Question

30+ points need it asap! :) pls show work.

Write a polynomial function f of least degree with the following zeros: 3, 4 + i

Answer 1

`f(x)=(x-3)(x^2-8x+17)`

Or:

`f(x)=x^3-11x^2+41x-51`

Explanation:

To find our factors, we can work backwards.

We know that 3 is a zero. This means that:

`x=3`

Subtract 3 from both sides:

`(x-3)=0`

So, (x-3) is one of our factors.

We also know that (4+i) is a zero. So:

`x=4+i`

First, let’s isolate the imaginary. So, subtract 4 from both sides:

`(x-4)=i`

Now, let’s square both sides:

`(x-4)^2=(i)^2`

Expand the left. Evaluate the right:

`(x^2-8x+16)=-1`

Add 1 to both sides. So, our factor is:

`(x^2-8x+17)=0`

Hence, our polynomial function is:

`f(x)=(x-3)(x^2-8x+17)`

This is also the least degree since we did the most minimum we can do to solve backwards.

Further Notes:

If we want to convert this to standard form, we can distribute:

`f(x)=(x^3-8x^2+17x)+(-3x^2+24x-51)`

Combining like terms will yield:

`f(x)=x^3-11x^2+41x-51`