Polynomial factoring

Using rational root theorem, since the first and last coefficients are $a_4=1$ and $a_0 =16 =2^4$ respectively, the factors can only involve multiples of $2$ and not $3$ . Since all the options involve $3$ except $(x+2)^4$ , it is the only possible solution.

We can confirm that $(x+2)^4$ is the solution by Vieta's formulas. The sum of roots $S_1=-2-2-2-2 =-8=-a_3$ , $S_2=6(-2)^2 =24=a_2$ , $S_3=4(-2)^3 =-32=-a_1$ and $S_4=(-2)^4 =32=a_0$ .

Therefore, the answer is $\boxed{(x+2)^4}$ .

$\begin{aligned} x^4 + 8x^3 + 24x^2 + 32x + 16 & = x^4 + (2x^3 + 6x^3) + (12x^2 + 12x^2) + (24x + 8x) + 16\\ & = x^3(x + 2) + 6x^2(x + 2) + 12x(x + 2) + 8(x + 2)\\ & = (x + 2)(x^3 + 6x^2 + 12x + 8)\\ & = (x + 2)(x^3 + (2x^2 + 4x^2) + (8x + 4x) + 8)\\ & = (x + 2)(x^2(x + 2) + 4x(x + 2) + 4(x+2))\\ & = (x + 2)(x + 2)(x^2 + 4x + 4)\\ & = (x + 2)(x + 2)(x^2 + (2x + 2x) + 4)\\ & = (x + 2)(x + 2)(x(x + 2) + 2(x + 2))\\ & = (x + 2)(x + 2)(x + 2)(x + 2)\\ & = \boxed{{(x + 2)}^4} \end{aligned}$

$\begin{matrix} x^4+8x^3+24x^2+32x+16\\\\=\left(x+2\right)\frac{x^4+8x^3+24x^2+32x+16}{x+2}\\\\ =\left(x+2\right)(x^3+\frac{6x^3+24x^2+32x+16}{x+2})\\\\=\left(x+2\right)(x^3+6x^2+\frac{12x^2+32x+16}{x+2})\\\\=\left(x+2\right)(x^3+6x^2+12x+\frac{8x+16}{x+2})\\\\=\left(x+2\right)(x^3+6x^2+12x+8)\\\mathrm{a^3+3a^2b+3ab^2+b^3=\left(a+b\right)^3}\\ \therefore (x+2)\left(x+2\right)^3\\=\left(x+2\right)^{1+3}\\=\mathbf{\left(x+2\right)^4} \end{matrix}$