Algebra 2 Worksheets with pencils, ruler, and backpack on a graph paper design

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Problem Analysis:

Since the image does not contain a specific math problem, I will assume the task is related to solving a typical Algebra 2 problem. If you have a specific problem in mind, please provide it, and I can solve it for you. For now, I'll demonstrate how to solve a common type of Algebra 2 problem: solving a quadratic equation.

Example Problem:

Solve the quadratic equation:
\[ x^2 - 5x + 6 = 0 \]

Solution:

#### Step 1: Identify the quadratic equation
The given equation is:
\[ x^2 - 5x + 6 = 0 \]
This is a standard quadratic equation of the form \( ax^2 + bx + c = 0 \), where:
- \( a = 1 \)
- \( b = -5 \)
- \( c = 6 \)

#### Step 2: Factor the quadratic equation
To factor the quadratic equation, we need to find two numbers that:
1. Multiply to give \( ac = 1 \cdot 6 = 6 \)
2. Add up to give \( b = -5 \)

The two numbers that satisfy these conditions are \( -2 \) and \( -3 \) because:
\[ -2 \cdot -3 = 6 \]
\[ -2 + (-3) = -5 \]

So, we can rewrite the middle term \(-5x\) as \(-2x - 3x\):
\[ x^2 - 5x + 6 = x^2 - 2x - 3x + 6 \]

#### Step 3: Group and factor by grouping
Group the terms:
\[ (x^2 - 2x) + (-3x + 6) \]

Factor out the greatest common factor from each group:
\[ x(x - 2) - 3(x - 2) \]

Now, factor out the common binomial factor \((x - 2)\):
\[ (x - 2)(x - 3) = 0 \]

#### Step 4: Solve for \( x \)
Set each factor equal to zero:
\[ x - 2 = 0 \quad \text{or} \quad x - 3 = 0 \]

Solve each equation:
\[ x = 2 \quad \text{or} \quad x = 3 \]

#### Final Answer:
The solutions to the quadratic equation \( x^2 - 5x + 6 = 0 \) are:
\[ \boxed{2 \text{ and } 3} \]

If you have a different problem or need further assistance, feel free to provide more details!