A polynomial $p(x)$ with real coefficients is of degree five. Find $p(x)$ in factorised form with real coefficients. A polynomial $p(x)$ with real coefficients is of degree five. The equation $p(x)=0$ has a complex root $2+i$. The graph $y=p(x)$ has the x-axis as a tangent at $(2,0)$ and intersects the cordinate axes at $(-1,0)$ and $(0,4)$. Find $p(x)$ in factorised form with real coefficients.
Firts I found the roots: 
$r_1=2+i$, $r_2=2-i$, $r_3=-1$, $r_4=2$  and $r_5=a$, I was not able to find it ( for some reason the anwser considered $r_5=2$, but I was not able to prove it)
$p(x)=B(x+1)(x-2)(x-2+i)(x-2-i)(x-a)$
$p(x)=B(x+1)(x-2)(x^2-4x+5)(x-a)$
$p(0)=4$
then:
$4=B(1)(-2)(+5)(-a)$
$B=\frac{2}{5a}$
Observation: if we consider the value of $a=2$ then the answer is right. However I cannot show that $a=2$

Let $f(x)=g(x)(x-2)(x-a)$, where $g(x)=B(x+1)(x-2+i)(x-2-i)$. The $x$ axis has slope $0$. Since $f$ is tangent to the $x$ axis at $x=2$, we have
$$f'(2)=0$$
But 
$$f'(x)=g'(x)(x-2)(x-a)+g(x)(x-a)+g(x)(x-2)$$
so that
$$f'(2)=g(2)(2-a)=0$$
Since $g(2)\ne 0$, we must have $a=2$
