91Ó°ÊÓ

In algebra, the zero product property is a fundamental principle that allows us to solve equations like the one in the exercise. This property asserts that if the product of two factors equals zero, then at least one of the factors must be zero. This is a crucial concept when solving polynomial equations, especially after factoring.

Let's break this down:

• If you have a factored equation like \( a \cdot b = 0 \), then either \( a = 0 \) or \( b = 0 \), or both.
• This principle helps to identify individual components of the equation that can be solved separately.

In our exercise, once we factored out the common term \( p \), we reached the equation \( p(4p^2 - 9) = 0 \). This implied that either \( p = 0 \) or \( 4p^2 - 9 = 0 \). By solving each factor separately, we efficiently find all possible solutions to the polynomial equation.

Cubic equations, characterized by the presence of a term raised to the third power, can sometimes seem daunting. However, by using techniques such as factoring, they become more manageable. In the current exercise, the equation is originally in cubic form: \(4p^3 - 9p = 0\).

Factoring is your first step. Look for the greatest common factor, which in this case is \(p\). Extracting this factor simplifies the equation to a more straightforward form, reducing it to quadratic equations or even simpler sums:

• The expression becomes \( p(4p^2 - 9) = 0 \).

Once factored, you can apply the zero product property, splitting it into manageable parts. This transformation turns a potentially complex problem into smaller, easier-to-solve equations. By addressing each resulting factor individually, solutions from cubic equations can be found much quicker.

Quadratic equations are equations of the form \( ax^2 + bx + c = 0 \) and are essential building blocks in algebra. In the exercise, after factoring out \(p\), we dealt with a quadratic equation: \( 4p^2 - 9 = 0 \).

To handle such a quadratic equation, follow these steps:

• First, isolate the squared term: \( 4p^2 = 9 \).
• Divide by the coefficient of \( p^2 \) to get \( p^2 = \frac{9}{4} \).

Next, apply the square root to both sides to solve for \( p \):

• This yields \( p = \pm \frac{3}{2} \), indicating there are two solutions derived from taking both the positive and negative roots.

Understanding how to solve quadratic equations is invaluable, as they frequently appear in various mathematical contexts, and mastering them fortifies your algebraic toolkit.