Factor completely. \( 5 x^{2}-45 \) A. \( (5 x+9)(x-5) \) B. \( 5(x-9)(x+9) \) C. \( 5(x+3)(x-3) \) D. \( 5\left(x^{2}-45\right) \)

To factor \( 5x^{2} - 45 \), first notice that you can factor out the common term of 5. This leaves you with: \[ 5(x^{2} - 9) \] Next, \( x^{2} - 9 \) is a difference of squares, which factors further to: \[ (x - 3)(x + 3) \] Putting it all together, the complete factorization is: \[ 5(x - 3)(x + 3) \] So, the correct option is: C. \( 5(x+3)(x-3) \) Now, let’s dive a bit deeper! The expression you've just factored is a classic example of utilizing the difference of squares formula, \( a^2 - b^2 = (a-b)(a+b) \). Here, \( x^2 - 9 \) is \( x^2 - 3^2 \). Recognizing these patterns can make factoring easier and faster! Factoring not only helps in simplifying equations, but it also plays a critical role in solving quadratic equations and optimization problems in calculus. By understanding how to factor polynomials, you gain tools to analyze and solve real-world problems, such as maximizing profit in business models or understanding projectile motion in physics!