Question

difficulty:
factorise completely $3x^2 + 3x - 6$
a $-3(x - 2)(x + 1)$
b $3(x - 2)(x + 1)$
c $-3(x + 2)(x - 1)$
d $3(x + 2)(x - 1)$

Explanation:

Step1: Factor out the GCF

First, find the greatest common factor (GCF) of the terms \(3x^2\), \(3x\), and \(-6\). The GCF of 3, 3, and -6 is 3. Factor out 3 from the expression:
\(3x^2 + 3x - 6 = 3(x^2 + x - 2)\)

Step2: Factor the quadratic

Now, factor the quadratic \(x^2 + x - 2\). We need two numbers that multiply to -2 and add to 1. The numbers are 2 and -1. So, we can factor \(x^2 + x - 2\) as \((x + 2)(x - 1)\).

Step3: Combine the factors

Substitute the factored quadratic back into the expression from Step 1:
\(3(x^2 + x - 2) = 3(x + 2)(x - 1)\)

Answer:

D. \(3(x + 2)(x - 1)\)