write step by stepili actorise the following expressions \( x^{3}-125 \) \( x^{3}-x^{2}-42 x \) \( \frac{1}{2} x^{3}+4 \) t) \( \frac{1}{2} x^{2}-2 x+2 \) 5) \( 5 x^{2}-3 \frac{1}{5} \) 6) \( y^{4}-15 y^{2}-16 \)

Sure! Let's factorize each of the given expressions step by step. --- ### 1) \( x^{3} - 125 \) **Step 1: Recognize the Form** The expression is a difference of cubes, since \( 125 = 5^3 \). **Step 2: Apply the Difference of Cubes Formula** The difference of cubes formula is: \[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \] Here, \( a = x \) and \( b = 5 \). **Step 3: Substitute into the Formula** \[ x^3 - 5^3 = (x - 5)(x^2 + x \cdot 5 + 5^2) \] \[ = (x - 5)(x^2 + 5x + 25) \] **Final Factorization:** \[ x^{3} - 125 = (x - 5)(x^2 + 5x + 25) \] --- ### 2) \( x^{3} - x^{2} - 42x \) **Step 1: Factor Out the Greatest Common Factor (GCF)** All terms have an \( x \), so factor out \( x \): \[ x(x^2 - x - 42) \] **Step 2: Factor the Quadratic** Factor \( x^2 - x - 42 \). Find two numbers that multiply to -42 and add to -1. These numbers are -7 and 6. \[ x^2 - x - 42 = (x - 7)(x + 6) \] **Final Factorization:** \[ x^{3} - x^{2} - 42x = x(x - 7)(x + 6) \] --- ### 3) \( \frac{1}{2}x^{3} + 4 \) **Step 1: Factor Out the GCF** Factor out \( \frac{1}{2} \): \[ \frac{1}{2}(x^3 + 8) \] **Step 2: Recognize the Sum of Cubes** Notice that \( 8 = 2^3 \). **Step 3: Apply the Sum of Cubes Formula** The sum of cubes formula is: \[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \] Here, \( a = x \) and \( b = 2 \). **Step 4: Substitute into the Formula** \[ \frac{1}{2}(x^3 + 2^3) = \frac{1}{2}(x + 2)(x^2 - 2x + 4) \] **Final Factorization:** \[ \frac{1}{2}x^{3} + 4 = \frac{1}{2}(x + 2)(x^2 - 2x + 4) \] --- ### 4) \( \frac{1}{2}x^{2} - 2x + 2 \) **Step 1: Factor Out the GCF** Factor out \( \frac{1}{2} \): \[ \frac{1}{2}\left( x^2 - 4x + 4 \right) \] **Step 2: Factor the Quadratic** Factor \( x^2 - 4x + 4 \). This is a perfect square trinomial: \[ x^2 - 4x + 4 = (x - 2)^2 \] **Final Factorization:** \[ \frac{1}{2}x^{2} - 2x + 2 = \frac{1}{2}(x - 2)^2 \] --- ### 5) \( 5x^{2} - 3 \frac{1}{5} \) **Step 1: Simplify the Expression** First, convert \( 3 \frac{1}{5} \) to an improper fraction: \[ 3 \frac{1}{5} = \frac{16}{5} \] So the expression becomes: \[ 5x^{2} - \frac{16}{5} \] **Step 2: Factor Out the GCF** Notice that \( 5x^2 \) and \( \frac{16}{5} \) have a common factor of \( \frac{1}{5} \): \[ 5x^{2} - \frac{16}{5} = \frac{1}{5}(25x^2 - 16) \] **Step 3: Recognize the Difference of Squares** \[ 25x^2 - 16 = (5x)^2 - 4^2 = (5x - 4)(5x + 4) \] **Final Factorization:** \[ 5x^{2} - 3 \frac{1}{5} = \frac{1}{5}(5x - 4)(5x + 4) \] --- ### 6) \( y^{4} - 15y^{2} - 16 \) **Step 1: Let’s Make a Substitution** Let \( z = y^{2} \). Then the expression becomes: \[ z^2 - 15z - 16 \] **Step 2: Factor the Quadratic in \( z \)** Find two numbers that multiply to -16 and add to -15. These numbers are -16 and 1. \[ z^2 - 15z - 16 = (z - 16)(z + 1) \] **Step 3: Substitute Back \( z = y^{2} \)** \[ (y^{2} - 16)(y^{2} + 1) \] **Step 4: Factor \( y^{2} - 16 \) Further** Recognize this as a difference of squares: \[ y^{2} - 16 = (y - 4)(y + 4) \] **Final Factorization:** \[ y^{4} - 15y^{2} - 16 = (y - 4)(y + 4)(y^{2} + 1) \] --- I hope these step-by-step factorizations help you understand the process better!