Factorise the following expressions: \( \begin{array}{l}2.3 .1 \quad 12 a^{2}-3 b^{2} \\ 2.3 .2 \quad 14-5 p-p^{2} \\ 2.3 .3 \quad 2 x^{3}-2-4 x^{2}+x \\ \text { Simplify the following expression to it }\end{array} \)

We will factorise each expression step by step. ────────────────────────────── 1. Factorise 12a² – 3b² Step 1. Notice that both terms have a common factor of 3:   12a² – 3b² = 3(4a² – b²) Step 2. The expression in parentheses is a difference of two squares since 4a² = (2a)² and b² = (b)². Recall that:   X² – Y² = (X – Y)(X + Y) Thus, with X = 2a and Y = b:   4a² – b² = (2a – b)(2a + b) So, the fully factorised form is:   3(2a – b)(2a + b) ────────────────────────────── 2. Factorise 14 – 5p – p² Step 1. Write the expression in a standard quadratic form (in descending powers of p):   14 – 5p – p² = –p² – 5p + 14 Step 2. Factor out –1 to make the quadratic easier to factorise:   –(p² + 5p – 14) Step 3. Now, find two numbers that multiply to –14 and add to 5. These numbers are 7 and –2:   p² + 5p – 14 = (p + 7)(p – 2) Thus, the factorised form becomes:   –(p + 7)(p – 2) An alternative way to write the answer is to absorb the negative sign into one factor:   (2 – p)(p + 7) since (2 – p) = –(p – 2). ────────────────────────────── 3. Factorise 2x³ – 2 – 4x² + x Step 1. Rearrange the terms in descending powers of x:   2x³ – 4x² + x – 2 Step 2. Use grouping:   Group the first two and the last two terms:   [2x³ – 4x²] + [x – 2] Step 3. Factor out common factors from each group:   From 2x³ – 4x², factor out 2x² to get:     2x²(x – 2)   From x – 2, the common factor is 1:     1(x – 2) Step 4. Now, factor out the common binomial (x – 2):   (2x² + 1)(x – 2) Thus, the fully factorised expression is:   (x – 2)(2x² + 1) ────────────────────────────── Summary of Answers: 1. 12a² – 3b² = 3(2a – b)(2a + b) 2. 14 – 5p – p² = –(p + 7)(p – 2)         or equivalently (2 – p)(p + 7) 3. 2x³ – 2 – 4x² + x = (x – 2)(2x² + 1)