Factor completely and solve \[ 16 x^{2}-36 \]

To factor the expression \( 16x^{2} - 36 \), we can notice that it’s a difference of squares. We rewrite it as: \[ (4x)^{2} - 6^{2} \] Now, apply the difference of squares formula, which states that \( a^{2} - b^{2} = (a - b)(a + b) \): \[ (4x - 6)(4x + 6) \] Next, we can simplify those factors by factoring out the greatest common factor (GCF) from both terms: \[ 2(2x - 3) \quad \text{and} \quad 2(2x + 3) \] Thus, the fully factored form is: \[ 2(2x - 3)2(2x + 3) = 4(2x - 3)(2x + 3) \] Now, let’s solve for \( x \) by setting each factor equal to zero: 1. From \( 2x - 3 = 0 \) we get: \[ 2x = 3 \quad \Rightarrow \quad x = \frac{3}{2} \] 2. From \( 2x + 3 = 0 \) we get: \[ 2x = -3 \quad \Rightarrow \quad x = -\frac{3}{2} \] So the completely factored form is \( 4(2x - 3)(2x + 3) \) and the solutions are \( x = \frac{3}{2} \) and \( x = -\frac{3}{2} \).