Question

trinomial: (6x^2 - x - 15)

Explanation:

Assuming the problem is to factor the trinomial \(6x^{2}-x - 15\), we can use the AC method for factoring quadratic trinomials of the form \(ax^{2}+bx + c\) (here \(a = 6\), \(b=- 1\), \(c = - 15\)).

Step 1: Find two numbers that multiply to \(a\times c\) and add up to \(b\)

First, calculate \(a\times c=6\times(-15)=-90\). We need two numbers that multiply to \(-90\) and add up to \(b=-1\). The numbers are \(-10\) and \(9\) because \(-10\times9=-90\) and \(-10 + 9=-1\).

Step 2: Rewrite the middle term using these two numbers

Rewrite the trinomial \(6x^{2}-x - 15\) as \(6x^{2}+9x-10x - 15\) (since \(-x=9x-10x\)).

Step 3: Group the terms and factor by grouping

Group the first two terms and the last two terms: \((6x^{2}+9x)+(-10x - 15)\)

Factor out the greatest common factor (GCF) from each group:

• From \(6x^{2}+9x\), the GCF is \(3x\), so \(3x(2x + 3)\)
• From \(-10x - 15\), the GCF is \(-5\), so \(-5(2x + 3)\)

Now we have \(3x(2x + 3)-5(2x + 3)\). Notice that \((2x + 3)\) is a common factor.

Step 4: Factor out the common binomial factor

Factor out \((2x + 3)\) from both terms: \((2x + 3)(3x-5)\)

Answer:

The factored form of \(6x^{2}-x - 15\) is \((2x + 3)(3x - 5)\)