Nicolae's polynomial division

Since $P(x)$ is of degree 7, it can be written as

$P(x) = Q(x) (x - 2) (x - 3) (x - 5) + U(x)$

where Q(x) is of degree 4 (not relevant here) and U(x) is of degree 2 (important).

By substituting the values of $x = 2, 3$ and $5$ in the above equation we get three values of the degree 2 polynomial $U(x)$ :

$U(2) = 35 \\ U(3) = 25 \\ U(5) = 23 \\$

Hence, $U(x) = 3x^2 - 25x + 73$ . Further, $U(1) = 51$ .

---

Well, there are many ways to "fit" a polynomial on a set of points. I am not sure if they are shorter/easier than solving the equations. One such method is "Lagrange's Polynomial". Fitting (2,35), (3,25), (5,23) on a degree 2 polynomial gives:

$\begin{aligned} U(x) &= 35 \frac{(x-3)(x-5)}{(2-3)(2-5)} + 25 \frac{(x-2)(x-5)}{(3-2)(3-5)} + 23 \frac{(x-2)(x-3)}{(5-2)(5-3)} \\ &= 3x^2 - 25x + 73 \end{aligned}$

Let $P(x)=(x-2)(x-3)(x-5)Q(x)+U(x)$ for some polynomials $Q(x),U(x)$ . As $\text{deg }P(x)=7$ and $\text{deg }(x-2)(x-3)(x-5)=3$ , $\text{deg }Q(x)$ is thus $7-3=4$ . Thus $U(x)$ has a maximum degree of $\text{min}(3-1,4-1)=2$ , so let $U(x)$ be a quadratic $ax^2+bx+c$ for constants $a,b,c$ .

We can solve for $a,b,$ and $c$ by plugging in our values of $P(x)$ . We find the following system of equations:

$P(2)=(2-2)(2-3)(2-5)Q(x)+4a+2b+c=4a+2b+c=35$

$P(3)=(3-2)(3-3)(3-5)Q(x)+9a+3b+c=9a+3b+c=25$

$P(2)=(5-2)(5-3)(5-5)Q(x)+25a+5b+c=25a+5b+c=23$

Solving this yields one solution, $(a,b,c)=(3,-25,73)$ . So,

$U(x)=3x^2-25x+73$

and

$U(1)=3(1^2)-25(1)+73=\boxed{51}$

Let $P(x)$ be polynomial of degree 7, and Let $D(x) = (x-2)(x-3)(x-5)$ , we've $D(x)$ divide $P(x)$ with a remainder $U(x)$ , so $P(x) = F(x)D(x) + U(x)$ (with $F(x)$ some polynomial)

we've the degree of $D(x)$ is $3$ then the degree of remainder is less than $D(x)$ so: $U(x) = ax^2 + bx + c$ , by solving system:

$a*(2^2) + b*2 + c = 35$

$a*(3^2) + b*3 + c = 25$

$a*(5^2) + b*5 + c = 23$

we get $a = 3, b = -25, c = 73$ where $U(x) = 3x^2 -25x + 73$ and $U(1) = 51$