\( ( x - 4 ) ^ { 2 } ( 2 x + y ) = 0 \)

To solve the equation \( ( x - 4 ) ^ { 2 } ( 2 x + y ) = 0 \), we can use the principle that for a product to be zero, at least one of the factors must be zero. This gives us two cases to consider: 1. \( ( x - 4 ) ^ { 2 } = 0 \) - From this, we find that \( x - 4 = 0 \) which leads us to \( x = 4 \). Since it was squared, this is the only solution for \( x \). 2. \( 2 x + y = 0 \) - Rearranging this gives us \( y = -2x \). This shows a linear relationship between \( x \) and \( y \), with \( y \) being dependent on \( x \). Combining these results, we have \( x = 4 \) and \( y = -8 \) (when substituting \( x \) into the linear equation). Thus, the solution is \( (4, -8) \). Now that we've solved the equation, let’s add some more context and fun facts! The famous quadratic equation \( (x - 4)^2 \) is a perfect square trinomial! It reflects the vertex form of a parabola that opens upwards and has its vertex at the point \( (4, 0) \). This means that at \( x = 4 \), you can expect the least value of the function (which is 0, as the other factor could be anything) – the steepness of the parabola is influenced by how it expands after that point! In real-world situations, you might come across equations like this in physics, especially in projectile motion where two paths might intersect. The expression could represent the trajectory of an object hitting a surface (like water) or two objects colliding. Understanding the relationships helps in predicting outcomes, like the maximum height or distance!