91Ó°ÊÓ

When we talk about a quadratic trinomial, we are referring to a polynomial with three terms where the highest degree is two. In the context of quadratic expressions, each term typically involves variables, usually expressed in the form of:

• ax² (where 'a' is a coefficient)
• bx (where 'b' is a coefficient)
• c (a constant term)

This form is called a trinomial because it consists of three parts. In our exercise, we started with the polynomial expression:

\[8x^2 + 14xy - 15y^2\] This is indeed a quadratic trinomial because of its structure:

• The term \(8x^2\) is the quadratic term.
• The term \(14xy\) is a single-degree term involving two variables.
• The term \(-15y^2\) is another quadratic term.

Understanding that we're dealing with a quadratic trinomial sets the stage for choosing the right method to factor it.

Factoring by grouping is a vital technique for solving polynomials, especially quadratic trinomials that don't easily factor in one step. When direct factoring is not straightforward, this approach stands as a reliable method. Here's how it was applied in our exercise:

The main idea of factoring by grouping involves splitting the middle term so that the trinomial can be broken into simpler "groups" that have a common factor. We initially wrote the trinomial:
\[8x^2 + 14xy - 15y^2\]
Finding two numbers, which in this case are 20 and -6, that multiply to the product of \(ac\) (which is -120) and add up to \(b\) (which is 14), allowed us to split the term 14xy:

• Split into 8x² + 20xy - 6xy - 15y²
• Group as (8x² + 20xy) and (-6xy - 15y²)
• Factor out the greatest common factor from each group:
• From the first group: Factor out \(4x\) to get \(4x(2x + 5y)\)
• From the second group: Factor out \(-3y\) to get \(-3y(2x + 5y)\)

Finally, notice both groups have a common factor \(2x + 5y\). This allows us to write the trinomial as \((4x - 3y)(2x + 5y)\). This method greatly simplifies complex quadratic trinomials by breaking them into manageable parts.

The standard form of a quadratic equation is crucial for analyzing and solving quadratic expressions effectively. It is often expressed as \(ax^2 + bx + c = 0\). For trinomials like the one in our exercise, the standard form helps us visualize and organize the expression in a way that clarifies the roles of each term.

Our original expression was \(8x^2 + 14xy - 15y^2\), which already fit into the standard form layout:

• The first term (8x²) is the quadratic term, indicating the highest degree.
• The middle term (14xy) is linear in nature when considering either x or y separately.
• The third term (-15y²) is another quadratic term involving the other variable y.

What's essential with the standard form is the coefficients: \(a = 8\), \(b = 14\), and \(c = -15\). Recognizing these parts allows us to systematically approach techniques like factoring by grouping. Moreover, having the expression in standard form also facilitates the use of other methods such as completing the square or the quadratic formula, should they become necessary. These approaches all rely on the clarity provided by the orderly arrangement of terms in the standard quadratic form.