Question

A monic quadratic polynomial $p\left( x \right)$ is such that \[p\left( x \right) = 0\] and $p\left( {p\left( {p\left( x \right)} \right)} \right) = 0$ have a common root, then find the value of $p\left( 0 \right) \cdot p\left( 1 \right)$ .

Answer 1

Hint: A monic quadratic means a quadratic expression with the coefficient of ${x^2}$ is $1$ . Use that to assume the definition for \[p\left( x \right)\] as $p\left( x \right) = {x^2} + bx + c$. Assume some variable $'m'$ for the value of the common root, i.e. $p\left( m \right) = {m^2} + bm + c = 0{\text{ and also }}p\left( {p\left( {p\left( m \right)} \right)} \right) = p\left( {p\left( 0 \right)} \right) = 0$ . Now take $p\left( {p\left( {p\left( x \right)} \right)} \right) = 0$ and start resolving each bracket from inside to out step by step using $p\left( x \right)$. Now, find values of $p\left( 0 \right)$ and $p\left( 1 \right)$. Multiply them to get the required answer.

Complete step-by-step answer:
Let’s first analyse the given information. For a monic quadratic polynomial $p\left( x \right)$ , it is given that $p\left( x \right) = 0$ and $p\left( {p\left( {p\left( x \right)} \right)} \right) = 0$ has a common root. With this information, we need to find the value for $p\left( 0 \right) \cdot p\left( 1 \right)$
$p\left( x \right)$ is a monic quadratic polynomial, i.e. is a single-variable polynomial in which the leading coefficient is equal to one. And for a quadratic monic polynomial, the leading coefficient will be the coefficient of the variable with the highest power, i.e. ${x^2}$ .
$ \Rightarrow p\left( x \right)$ can be represented by ${x^2} + bx + c$ , where $b \ne 0{\text{ and }}c \ne 0$ .
Therefore, we can write: $p\left( x \right) = {x^2} + bx + c$ (i)
Now, let’s assume that the common root of $p\left( x \right) = 0$ and $p\left( {p\left( {p\left( x \right)} \right)} \right) = 0$ be some value $'m'$ .
$ \Rightarrow p\left( m \right) = {m^2} + bm + c = 0{\text{ and also }} \Rightarrow p\left( {p\left( {p\left( m \right)} \right)} \right) = 0$
Since we know $p\left( m \right) = 0$ , the above can expressions can also be written as:
$ \Rightarrow p\left( m \right) = {m^2} + bm + c = 0{\text{ and also }} \Rightarrow p\left( {p\left( {p\left( m \right)} \right)} \right) = p\left( {p\left( 0 \right)} \right) = 0$ (ii)
So form (i), we can write $p\left( 0 \right)$ as:
$ \Rightarrow p\left( 0 \right) = {0^2} + b \times 0 + c = c \Rightarrow p\left( 0 \right) = c$ (iii)
Now, we use this value of $p\left( 0 \right)$ in equation (ii)
$ \Rightarrow p\left( {p\left( {p\left( m \right)} \right)} \right) = p\left( {p\left( 0 \right)} \right) = p\left( c \right) = 0$ (iv)
Again using (i) and (iv), we can define as:

Now from the above expression, we can define the value of $c$ as:
$ \Rightarrow {c^2} + bc + c = 0 \Rightarrow c\left( {c + b + 1} \right) = 0$
Since we know that $c$ cannot be equal to zero. Then we can say:
$ \Rightarrow c\left( {c + b + 1} \right) = 0 \Rightarrow c + b + 1 = 0 \Rightarrow c = - 1 - b$ (v)
Thus, from (iii) we know that $p\left( 0 \right) = c$ and also $c \ne 0$ . Therefore, $p\left( 0 \right) \ne 0$
Also, from (i) we have: $p\left( 1 \right) = {1^2} + b \times 1 + c = 1 + b + c$
And from (v), we can say $p\left( 1 \right) = 1 + b + c = 0 \Rightarrow p\left( 1 \right) = 0$
Therefore, the required value of $p\left( 0 \right) \times p\left( 1 \right)$ can be represented as:
$ \Rightarrow p\left( 0 \right) \times p\left( 1 \right) = c \times 0 = 0$
Hence, we got the required value $p\left( 0 \right) \times p\left( 1 \right) = 0$.

Note: Follow a step by step procedure while solving this problem. Write all the found results to avoid any confusion. Notice that while solving, we have started from inside braces of $p\left( {p\left( {p\left( m \right)} \right)} \right)$ and step by step made our way out by solving one by one. Assume a monic quadratic expression was a crucial part of the solution.