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 $a=2$

As the excellent hint in the comment stated, an $x$-axis tangent means a repeated root. That means that $(x-2)$ is a repeated factor of $p(x)$, i.e that $(x-1)^m, m \geq 2$ is a factor of $p(x)$.
The other obvious factors are $(x+1)$ and $(x-(2+i))(x-(2-i)) = (x^2 - 4x + 5)$, which, when multiplied together, gives you a cubic.
That means that the only remaining factor to achieve a fifth degree polynomial is of degree two, and therefore it must be $(x-2)^2$.
So $p(x) = k(x-2)^2(x+1)(x^2-4x+5)$
Apply the last piece of information, that $p(0) = 4$, which gives $k(4)(1)(5) = 4$ or $k = \frac 15$, so $p(x) = \frac 15(x-2)^2(x+1)(x^2-4x+5)$
