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The expression you provided is the quadratic trinomial:

\[
3x^2 + 8x - 3
\]

To factor this trinomial, we'll follow these steps:

### Step 1: Multiply the leading coefficient and the constant
Multiply the coefficient of \(x^2\) (which is 3) by the constant term (which is -3):

\[
3 \times -3 = -9
\]

### Step 2: Find two numbers that multiply to give -9 and add to give the middle coefficient, 8
We need two numbers that multiply to give -9 and add to give 8. These numbers are 9 and -1 because:

\[
9 \times (-1) = -9 \quad \text{and} \quad 9 + (-1) = 8
\]

### Step 3: Rewrite the middle term using these numbers
We rewrite the expression \(3x^2 + 8x - 3\) by splitting the middle term (8x) into two terms using 9 and -1:

\[
3x^2 + 9x - x - 3
\]

### Step 4: Factor by grouping
Now group the terms:

\[
(3x^2 + 9x) - (x + 3)
\]

Factor each group:

\[
3x(x + 3) - 1(x + 3)
\]

### Step 5: Factor the common binomial
Both terms have a common factor of \((x + 3)\), so factor that out:

\[
(3x - 1)(x + 3)
\]

Thus, the factored form of the trinomial \(3x^2 + 8x - 3\) is:

\[
(3x - 1)(x + 3)
\]

Let me know if you'd like more details on any step, and here are some follow-up questions that can further your understanding:

1. How do you factor a trinomial when the leading coefficient is not 1?
2. What are some other methods for factoring quadratic trinomials?
3. How can you check if your factorization is correct?
4. What is the role of factoring in solving quadratic equations?
5. What happens if a quadratic trinomial is not factorable over the integers?

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

Learn how to factor the quadratic trinomial 3x^2 + 8x - 3 with a step-by-step solution. This problem covers key algebraic concepts such as factoring by grouping, using the ac-method to split the middle term, and finding common factors. Suitable for students in Grades 9-11.

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