Answer:
The answer is {[tex]x^{3}[/tex]-(5)[tex]x^{2}[/tex] (36)× [tex]x[/tex]-(180)
Step-by-step explanation:
A polynomial with rational coefficients has roots 5 and -6i (There i is the
imaginary number not variable )
As we knew that imaginary no comes in pair . This means that if -6i is
one root then the other root will be 6i
So if we assume the lowest polynomial that is possible is given as
{[tex]x^{3}[/tex]-(sum of all roots)[tex]x^{2}[/tex] +(roots taken two at a time)
[tex]x[/tex]-(product of all roots)
{[tex]x^{3}[/tex]-(5+6i+(-6i))[tex]x^{2}[/tex] +(5×6i +5×(-6i) +6i×(-6i))
[tex]x[/tex]-(5×6i×(-6i))
[tex]i^2[/tex] = -1
{[tex]x^{3}[/tex]-(5)[tex]x^{2}[/tex] (36)× [tex]x[/tex]-(180)
The general case we assume that the three previous roots and rest roots
a4,a5,a6 ...............an
[tex]x^n[/tex]-(sum of all roots)×[tex]x^{n-1}[/tex] +(roots taken two at a
time) ×[tex]x^{n-2}[/tex]- .......................................... (product of all roots)
