Respuesta :
Answer:
f(x)=x³-2x²+9x-18
Step-by-step explanation:
complex zeros always occur in pairs.
zeros are 2,3i,-3i
f(x)=(x-2)(x-3i)(x+3i)=(x-2)((x)^2-(3i)^2)
=(x-2)(x^2-9i^2)
=(x-2)(x^2+9)
=x^3-2x^2+9x-18
A degree 3 polynomial with real coefficients having zeros 2 and 3i and a lead coefficient of 1 is
[tex]P(x)=x^3+9x-2x^2-18[/tex]
Given :
zeros of polynomial are 2 and 3i and a lead coefficient of 1.
zeros of polynomial are occurs in pairs
3i is one of the zero . Another zero is -3i
So zeros are 2,3i,-3i
Now we write the zeros in factor form
If 'a' is a zero then (x-a) is a factor
zeros are 2,3i,-3i
the factor form is
[tex]\left(x-2\right)\left(x-3i\right)\left(x-\left(-3i\right)\right)[/tex]
Now we multiply it to get the polynomial
[tex]\left(x-2\right)\left(x-3i\right)\left(x-\left(-3i\right)\right)\\\left(x-2\right)\left(x-3i\right)\left(x+3i\right)\\\left(x-2\right)\left(x^2+9\right)\\x^3+9x-2x^2-18[/tex]
The polynomial P(x) can be written as
[tex]P(x)=a(x^3+9x-2x^2-18)[/tex]
where 'a' is the leading coefficient
we know that leading coefficient is 1
[tex]P(x)=x^3+9x-2x^2-18[/tex]
Learn more: brainly.com/question/15409950
