Respuesta :
2 real roots and 2 imaginary roots
f(x) = (x - 3)(x + 3 )(x-2 + i)(x - 2 - i)
(x - 3) and (x + 3) give real roots x = ± 3
(x - 2 + i)(x - 2 - i) give 2 imaginary roots x = 2 ± i
Answer:
Real roots: [tex]x=\pm3[/tex]
Imaginary roots: [tex]x=2\pm i[/tex]
Step-by-step explanation:
We have been given fully form of a polynomial: [tex]f(x)=(x-3)(x+3)[x-(2-i)][x-(2+i)][/tex]. We are asked to find the real and imaginary roots of our given polynomial.
We can see that our given polynomial has two real and two imaginary roots.
Upon equating our given factor to 0 we will get roots of our given polynomial as:
[tex](x-3)(x+3)=0[/tex]
[tex](x-3)=0\text{ or }(x+3)=0[/tex]
[tex]x=3\text{ or }x=-3[/tex]
[tex]x=\pm3[/tex]
Therefore, [tex]x=\pm 3[/tex] are real roots of our given polynomial.
[tex][x-(2-i)][x-(2+i)]=0[/tex]
[tex][x-(2-i)]=0\text{ or }[x-(2+i)]=0[/tex]
[tex]x-(2-i)=0\text{ or }x-(2+i)=0[/tex]
[tex]x=(2-i)\text{ or }x=(2+i)[/tex]
[tex]x=2\pm i[/tex]
Therefore, [tex]x=2\pm i[/tex] are imaginary roots of our given polynomial.
