Question

PLEASE HELP ME!!!!! (Urgent)

Find a polynomial function​ f(x) of least degree having only real coefficients with zeros of 0, 2i, and 3+i.


The polynomial function is ​f(x)= _____

Answer 1

`f(x)=x^5-6x^4+14x^3-24x^2+40x`

Explanation:

Given information:

• Polynomial function with real coefficients.
• Zeros: 0, 2i and (3+i).

For any complex number
`z=a+bi` , the complex conjugate of the number is defined as
`z^*=a-bi`.

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 f(x) is a polynomial with real coefficients, and 2i is a root of f(x)=0, then its complex conjugate -2i is also a root of f(x)=0.

Similarly, if (3+i) is a root of f(x)=0, then its complex conjugate (3-i) is also a root of f(x)=0.

Therefore, the polynomial in factored form is:

`f(x)=ax(x-2i)(x-(-2i))(x-(3+i))(x-(3-i))`

`f(x)=ax(x-2i)(x+2i)(x-3-i)(x-3+i)`

As we have not been given a leading coefficient, assume a = 1:

`f(x)=x(x-2i)(x+2i)(x-3-i)(x-3+i)`

Expand the polynomial:

`f(x)=x(x^2+2ix-2ix-4i^2)(x^2-3x+xi-3x+9-3i-xi+3i-i^2)`

`f(x)=x(x^2-4i^2)(x^2-6x+9-i^2)`

`f(x)=x(x^2-4(-1))(x^2-6x+9-(-1))`

`f(x)=x(x^2+4)(x^2-6x+10)`

`f(x)=(x^3+4x)(x^2-6x+10)`

`f(x)=x^5-6x^4+10x^3+4x^3-24x^2+40x`

`f(x)=x^5-6x^4+14x^3-24x^2+40x`