Question

1. $3x^2 - 11x - 4$

Explanation:

Step1: Find two numbers for splitting the middle term

We need two numbers \(a\) and \(b\) such that \(a\times b = 3\times(-4)=-12\) and \(a + b=-11\). The numbers are \(-12\) and \(1\) since \(-12\times1 = -12\) and \(-12 + 1=-11\).
So, rewrite the middle term: \(3x^{2}-12x + x-4\)

Step2: Group the terms

Group the first two and last two terms: \((3x^{2}-12x)+(x - 4)\)

Step3: Factor out the GCF from each group

From the first group, GCF of \(3x^{2}\) and \(-12x\) is \(3x\), so \(3x(x - 4)\). From the second group, GCF of \(x\) and \(-4\) is \(1\), so \(1(x - 4)\). Now we have \(3x(x - 4)+1(x - 4)\)

Step4: Factor out the common binomial factor

Factor out \((x - 4)\) from both terms: \((3x + 1)(x - 4)\)

(For the box method, the four terms in the box would be \(3x^{2}\), \(-12x\), \(x\), and \(-4\) arranged such that we can factor by grouping. The top - left: \(3x^{2}\), top - right: \(-12x\), bottom - left: \(x\), bottom - right: \(-4\))

Answer:

The factored form of \(3x^{2}-11x - 4\) is \(\boldsymbol{(3x + 1)(x - 4)}\)