Answer:
[tex]P(x)=x^3-3x^2+16x-48[/tex]
Step-by-step explanation:
Given information:
- Polynomial function with real coefficients.
- Zeros: 3 and 4i
- Lead coefficient of 1.
For any complex number [tex]z=a+bi[/tex], the complex conjugate of the number is defined as [tex]z^*=a-bi[/tex].
If f(z) is a polynomial with real coefficients, and z₁ is a root of f(z)=0, then its complex conjugate z₁* is also a root of f(z)=0.
Therefore, if P(x) is a polynomial with real coefficients, and 4i is a root of f(x)=0, then its complex conjugate -4i is also a root of P(x)=0.
Therefore, the polynomial in factored form is:
[tex]P(x)=1(x-3)(x-4i)(x-(-4i))[/tex]
[tex]P(x)=(x-3)(x-4i)(x+4i)[/tex]
Expand the polynomial:
[tex]P(x)=(x-3)(x^2+4ix-4ix-16i^2)[/tex]
[tex]P(x)=(x-3)(x^2-16(-1))[/tex]
[tex]P(x)=(x-3)(x^2+16)[/tex]
[tex]P(x)=x^3+16x-3x^2-48[/tex]
[tex]P(x)=x^3-3x^2+16x-48[/tex]
