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The problem given is to factor the quadratic expression \( 10x^2 - 19x + 6 \).

Let's first attempt factoring the quadratic expression by the method of trial and error:

1. The expression can be written as \( 10x^2 - 19x + 6 \).
2. We are looking for two numbers that multiply to \( 10 \times 6 = 60 \) (the product of the coefficient of \( x^2 \) and the constant term) and add up to \( -19 \) (the coefficient of \( x \)).

These two numbers are \( -15 \) and \( -4 \), since:
\[
-15 \times -4 = 60 \quad \text{and} \quad -15 + -4 = -19.
\]

3. Now, we can rewrite the middle term using these two numbers:
\[
10x^2 - 15x - 4x + 6.
\]

4. Factor by grouping:
\[
5x(2x - 3) - 2(2x - 3).
\]

5. Factor out the common binomial \( (2x - 3) \):
\[
(5x - 2)(2x - 3).
\]

So, the factored form of \( 10x^2 - 19x + 6 \) is \( (5x - 2)(2x - 3) \), which matches option **A**.

### Answer: A) \( (5x - 2)(2x - 3) \).

Do you want any further details or explanations?

Here are 5 additional related questions you can explore:
1. How do you know when a quadratic expression is factorable versus prime?
2. What methods are used to factor quadratic equations when trial and error is difficult?
3. How can you check if your factored form of a quadratic is correct?
4. What happens if the quadratic has complex roots? How would the factoring change?
5. How does completing the square compare to factoring in solving quadratic equations?

**Tip:** Always verify your factorization by expanding the factors to make sure you return to the original quadratic expression.

Factor the quadratic expression 10x^2 - 19x + 6.

Learn how to factor the quadratic expression 10x^2 - 19x + 6 using the trial and error method. The detailed step-by-step process helps students in grades 9-11 understand the key algebraic concepts behind factoring.

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