Given f(x) = -3x³ + 30x and a = -3.

The linearization of $f(x)$ at $x=a$ is given by $L(x) = f(a) + f'(a)(x-a)$.
Given $f(x) = -3x^3 + 30x$ and $a = -3$.

First, calculate $f(a)$:

$$f(-3) = -3(-3)^3 + 30(-3) = -3(-27) - 90 = 81 - 90 = -9$$

Next, find the first derivative $f'(x)$:

$$f'(x) = \frac{d}{dx}(-3x^3 + 30x) = -9x^2 + 30$$

Substitute $a=-3$ into $f'(x)$:

$$f'(-3) = -9(-3)^2 + 30 = -9(9) + 30 = -81 + 30 = -51$$

Substitute $f(-3)$ and $f'(-3)$ into the linearization formula:

$$L(x) = f(-3) + f'(-3)(x - (-3))$$ $$L(x) = -9 + (-51)(x + 3)$$ $$L(x) = -9 - 51x - 153$$ $$L(x) = -51x - 162$$

The linearization of $f(x)$ at $x=-3$ is:

$$\boxed{L(x) = -51x - 162}$$

$$f(2) = 2(2)^3 - (2)^2 + 5(2) = 2(8) - 4 + 10 = 16 - 4 + 10 = 22$$

This matches the constant term in $R(x)$.

First, find the first derivative $f'(x)$:

$$f'(x) = \frac{d}{dx}(2x^3 - x^2 + 5x) = 6x^2 - 2x + 5$$

Now, evaluate $f'(2)$:

$$f'(2) = 6(2)^2 - 2(2) + 5 = 6(4) - 4 + 5 = 24 - 4 + 5 = 25$$

This matches the coefficient of $(x-2)$ in $R(x)$.

First, find the second derivative $f''(x)$:

$$f''(x) = \frac{d}{dx}(6x^2 - 2x + 5) = 12x - 2$$

Now, evaluate $f''(2)$:

$$f''(2) = 12(2) - 2 = 24 - 2 = 22$$

The Taylor coefficient for $(x-a)^2$ is $\frac{f''(a)}{2!}$:

$$\frac{f''(2)}{2!} = \frac{22}{2} = 11$$

This matches the coefficient of $(x-2)^2$ in $R(x)$.

First, find the third derivative $f'''(x)$:

$$f'''(x) = \frac{d}{dx}(12x - 2) = 12$$

Now, evaluate $f'''(2)$:

$$f'''(2) = 12$$

The Taylor coefficient for $(x-a)^3$ is $\frac{f'''(a)}{3!}$:

$$\frac{f'''(2)}{3!} = \frac{12}{6} = 2$$

This matches the coefficient of $(x-2)^3$ in $R(x)$.

Since all calculated Taylor coefficients $f(2)$, $f'(2)$, $\frac{f''(2)}{2!}$, and $\frac{f'''(2)}{3!}$ match the coefficients in $R(x)$, $R(x)$ is indeed the Taylor polynomial for $f(x)$ centered at $a=2$.