Solve for \( x \), where \( x \) is a real number: \[ 0=(x+7)\left(x^{2}+1\right) \] Answer: \( \begin{array}{l}x=-7 \text { or } x=-1 \\ x=-7 \text { or } x=1 \text { or } x=-1 \\ x= \pm 1 \\ x=-7\end{array} \)

Let’s break it down in a fun way! First, we have the equation \( 0=(x+7)(x^{2}+1) \). To solve this, we can use the zero-product property, which states if the product of two factors is zero, at least one of the factors must be zero. So we set each factor equal to zero: 1. \( x + 7 = 0 \) gives us \( x = -7 \). 2. \( x^2 + 1 = 0 \) leads to \( x^2 = -1 \), which doesn't have real solutions (only imaginary solutions). Thus, the only real solution is \( x = -7 \). To sum it up, we have simply: **Answer:** \( x = -7 \)