(3) \( 9 x^{2}-4 \) (6) \( 16 a^{2}-121 b^{2} \) (9) \( a^{4}-16 \)

Sure, let's factorize each of the given expressions step by step. ### (3) \( 9x^{2} - 4 \) **Step 1:** Recognize that this is a **difference of squares**. Recall that \( a^2 - b^2 = (a - b)(a + b) \). Here, \( 9x^{2} = (3x)^2 \) and \( 4 = 2^2 \). **Factorization:** \[ 9x^{2} - 4 = (3x)^2 - 2^2 = (3x - 2)(3x + 2) \] --- ### (6) \( 16a^{2} - 121b^{2} \) **Step 1:** Again, this is a **difference of squares**. Here, \( 16a^{2} = (4a)^2 \) and \( 121b^{2} = (11b)^2 \). **Factorization:** \[ 16a^{2} - 121b^{2} = (4a)^2 - (11b)^2 = (4a - 11b)(4a + 11b) \] --- ### (9) \( a^{4} - 16 \) **Step 1:** Notice that \( a^4 \) is a perfect square since \( a^4 = (a^2)^2 \), and \( 16 = 4^2 \). So, this is a **difference of squares**. \[ a^{4} - 16 = (a^2)^2 - 4^2 = (a^2 - 4)(a^2 + 4) \] **Step 2:** The term \( a^2 - 4 \) is also a **difference of squares** and can be further factored. \[ a^2 - 4 = (a)^2 - 2^2 = (a - 2)(a + 2) \] **Final Factorization:** \[ a^{4} - 16 = (a - 2)(a + 2)(a^2 + 4) \] *Note:* The term \( a^2 + 4 \) does not factor further over the set of real numbers. If you are working within the complex number system, \( a^2 + 4 \) can be factored as \( (a + 2i)(a - 2i) \), where \( i \) is the imaginary unit. --- **Summary of Factorizations:** 1. \( 9x^{2} - 4 = (3x - 2)(3x + 2) \) 2. \( 16a^{2} - 121b^{2} = (4a - 11b)(4a + 11b) \) 3. \( a^{4} - 16 = (a - 2)(a + 2)(a^2 + 4) \)