Let's factor and solve the equation \( 2(x^2 + 1) = 5x \).

### Step 1: Simplify the equation
First, expand the left-hand side:
\[
2(x^2 + 1) = 2x^2 + 2
\]
So the equation becomes:
\[
2x^2 + 2 = 5x
\]

### Step 2: Rearrange the equation
Move all terms to one side to set the equation equal to zero:
\[
2x^2 - 5x + 2 = 0
\]

### Step 3: Factor the quadratic equation
We now factor \( 2x^2 - 5x + 2 \).

Find two numbers that multiply to \( 2 \times 2 = 4 \) and add to \( -5 \). These numbers are \( -4 \) and \( -1 \).

We rewrite the middle term using these values:
\[
2x^2 - 4x - x + 2 = 0
\]

### Step 4: Factor by grouping
Group the terms in pairs:
\[
(2x^2 - 4x) - (x - 2) = 0
\]

Factor each group:
\[
2x(x - 2) - 1(x - 2) = 0
\]

Factor out \( (x - 2) \):
\[
(2x - 1)(x - 2) = 0
\]

### Step 5: Solve for \( x \)
Now, solve the factored equation:
\[
2x - 1 = 0 \quad \text{or} \quad x - 2 = 0
\]

For \( 2x - 1 = 0 \):
\[
x = \frac{1}{2}
\]

For \( x - 2 = 0 \):
\[
x = 2
\]

### Final Answer:
The solutions are:
\[
x = \frac{1}{2} \quad \text{or} \quad x = 2
\]

Would you like more details or have any questions?

Here are 5 related questions to explore:

1. How can you verify the solutions to a factored equation?
2. What is the quadratic formula, and how can it solve any quadratic equation?
3. How can you factor quadratics when the leading coefficient is not 1?
4. What are other methods to solve quadratic equations besides factoring?
5. What are the steps to complete the square for solving a quadratic equation?

**Tip:** Always check your factored solutions by substituting them back into the original equation.

Learn how to factor and solve the quadratic equation 2(x^2 + 1) = 5x with a detailed step-by-step solution. This problem involves expanding, factoring, and solving using the zero product property, and is suitable for students in Grades 9-10.

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