Question

Now, suppose one of the roots of the polynomial function is complex. The roots of the function are 2 + i, and 5. Write the equation for this polynomial function.
Which of the following must also be a root of the function?

–3

–5

2 – i

2i

Answer 1

The other commentor is right. The correct answer is 2-i
The second follow-up question is the second option
f(x) = (x – (2 + i))(x – (2 – i))(x – 5)
The third follow-up question has these three answers. 
9x^2
25x
25

Answer 2

Answer-

The equation of polynomial- $x^3-9x^2+25x-25=0$

$(2-i)$ must also be a root of the function.

Solution-

Complex Conjugate Root Theorem-

For a real coefficient polynomial P(x), if $(a+bi)$ is a root then $(a-bi)$ will also be a root of P(x).

As $2+i$ is a root of the function, so $2-i$ will also a root of the function.

All the roots of the functions are,

$5,\ 2+i,\ 2-i$

So,

$\Rightarrow P(x)=(x-5)(x-(2+i))(x-(2-i))=0$

$\Rightarrow (x-5)(x-2-i)(x-2+i)=0$

$\Rightarrow (x-5)((x-2)+(i))((x-2)-(i))=0$

$\Rightarrow (x-5)((x-2)^2-(i)^2)=0$

$\Rightarrow (x-5)(x^2-4x+4+1)=0$

$\Rightarrow (x-5)(x^2-4x+5)=0$

$\Rightarrow xx^2+x\left(-4x\right)+x\cdot \:5+\left(-5\right)x^2+\left(-5\right)\left(-4x\right)+\left(-5\right)\cdot \:5=0$

$\Rightarrow x^2x-4xx+5x-5x^2+5\cdot \:4x-5\cdot \:5=0$

$\Rightarrow x^3-9x^2+25x-25=0$