Question

I really need help with this. Can anyone help me please?

Answer 1

Answer:

$f(x)=(x-3)(x^2+4x+8)$

Step-by-step explanation:

A polynomial function is given by the form:

$f(x)=a(x-p)(x-q)...$

Where a is the leading coefficient, and p and q are the roots (more can be added if necessary).

Our zeros are 3 and (-2 + 2i).

And our leading coefficient a = 1.

Furthermore, by the Complex Root Theorem, if (-2 + 2i) is a zero, then (-2 - 2i) must also be a zero.

So, by substitution, we acquire:

$f(x)=1(x-(3))(x-(-2+2i))(x-(-2-2i))$

Simplify:

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

Expand the second and third factors:

$\begin{aligned} &=(x+2-2i)x+(x+2-2i)(2)+(x+2-2i)(2i)\\&= (x^2+2x-2ix)+(2x+4-4i)+(2ix+4i-4i^2)\\&=(x^2)+(2x+2x)+(-2ix+2ix)+(-4i+4i)+(4-4i^2)\\&=x^2+4x+(4+4)\\&=x^2+4x+8 \end{aligned}$

Therefore, our polynomial function of least degree and the given zeros will be:

$f(x)=(x-3)(x^2+4x+8)$