X(x^2+6x+9)=4(x+3)

To solve the equation \( X(x^2 + 6x + 9) = 4(x + 3) \), we can start by simplifying both sides. 1. Recognize that \( x^2 + 6x + 9 \) is a perfect square trinomial: \[ x^2 + 6x + 9 = (x + 3)^2 \] So we can rewrite the left side: \[ X(x + 3)^2 = 4(x + 3) \] 2. Next, we can divide both sides by \( (x + 3) \) as long as \( x + 3 \neq 0 \) (i.e., \( x \neq -3 \)): \[ X(x + 3) = 4 \] 3. We can then solve for \( X \): \[ X = \frac{4}{x + 3} \] This gives us a general expression for \( X \) in terms of \( x \) where \( x \neq -3 \). 4. If we want to consider the case when \( x + 3 = 0 \) (i.e., \( x = -3 \)), we should substitute \( x = -3 \) back into the original equation to see if it holds: \[ X(0) = 4(0) \implies 0 = 0 \] This is true, which means \( X \) can take any value when \( x = -3 \). In summary, the solution of the equation is: \[ X = \frac{4}{x + 3} \quad \text{for } x \neq -3 \] And for \( x = -3 \), \( X \) can be any value.