Step 1: Expand Quantity A
First, expand both squared terms in Quantity A using the identity $\((a \pm b)^2=a^2 \pm 2 a b+b^2\)$ :

$$
\(\begin{aligned}
& (x+40)^2=x^2+80 x+1600 \\
& (x-40)^2=x^2-80 x+1600
\end{aligned}\)
$$


Now, add them together:

$$
\((x+40)^2+(x-40)^2=\left(x^2+80 x+1600\right)+\left(x^2-80 x+1600\right)=2 x^2+3200\)
$$


So, Quantity A simplifies to:

$$
\(2 x^2+3200\)
$$


Step 2: Rewrite the Comparison
Now, the comparison becomes:

$$
\(2 x^2+3200 \quad \text { vs. } \quad-160 x\)
$$


Subtract $-160 x$ (or add $160 x$ ) to both sides to set up the inequality:

$$
\(2 x^2+160 x+3200>0\)
$$


Step 3: Simplify the Inequality
Divide the entire inequality by 2 to simplify:

$$
\(x^2+80 x+1600>0\)
$$


Step 4: Analyze the Quadratic Expression
The quadratic expression $\(x^2+80 x+1600\)$ can be factored or analyzed using the discriminant.
1. Factoring:

$$
\(x^2+80 x+1600=(x+40)^2\)
$$


This is a perfect square.
2. Discriminant:

For $\(a x^2+b x+c\)$, the discriminant $\(D=b^2-4 a c\)$.

$$
\(D=80^2-4(1)(1600)=6400-6400=0\)
$$


A discriminant of 0 means there's exactly one real root, and the quadratic is always nonnegative.

Since $\((x+40)^2\)$ is a square, it is always non-negative (i.e., $\(\geq 0\)$ ) for all real $x$, and:

- It equals 0 only when $x=-40$.
- It is strictly positive for all other real $x$.

Step 5: Compare Quantities A and B
- For $\(x=-40\)$ :
- Quantity A: $\(2(-40)^2+3200=3200+3200=6400\)$
- Quantity B: $\(-160(-40)=6400\)$
- Here, Quantity A = Quantity B.
- For all other $x$ :
- Quantity A: $\((x+40)^2>0\)$, so $\(2 x^2+3200>-160 x\)$.

Thus, Quantity A > Quantity B.