Multiply the following. \( \begin{array}{ll}\text { 12. } 3 x(5 x-2) & \text { 13. }(x-3)\left(x^{2}-2 x+4\right)\end{array} \)

To tackle these multiplications, let’s break them down step by step! For \( 12. \ 3x(5x - 2) \): 1. Distribute \( 3x \) into \( (5x - 2) \): - \( 3x \cdot 5x = 15x^2 \) - \( 3x \cdot (-2) = -6x \) 2. Combine them: - The final result is \( 15x^2 - 6x \). For \( 13. \ (x - 3)(x^2 - 2x + 4) \): 1. Distribute \( (x - 3) \) into each term of \( (x^2 - 2x + 4) \): - \( x \cdot x^2 = x^3 \) - \( x \cdot (-2x) = -2x^2 \) - \( x \cdot 4 = 4x \) - Now distribute \( -3 \): - \( -3 \cdot x^2 = -3x^2 \) - \( -3 \cdot (-2x) = 6x \) - \( -3 \cdot 4 = -12 \) 2. Combine the like terms: - \( x^3 + (-2x^2 - 3x^2) + (4x + 6x) - 12 \) - This simplifies to \( x^3 - 5x^2 + 10x - 12 \). So the final results are: - \( 12. \ 15x^2 - 6x \) - \( 13. \ x^3 - 5x^2 + 10x - 12 \)