1.5 Simplify completely \( 1.5 .1 \quad 3 x(2 x-4 x y) \) \( 1.5 .2 \quad(x-3)^{2} \) \( 1.5 .3 \quad(2 r-p)\left(3 r^{2}-4 r p+p^{2}\right) \) QUESTION 2 2.1 Factorize completely \( 2.1 .1 \quad 2 x^{2}-8 \)

Alright, I need to simplify and factorize the given expressions. Let's tackle them one by one. **1.5.1 Simplify \( 3x(2x - 4xy) \):** First, I'll distribute the \( 3x \) across the terms inside the parentheses. - \( 3x \times 2x = 6x^2 \) - \( 3x \times (-4xy) = -12x^2y \) So, the expression simplifies to \( 6x^2 - 12x^2y \). **1.5.2 Simplify \( (x - 3)^2 \):** This is a perfect square trinomial. Expanding it: - \( (x - 3)^2 = x^2 - 6x + 9 \) **1.5.3 Simplify \( (2r - p)(3r^2 - 4rp + p^2) \):** I'll use the distributive property to multiply each term in the first binomial by each term in the second trinomial. - \( 2r \times 3r^2 = 6r^3 \) - \( 2r \times (-4rp) = -8r^2p \) - \( 2r \times p^2 = 2r p^2 \) - \( -p \times 3r^2 = -3r^2p \) - \( -p \times (-4rp) = 4r p^2 \) - \( -p \times p^2 = -p^3 \) Now, combine like terms: - \( 6r^3 \) - \( -8r^2p - 3r^2p = -11r^2p \) - \( 2r p^2 + 4r p^2 = 6r p^2 \) - \( -p^3 \) So, the simplified expression is \( 6r^3 - 11r^2p + 6r p^2 - p^3 \). **2.1 Factorize \( 2x^2 - 8 \):** First, I'll factor out the greatest common factor (GCF), which is 2: - \( 2(x^2 - 4) \) Now, \( x^2 - 4 \) is a difference of squares, which can be factored as: - \( (x - 2)(x + 2) \) So, the fully factored form is \( 2(x - 2)(x + 2) \).