An algebra problem by Wen Z

Firstly, we find the derivative of the expression and find the values of $x$ for which the slope is 0. The derivative we get is $4x^3+6x^2+10x+8$ . We now have

$\begin{aligned} 4x^3+6x^2+10x+8&=0\\ 2(x+1)(2x^2+x+4)&=0 \end{aligned}$

The second bracket has complex roots (can be checked through finding the discriminant). Since we're looking only at real roots, we can ignore this bracket. Thus, this quartic must have only 1 minima (doesn't have a maxima since the leading coefficient is positive). This value is attained at $x=-1$ , so the value of the expression is 13.

Therefore, the minimum value is 13.

Observe that the first two terms almost form a perfect square.

Repartition and factor the expression as:

x²(x+1)² + 4x² + 8x + 17

Now observe that these remaining terms also nearly form a perfect square.

Repartition and refactor again to obtain:

(x²+4)(x+1)² + 13

Thus the minimum is 13, which holds when x=-1