Monic cubic polynomial

Let $Q(x)$ be another monic cubic polynomial defined as $Q(x)=P(x)-(x^2+(x+1)^2)=P(x)-(2x(x+1)+1)$

Then, $Q(x)=0$ has the 3 roots 1, 4 and 9. Therefore, $Q(x)=(x-1)(x-4)(x-9)$ .

Hence, $P(x)=(x-1)(x-4)(x-9)+2x(x+1)+1=x^3-12x^2+51x-35$

Using Vieta's formula, we get the sum of roots of $P(x)=0$ as $\boxed{12}$ .

"a monic cubic polynomial" means a polynomial of the form x^3 + ax^2 + bx + c (the monic part is why the coefficient of the cubed term is just 1)

If we use this to expand P(1) = 5 we get 1 + a + b + c = 5

Doing the same with P(4) = 41 and P(9) = 181 we get

1 + a + b + c = 5

64 + 16a + 4b + c = 41

729 + 81a + 9b + c = 181

which is a system of linear equations which can be simplified down to

a=-12 b=51 c=-35

So our polynomial is

P(x) = x^3 - 12x^2 + 51x - 35

Now, we could try to factor that to get its roots, but that sounds hard. What we really want is the sum of its roots. Maybe there's an easy way to get that.

Well, it's a cubic, it's got three roots, let's name them i, j, and k. Then we can assume that the polynomial can be broken down into

(x - i)(x - j)(x - k) = P(x)

Now if we multiply out that left hand expression we get

x^3 - (i + j + k)x^2 + (ij +jk + ik)x - ijk = P(x) = x^3 - 12x^2 + 51x - 35

and setting like terms alike we get

i + j + k = 12

which is what we want.