Find polynomials that satisfy some conditions

Let $p$ be a polynomial of degree less than or equal to 2. Find all $p$ that satisfy the given conditions.

Since $p$ is a polynomial of degree less than or equal to 2 we have

$p(x) = ax^2 + bx + c$

for constants $a,b,c \in \mathbb{R}$ for all of the below.

Let $f(x) = p(x) - 1$ , then $f$ is of degree at most 2, and we have three points at which $f(x) = 0$ (since $p(x) = 1$ implies $f(x) = 0$ ). By part (d) of the previous exercise we have all of the coefficients of $f$ are 0 and $f(x) = 0$ for all $x \in \mathbb{R}$ . Thus,

$p(x) -1 = 0 \qquad \implies \qquad p(x) = 1 \qquad \text{for all } x \in \mathbb{R}.$
We have

$p(0) = 1 \quad \implies \quad c = 1$

Then, with $c = 1$ ,

$p(1) = 1 \quad \implies \quad a + b = 0 \quad \implies \quad b = -a$

Finally, with $c=1$ and $b = -a$ , we have,

$p(2) = 2 \quad \implies \quad 4a - 2a = 1 \quad \implies \quad a = \frac{1}{2}, b = -\frac{1}{2}.$

Hence,

$p(x) = \frac{1}{2} x^2 - \frac{1}{2} x + 1 = \frac{1}{2} x (x-1) + 1.$
Again, from $p(0) = 1$ we have,

$0a + 0b + c = 1 \quad \implies \quad c = 1.$

Then, with $c = 1$ and $p(1) = 1$ we have,

$a + b = 0 \quad \implies \quad b = -a.$

Thus,

$p(x) = ax^2 - ax + 1 = ax (x-1) + 1.$
Just substituting these values we have,

$p(0) = p(1) \quad \implies \quad c = a + b + c \quad \implies \quad b = -a.$

So,

$p(x) = ax^2 - ax + c = ax(x-1) + c.$

Where $a,c$ are arbitrary real numbers.