\cdot (3)+27[/tex]
[tex]f(3) = 2109[/tex]
x = -i
[tex]f(-i) = (-i)^{5}+9\cdot (-i)^{4}+28\cdot (-i)^{3}+36\cdot (-i)^{2}+27\cdot (-i)+27[/tex]
[tex]f(-i) = -i+9 -28\cdot i +36-27\cdot i +27[/tex]
[tex]f(-i) = -56\cdot i +64[/tex]
[tex]f(-i) = 64 -56\cdot i[/tex]
C. [tex]f(x) = x^{5}-9\cdot x^{4}+28\cdot x^{3} - 36\cdot x^{2}+27\cdot x -27[/tex]
x = 3
[tex]f(3) = (3)^{5}-9\cdot (3)^{4}+28\cdot (3)^{3}-36\cdot (3)^{2}+27\cdot (3)-27[/tex]
[tex]f(3) = 0[/tex]
x = -i
[tex]f(-i) = (-i)^{5}-9\cdot (-i)^{4}+28\cdot (-i)^{3}-36\cdot (-i)^{2}+27\cdot (-i)-27[/tex]
[tex]f(-i) = -i - 9+28\cdot i+36-27\cdot i-27[/tex]
[tex]f(-i) = 0[/tex]
D. [tex]f(x) = x^{3}+3\cdot x^{2}+x+3[/tex]
x = 3
[tex]f(3) = (3)^{3}+3\cdot (3)^{2}+(3)+3[/tex]
[tex]f(3) = 60[/tex]
x = -i
[tex]f(-i) = (-i)^{3}+3\cdot (-i)^{2}+(-i)+3[/tex]
[tex]f(-i) = -i+3-i+3[/tex]
[tex]f(-i) = 6-i\,2[/tex]
The first and third polynomials have roots in x = 3 and x = -i. (A, C)
