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The given trinomial is:

\[
5x^2 + 41x + 42
\]

We need to factor this expression. Here's the step-by-step process:

### Step 1: Multiply the leading coefficient by the constant term
Multiply the coefficient of \(x^2\) (which is 5) by the constant term (which is 42):

\[
5 \times 42 = 210
\]

### Step 2: Find two numbers that multiply to give 210 and add up to 41
We need two numbers that multiply to 210 and add to 41. After checking pairs of factors, we find:

\[
6 \times 35 = 210 \quad \text{and} \quad 6 + 35 = 41
\]

### Step 3: Rewrite the middle term using these two numbers
Now we rewrite the middle term \(41x\) as \(6x + 35x\):

\[
5x^2 + 6x + 35x + 42
\]

### Step 4: Factor by grouping
Group the terms in pairs and factor each pair:

\[
(5x^2 + 6x) + (35x + 42)
\]

Factor out the common factors from each group:

\[
x(5x + 6) + 7(5x + 6)
\]

### Step 5: Factor out the common binomial
Now, factor out the common binomial \((5x + 6)\):

\[
(x + 7)(5x + 6)
\]

### Final Answer:
The factorization of \(5x^2 + 41x + 42\) is:

\[
(x + 7)(5x + 6)
\]

Would you like a more detailed explanation of any step? 

Here are some related questions to expand on this:

1. How do you factor trinomials where the leading coefficient is not 1?
2. What are the different methods of factoring quadratic expressions?
3. How do you verify if your factorization is correct?
4. What happens if no real factors exist for a trinomial?
5. Can all quadratic expressions be factored using integers?

**Tip**: Always verify your factorization by expanding the factors to check if they multiply back to the original expression.

Learn how to factor the trinomial 5x^2 + 41x + 42 step by step. This detailed solution covers factoring using grouping methods and is ideal for students in Grades 8-10.

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