a) \( x^{2}-2 x \) b) \( x^{2}-4 \) c) \( x^{2}+2 x-15 \) d) \( 6 x^{2}-2 \cdot a x-3 x y+a y \) e) \( 27 x^{3}+y^{3} \) f) \( 54 x^{3}-16 \)

Certainly! Let's factor each of the given expressions step by step. --- ### a) \( x^{2} - 2x \) **Factoring:** \[ x^{2} - 2x = x(x - 2) \] --- ### b) \( x^{2} - 4 \) **Factoring:** This is a **difference of squares**, which can be factored as: \[ x^{2} - 4 = (x - 2)(x + 2) \] --- ### c) \( x^{2} + 2x - 15 \) **Factoring:** We look for two numbers that multiply to **-15** and add to **2**. These numbers are **5** and **-3**. \[ x^{2} + 2x - 15 = (x + 5)(x - 3) \] --- ### d) \( 6x^{2} - 2a x - 3x y + a y \) **Factoring:** Group the terms to factor by grouping: \[ \begin{align*} 6x^{2} - 2a x - 3x y + a y &= (6x^{2} - 2a x) + (-3x y + a y) \\ &= 2x(3x - a) - y(3x - a) \\ &= (3x - a)(2x - y) \end{align*} \] --- ### e) \( 27x^{3} + y^{3} \) **Factoring:** This is a **sum of cubes**. Recall that \( A^{3} + B^{3} = (A + B)(A^{2} - AB + B^{2}) \). Let \( A = 3x \) and \( B = y \): \[ \begin{align*} 27x^{3} + y^{3} &= (3x)^{3} + y^{3} \\ &= (3x + y)\left((3x)^{2} - (3x)(y) + y^{2}\right) \\ &= (3x + y)(9x^{2} - 3xy + y^{2}) \end{align*} \] --- ### f) \( 54x^{3} - 16 \) **Factoring:** First, factor out the greatest common factor (GCF), which is **2**: \[ 54x^{3} - 16 = 2(27x^{3} - 8) \] Now, recognize that \( 27x^{3} - 8 \) is a **difference of cubes**: \[ 27x^{3} - 8 = (3x)^{3} - 2^{3} = (3x - 2)\left((3x)^{2} + (3x)(2) + 2^{2}\right) = (3x - 2)(9x^{2} + 6x + 4) \] Combining both steps: \[ 54x^{3} - 16 = 2(3x - 2)(9x^{2} + 6x + 4) \] --- ### **Summary of Factored Forms:** a) \( x^{2} - 2x = \mathbf{x(x - 2)} \) b) \( x^{2} - 4 = \mathbf{(x - 2)(x + 2)} \) c) \( x^{2} + 2x - 15 = \mathbf{(x + 5)(x - 3)} \) d) \( 6x^{2} - 2a x - 3x y + a y = \mathbf{(3x - a)(2x - y)} \) e) \( 27x^{3} + y^{3} = \mathbf{(3x + y)(9x^{2} - 3xy + y^{2})} \) f) \( 54x^{3} - 16 = \mathbf{2(3x - 2)(9x^{2} + 6x + 4)} \) --- If you have any further questions or need additional explanations, feel free to ask!