Find $a\in \mathbb{R}$ such that a root of $ax^3-13x^2+(15a)x-25$ is $2+i$ Find $a\in \mathbb{R}$ such that a root of the polynomial $$p(x)=ax^3-13x^2+(15a)x-25$$ is $2+i$
Solution:
$q(x)=\frac{p(x)}{a}$
$=x^3-\frac{13}{a}x^2+15x-\frac{25}{a}$
$p(x)$ and $q(x)$ have the same roots, call them $r_1, r_2, r_3$
$r_1=2+i$. Since the coefficients are all real we know $r_2=2-i$
$(x-r_1)(x-r_2)(x-r_3)= x^3-(r_1+r_2+r_3)x^2+(r_1r_2 + r_1r_3 +r_2r_3)x-r_1r_2r_3$
$r_1+r_2+r_3=\frac{13}{a}$
$(2+i)(2-i)+r_3=\frac{13}{a}$
$5+r_3=\frac{13}{a}$
We also have:
$r_1r_2r_3=\frac{25}{a}$
$(2+i)(2-i)r_3=\frac{25}{a}$
$5r_3=\frac{25}{a}$
$r_3=\frac{5}{a}$
Going back to $5+r_3=\frac{13}{a}$
We have $5+\frac{5}{a}=\frac{13}{a}$
Thus $a=\frac{8}{5}$
The statement I am having trouble with is
"$r_1=2+i$. Since the coefficients are all real we know $r_2=2-i$"
Why do we know what $r_2$ is? I understand it's the conjugate and that $r_1r_2=5$ is a real number. But I don't understand why the coefficients being real implies that $r_2=2-i$

The coefficients of the polynomial is real, hence complex roots comes in conjugate pairs.
If $$\sum_{i=0}^n a_i z^i = 0$$
We can take conjugate and see that
$$\sum_{i=0}^n a_i \bar{z}^i = 0$$
