The expression \( 8x^2 - 2x - 3 \) is a quadratic trinomial. To factor it, we need to find two binomials that multiply to give this expression.

The general approach for factoring a quadratic of the form \( ax^2 + bx + c \) is:

1. Multiply \( a \) and \( c \) (the coefficients of \( x^2 \) and the constant term, respectively).
2. Find two numbers that multiply to \( ac \) and add to \( b \) (the coefficient of \( x \)).
3. Use these two numbers to split the middle term and factor by grouping.

Let's apply this to \( 8x^2 - 2x - 3 \):

1. \( a = 8 \), \( b = -2 \), and \( c = -3 \).
2. Multiply \( a \) and \( c \): \( 8 \times -3 = -24 \).
3. We need two numbers that multiply to \(-24\) and add to \(-2\). The numbers are \( -6 \) and \( 4 \) because \( -6 \times 4 = -24 \) and \( -6 + 4 = -2 \).
4. Split the middle term using these numbers: 
   \[
   8x^2 - 6x + 4x - 3
   \]
5. Factor by grouping:
   \[
   (8x^2 - 6x) + (4x - 3)
   \]
   Factor out the common factors in each group:
   \[
   2x(4x - 3) + 1(4x - 3)
   \]
6. Factor out the common binomial factor:
   \[
   (4x - 3)(2x + 1)
   \]

So, the factorized form of \( 8x^2 - 2x - 3 \) is \( (4x - 3)(2x + 1) \).

Would you like more details or have any questions?

Here are some related questions:
1. How do you factor a trinomial when the leading coefficient is not 1?
2. What if the quadratic was \( 8x^2 + 2x - 3 \)?
3. How would you solve the equation \( 8x^2 - 2x - 3 = 0 \)?
4. Can the quadratic formula be used to factor this expression?
5. What is the significance of the discriminant in factoring?
6. How can you check if your factorization is correct?
7. How would this expression change if \( c = 3 \) instead of \( -3 \)?
8. What are the roots of the quadratic \( 8x^2 - 2x - 3 \)?

**Tip:** When factoring trinomials, double-check the middle term to ensure that the factors correctly add to it.

Learn how to factorize the quadratic expression 8x^2 - 2x - 3 step-by-step. Understand the process of factoring quadratic trinomials and apply it to solve similar problems. Suitable for high school students.

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