Vieta's Formula - Hard Problem 
Let $f(x)=(x^2+6x+9)^{50}-4x+3$, and let $r_1,r_2,\ldots,r_{100}$ be the roots of >$f(x)$.
Compute $(r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100}$.

How would I compute this? How to factor it?
 A: Note that for any $r$ that is the root of $(x^2+6x+9)^{50}-4x+3$, note that it satisfies $$(r^2+6r+9)^{50}=4r-3 \iff (r+3)^{100}=4r-3$$
So $$\sum_{i=1}^{100} (r_{i}+3)^{100}=\sum_{i=1}^{100} (4r_{i}-3)=4 \times  \text{(Sum of roots)}-300$$Expanding $(x+3)^{100}-4x+3$ using the binomial theorem, we get $$f(x)=x^{100}+\color{red}{300}x^{99}+\dots$$
So the sum of roots is equal to $-300$. The answer is $-1500$. 
A: Observe that $(x^2+6x+9)=(x+3)^2$, so $f(x)=(x+3)^{100}-4x+3$.  Since each root satisfies $f(r_i)=0$, we know that $(r_i+3)^{100}=4r_i-3$.  Therefore, you're interested in 
$$
4\sum r_i-300.
$$
The sum of the roots of a polynomial of degree $n$ is the negative of the coefficient of $x^{n-1}$.  In this case, the $x^{99}$ term has coefficient $3\binom{100}{99}$ by the binomial theorem.  Therefore, the sum is
$$
-1200-300=-1500.
$$
A: We begin by observing that $x^2+6x+9 = (x+3)^2$ 
Denote $f(x) = (x^2+6x+9)^{50} - 4x +3 = (x+3)^{100}-4x+3$
$\{r_i\}_{1\leq i \leq 100}$ are roots of this polynomial so $f(r_i)= (r_i+3)^{100}-4r_i+3 = 0$ $\forall 1 \leq i \leq 100$
By Vieta's formula, the sum of roots is negative of coefficient of $x^{99}$ term in $f(x)$ which by Binomial Expansion turns out to be $-\binom{100}{1}3^1 = -300$
Now, we just need to make some manipulations, as follows:
$$(r_i+3)^{100} - 4r_i + 3 = 0 \implies (r_i+3)^{100} = 4r_i - 3$$
Therefore, 
$$\sum_{i=1}^{100} (r_i+3)^{100} = \sum_{i=1}^{100} 4r_i - 3 = 4 \left(\sum_{i=1}^{100} r_i\right) -300 = -1500$$
A: Hint:
The equation is none other than
$$\bigl((x+3)^2\bigr)^{50}-4x+3=(x+3)^{100}-4x+3=0.$$
Hence the roots $r_i$ satisfy
$$(r_i+3)^{100}=4r_i-3.$$
