The quadratic polynomial $P(x)$ has a zero at $x=2$. The polynomial $P(P(x))$ has only one real zero at $x=5.$ Compute $P(0).$
If we have that $P(x) = ax^2 +bx+ c$, we get from the first condition that $P(x) = (x-2)(bx+c).$
From here $P(P(x)) = (ax+bx+c -2)(b(ax+bx+c)+c)$, but this just looks very messy and doesn't seem to be helpful at all. Is there some other trick here I'm missing?
On
Write $$P(x)=a(x-2)(x-b)\implies P(P(x))=a(P(x)-2)(P(x)-b)=a(a(x-2)(x-b)-2)(a(x-2)(x-b)-b)=a^3(x-5)^2(x-u)(x-\bar{u})$$ since $\deg(P(P(x))=4$, which means it must have at least a double root of 5, while the other two roots are complex conjugates (not necessarily $\ne 5$). So we have $(3a(5-b)-2)(3a(5-b)-b)=0$.
Since 5 is a double root of $P(P(x))$, it is also a root of its derivative function. Hence $(2x-2-b)(a(x-2)(x-b)-b)+(a(x-2)(x-b)-2)(2x-2-b)=0$ when $x=5$, i.e. $(8-b)(3a(5-b)-b)+(3a(5-b)-2)(8-b)=0$. Clearly, $b=8$ solves this second equation. If we plug this value into the first equation, we get
$$(-9a-2)(-9a-8)=0\implies a=-\dfrac 29 \;\text{or}\;-\dfrac 89$$
However by plugging into the original equation we see that both cases doesn't produce the desired $P(x)$. Note that there's no need to solve the quartic equation, it suffices to check that 5 doesn't solve the equation we get here.
Thus, $6a(5-b)-b-2=0\implies 3a(5-b)=\dfrac{b+2}2$. This gives $(\dfrac{b+2}2-2)(\dfrac{b+2}2-b)=0\implies b=2\;\text{or}\; -1$, corresponding to $a=\dfrac 29$ and $a=\dfrac 1{36}$ respectively.
By checking in the same way as before, we see the only fitting solution is $a=\dfrac 29, b=2$, and therefore $$P(x)=\dfrac 29(x-2)^2$$
which means that $$P(0)=\dfrac 89$$.
On
We can consider $ \ P(x) \ $ and $ \ P \circ P(x) \ $ in terms of how a quadratic polynomial $ \ a·(x - 2)·(x - r) \ $ "map" the real numbers: $$ \large{ \begin{array}{ccc} \mathbf{x} & \mathbf{P(x)} & \mathbf{ P \circ P(x)} \\ 0 & 2ar & P(2ar) \\ 2 \ , \ r & 0 & 2ar \\ \frac{2 \ + \ r}{2} \ = \ 1 + \frac{r}{2} & -a·(\frac{r}{2} - 1)^2 & P \ ( \ -a·[\frac{r}{2} - 1]^2 \ ) \\ 5 & 3a·(5 - r) & P \ ( \ 15a - 3ar \ ) \end{array} } \ \ . $$
The problem statement leaves open whether the second zero of $ \ P(x) \ $ is also $ \ 2 \ $ (making it a "double zero") or $ \ r \ \neq \ 2 \ \ . \ $ In the table for the map, the second line shows the zeroes of $ \ P(x) \ $ being mapped to zero; $ \ P \circ P(x) \ $ in turn "takes" this to the intercept $ \ y \ = \ 2ar \ \ . \ $ The third line tells us that the vertex of the parabola representing $ \ P(x) \ $ at $ \ x_v \ = \ 1 + \frac{r}{2} \ $ is "sent to" $$ \ P(x_v) \ \ = \ \ a· \left(1 \ + \ \frac{r}{2} \ - \ 2 \right)·\left(1 \ + \ \frac{r}{2} \ - \ r \right) \ \ = \ \ a· \left( \frac{r}{2} \ - \ 1 \right)·\left(1 \ - \ \frac{r}{2} \right) $$ $$ = \ \ -a· \left( \frac{r}{2} \ - \ 1 \right)^2 \ \ . $$ Finally, $ \ x \ = \ 5 $ is mapped to $ \ a·(5 - 2)·(5 - r) \ = \ 15a - 3ar \ \ ; \ $ it is $ \ P(15a - 3ar) \ $ that we want to be equal to zero.
If $ \ P(x) \ $ has a double zero with $ \ r \ = \ 2 \ \ , \ $ then naturally $ \ x_v \ = \ \frac{2 \ + \ 2}{2} \ = \ 2 \ \ $ and we confirm that $ \ P(x_v) \ = \ -a· \left( \frac{2}{2} \ - \ 1 \right)^2 \ = \ 0 \ \ . \ $ In order for $ \ P \circ P(5) \ $ to equal zero then, we must have $ \ P(5) \ = \ 2 \ \Rightarrow \ 15a - 3a·2 \ = \ 9a \ = \ 2 \ \ . $ We thereby would conclude that $ \ P(x) \ = \ \frac29·(x - 2)^2 \ \ . \ $ However, we find that for this polynomial, $$ \frac29·(x \ - \ 2)^2 \ \ = \ \ 2 \ \ \Rightarrow \ \ (x \ - \ 2)^2 \ \ = \ \ 9 \ \ \Rightarrow \ \ | \ x \ - \ 2 \ | \ \ = \ \ 3 \ \ \Rightarrow \ \ x \ \ = \ \ 5 \ \ , \ \ -1 \ \ . $$ So $ \ x \ = \ -1 \ $ is also mapped to $ \ 2 \ \ , \ $ which yields the composition $ \ P \circ P(-1) \ = \ 0 \ \ , \ $ in contradiction to the stated requirement that $ \ P \circ P(x) \ $ has the sole real zero $ \ x \ = \ 5 \ \ . $
[Indeed, if we carry out the computation of $ \ \frac29·\left( \ \left[ \ \frac29·(x - 2)^2 \ \right] - 2 \ \right)^2 \ \ , \ $ we obtain $$ P \circ P(x) \ = \ \frac{8}{729}·(x^4 \ - \ 8 x^3 \ + \ 6 x^2 \ + \ 40 x + \ 25) \ \ = \ \ \frac{8}{729}·(x \ - \ 5)^2 · (x \ + \ 1)^2 \ \ . $$ We should expect to see something of this sort: the "tails" of the curve for the quartic $ \ P \circ P(x) \ $ "run to positive infinity" and $ \ [ \ P \circ P(x) \ ]' \ = \ P'( \ P(x) \ )·P'(x) \ = \ \frac49·( \ P(x) - 2 \ )·\frac49·(x - 2) \ = \ 0 \ $ at $ \ x \ = \ 2 \ $ and the two values of $ \ x \ $ for which $ \ P(x) \ = \ 2 \ \ , \ \ x \ = \ 5 \ $ and $ \ x \ = \ -1 \ \ , \ $ giving $ \ P \circ P(x) \ $ three "turning points".]
So we have shown that $ \ P(x) \ $ must have two distinct zeroes. The difficulty we had with having the double zero $ \ x \ = \ 2 \ $ as the vertex is that the quadratic polynomial permits two values of $ \ x \ $ for which $ \ P(x) \ = \ 2 \ \ . \ $ We can prevent this by requiring that $ \ x \ = \ \mathbf{5} \ $ be the vertex for $ \ P(x) \ $ so that it is the only root of $ \ P(x) \ = \ 2 \ \ . \ $ The second zero of $ \ P(x) \ $ is then found from $ \ x_v \ = \ 1 + \frac{r}{2} \ = \ 5 \ \Rightarrow \ r \ = \ 8 \ \ . \ $ From this, we find $ \ P(5) \ = \ 3a·(5 - 8) \ = \ -9a \ \ ; \ $ our vertex requirement thus yields $ \ -9a \ = \ 2 \ \Rightarrow \ a \ = \ -\frac29 \ \ . \ $
This gives us $ \ \boxed{ \ P(x) \ = \ -\frac29·(x - 2)·(x - 8) \ = \ -\frac29·(x^2 - 10x + 16) \ } \ $ as the quadratic polynomial sought. The value requested by the problem is thus $ \ \mathbf{P(0)} \ = \ 2·\left(-\frac29 \right)·8 \ \mathbf{= \ -\frac{32}{9} } \ \ . $
We now have $$ P'(x) \ \ = \ \ -\frac29·[ \ (x - 8) \ + \ (x - 2) \ ] \ \ = \ \ -\frac49·(x \ - \ 5) $$ $$ \Rightarrow [ \ P \circ P(x) \ ]' \ \ = \ \ \left(-\frac49 \right)·( \ P(x) - 5 \ )·\left(-\frac49 \right)·(x - 5) \ \ . $$ Since $ \ P(x) \ = \ 2 - \frac29·(x - 5)^2 \ \ , \ $ there is no value of $ \ x \ $ for which $ \ P(x) \ = \ 5 \ \ , \ $ so $ \ [ \ P \circ P(x) \ ]' \ \ = \ 0 \ $ only at $ \ x \ = \ 5 \ \ , \ $ thus the composite function has only one turning point. With perhaps some computational aid, we can obtain the expression $$ P \circ P(x) \ \ = \ \ \left(-\frac{8}{729} \right) · [ \ x^4 \ - \ 20 x^3 \ + \ 177 x^2 \ - \ 770 \ + \ 1300 \ ] $$ $$ = \ \ \left(-\frac{8}{729} \right) · (x \ - \ 5)^2·(x^2 \ - \ 10x \ + \ 52) \ \ , \ $$ the quadratic factor being irreducible over the real numbers, with complex zeroes $ \ 5 \ \pm \ 3\sqrt3·i \ \ . \ $
$P(x) $ will be of the form
$$P(x)=(x-2)(ax+b).$$ Let
$$c=P(5)=3(5a+b).$$ then the equation
$$P(P(5))=(c-2)(ac+b)$$
$$=ac^2+(b-2a)c-2b=0$$
has only one root if the discriminant is zero.
$$\Delta=(b+2a)^2=0\iff b=-2a$$
and $$c=\frac{2a-b}{2a}=2$$ but $$c=3(5a+b)=9a=2$$ finally $$a=\frac 29\; \;,\;\; b=-\frac 49\;\;$$ $$\boxed{\;P(x)=\frac 29(x-2)^2}$$ $$P(0)=\frac 89$$