Question

use the zero product property to find the solutions to the equation (x + 2) (x + 3) = 12.\
\\(\circ\\) \\(x = -6\\) or \\(x = 1\\)\
\\(\circ\\) \\(x = -3\\) or \\(x = -2\\)\
\\(\circ\\) \\(x = -1\\) or \\(x = 6\\)\
\\(\circ\\) \\(x = 2\\) or \\(x = 3\\)

Explanation:

Step1: Expand the left side

First, we expand \((x + 2)(x + 3)\) using the distributive property (FOIL method).
\((x + 2)(x + 3)=x^2+3x + 2x+6=x^2 + 5x+6\)
So the equation becomes \(x^2+5x + 6=12\)

Step2: Rearrange to standard quadratic form

Subtract 12 from both sides to set the equation to zero:
\(x^2+5x+6 - 12=0\)
Simplify to get \(x^2+5x - 6 = 0\)

Step3: Factor the quadratic

We need two numbers that multiply to \(-6\) and add to \(5\). The numbers are \(6\) and \(-1\).
So, \(x^2+5x - 6=(x + 6)(x - 1)=0\)

Step4: Apply zero product property

The zero product property states that if \(ab = 0\), then either \(a = 0\) or \(b = 0\).
So, \(x+6 = 0\) or \(x - 1=0\)
Solving \(x+6 = 0\) gives \(x=-6\)
Solving \(x - 1=0\) gives \(x = 1\)

Answer:

\(x=-6\) or \(x = 1\) (corresponding to the first option: \(x=-6\) or \(x = 1\))