\( \begin{array}{ll}\text { 0. } 4 x^{2}-(4 x-3 y)^{2} & \text { 11. } 16 a^{2}-(3 b+4 c)^{2} \\ \text { 3. } 4(a-3)^{2}-49(4 a-5) & \text { 14. } 16 k^{2}-4 \\ \text { 6. } \frac{1}{9} a^{2}-4 b^{2} & \text { 17. } \frac{1}{2} x^{2}-2\end{array} \)

Certainly! Let's factor each of the given expressions. We'll use techniques such as the **difference of squares** and **factoring out common factors** where applicable. --- ### **0. \( 4x^{2} - (4x - 3y)^{2} \)** **Method:** Difference of Squares \[ \begin{aligned} 4x^{2} - (4x - 3y)^{2} &= (2x)^2 - (4x - 3y)^2 \\ &= [2x - (4x - 3y)][2x + (4x - 3y)] \quad \text{(Using } a^2 - b^2 = (a - b)(a + b)\text{)} \\ &= (-2x + 3y)(6x - 3y) \\ &= -1 \times (2x - 3y)(6x - 3y) \\ \end{aligned} \] **Factored Form:** \[ - (2x - 3y)(6x - 3y) \] --- ### **11. \( 16a^{2} - (3b + 4c)^{2} \)** **Method:** Difference of Squares \[ \begin{aligned} 16a^{2} - (3b + 4c)^{2} &= (4a)^2 - (3b + 4c)^2 \\ &= [4a - (3b + 4c)][4a + (3b + 4c)] \\ &= (4a - 3b - 4c)(4a + 3b + 4c) \end{aligned} \] **Factored Form:** \[ (4a - 3b - 4c)(4a + 3b + 4c) \] --- ### **3. \( 4(a - 3)^{2} - 49(4a - 5) \)** **Method:** Expand and Simplify First, expand each term: \[ \begin{aligned} 4(a - 3)^2 &= 4(a^2 - 6a + 9) = 4a^2 - 24a + 36 \\ 49(4a - 5) &= 196a - 245 \\ \end{aligned} \] Subtract the second expression from the first: \[ \begin{aligned} 4a^2 - 24a + 36 - (196a - 245) &= 4a^2 - 24a + 36 - 196a + 245 \\ &= 4a^2 - 220a + 281 \end{aligned} \] **Factored Form:** The quadratic \(4a^2 - 220a + 281\) does not factor neatly over the integers. However, it can be expressed using the quadratic formula or left in its simplified form: \[ 4a^2 - 220a + 281 \] --- ### **14. \( 16k^{2} - 4 \)** **Method:** Difference of Squares and Factoring Out Common Factors \[ \begin{aligned} 16k^2 - 4 &= 4(4k^2 - 1) \quad \text{(Factor out 4)} \\ &= 4[(2k)^2 - 1^2] \\ &= 4(2k - 1)(2k + 1) \quad \text{(Using } a^2 - b^2 = (a - b)(a + b)\text{)} \end{aligned} \] **Factored Form:** \[ 4(2k - 1)(2k + 1) \] --- ### **6. \( \frac{1}{9}a^{2} - 4b^{2} \)** **Method:** Difference of Squares \[ \begin{aligned} \frac{1}{9}a^2 - 4b^2 &= \left(\frac{a}{3}\right)^2 - (2b)^2 \\ &= \left(\frac{a}{3} - 2b\right)\left(\frac{a}{3} + 2b\right) \quad \text{(Using } a^2 - b^2 = (a - b)(a + b)\text{)} \end{aligned} \] **Factored Form:** \[ \left(\frac{a}{3} - 2b\right)\left(\frac{a}{3} + 2b\right) \] Alternatively, to eliminate the fraction: \[ \frac{1}{9}(a - 6b)(a + 6b) \] --- ### **17. \( \frac{1}{2}x^{2} - 2 \)** **Method:** Difference of Squares and Factoring Out Constants First, factor out the constant \(\frac{1}{2}\): \[ \begin{aligned} \frac{1}{2}x^2 - 2 &= \frac{1}{2}(x^2 - 4) \\ &= \frac{1}{2}(x - 2)(x + 2) \quad \text{(Using } a^2 - b^2 = (a - b)(a + b)\text{)} \end{aligned} \] **Factored Form:** \[ \frac{1}{2}(x - 2)(x + 2) \] Alternatively, expressing as a difference of squares directly: \[ \frac{1}{2}x^2 - 2 = \left(\frac{x}{\sqrt{2}}\right)^2 - (\sqrt{2})^2 = \left(\frac{x}{\sqrt{2}} - \sqrt{2}\right)\left(\frac{x}{\sqrt{2}} + \sqrt{2}\right) \] --- If you have specific questions about any of these factorizations or need further assistance, feel free to ask!