Answer:
The roots of the polynomial equation are:
-1,3+2i and 3-2i
Step-by-step explanation:
We are given a polynomial equation as:
[tex]x^3-4x^2+2x+10=x^2-5x-3[/tex]
this equation could also be written as:
[tex]y=x^3-4x^2+2x+10----(1)[/tex]
and [tex]y=x^2-5x-3-----(2)[/tex]
Now if this equation has a complex root as:
[tex]3+2i[/tex]
and we know that for any polynomial equation witth real coefficients the complex roots always appear in pair.
if a+ib is some root of a polynomial equation with real coefficients then its conjugate a-ib is also a root of the equation.
Hence, the other root of this polynomial equation will be:[tex]3-2i[/tex]
Now we will graph these system of equations i.e. equation (1) and equation (2).
on solving the equation we have:
[tex]x^3-5x^2+7x+13=0\\\\(x+1)(x^2-6x+13)=0[/tex]
(since
[tex]x^3-4x^2+2x+10=x^2-5x-3\\\\x^3-4x^2+2x+10-x^2-(-5x)-(-3)=0\\\\x^3-4x^2-x^2+2x+5x+10+3=0\\\\x^3-5x^2+7x+13=0[/tex]
)
Hence on solving the equation we have:
x=-1,3+2i and 3-2i are the roots of the equation.
