Question

(b) factorise $8ab - 3 + 12a - 2b$.

Explanation:

Step1: Group terms with common factors

Group the first two terms and the last two terms: \((8ab - 3)+(12a - 2b)\) is not helpful. Instead, look for common factors in all terms. Wait, let's re - examine the terms: \(8ab\), \(- 3\), \(12a\), \(-2b\). Let's try to factor by grouping. Rearrange the terms: \(8ab+12a - 2b - 3\). Now group as \((8ab + 12a)+(-2b - 3)\).

Step2: Factor out common factors from each group

For the first group \(8ab + 12a\), the common factor is \(4a\). So \(8ab+12a=4a(2b + 3)\). For the second group \(-2b-3=-(2b + 3)\).

Step3: Factor out the common binomial factor

Now we have \(4a(2b + 3)-(2b + 3)\). The common factor here is \((2b + 3)\). So we can factor it out: \((2b + 3)(4a - 1)\)

Answer:

\((2b + 3)(4a - 1)\)