Find a quintic polynomial meeting given conditions

Find a polynomial $P$ of degree $\leq 5$ satisfying the following conditions:

$\begin{align*} P(0) &= 1, \\ P(1) &= 2, \\ P'(0) &= P''(0) = P'(1) = P''(1) = 0. \end{align*}$

Since $P$ must be a polynomial of degree $\leq 5$ we may write

$P(x) = a_5 x^5 + a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x + a_0$

where any of the $a_i$ may be 0 (since we could have $P$ a polynomial of degree strictly less than 5). First, let’s apply the condition $P(0) = 1$ to obtain

$P(0) = a_0 = 1.$

Now, let’s take the first two derivatives since we have conditions on $P'$ and $P''$ .

$\begin{align*} P(x) &= a_5 x^5 + a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x + 1 \\ P'(x) &= 5a_5 x^4 + 4a_4 x^3 + 3a_3 x^2 + 2a_2 x + a_1 \\ P''(x) &= 20a_5 x^3 + 12a_4 x^2 + 6a_3 x + 2a_2. \end{align*}$

We can then apply the conditions $P'(0) = 0$ and $P''(0) = 0$ to obtain

$\begin{align*} P'(0) &= a_1 = 0 \\ P''(0) &= 2a_2 = 0. \end{align*}$

So now we have $a_0 = 1$ and $a_1 = a_2 = 0$ and so

$P(x) = a_5 x^5 + a_4 x^4 + a_3 x^3 + 1.$

Now we need to use the other three conditions

$\begin{align*} P(1) &= 2 & \implies && a_5 + a_4 + a_3 &= 1 \\ P'(1) &= 0 & \implies && 5a_5 + 4a_4 + 3a_3 &= 0 \\ P''(1) &= 0 & \implies && 20a_5 + 12a_4 + 6a_3 &= 0. \end{align*}$