Problem 61 - Introductory Algebra for College Students

Question

Factor completely. $$2 w^{4}-26 w^{3}-96 w^{2}$$

Step-by-Step Solution

Answer

The completely factored form of the polynomial \(2w^{4} - 26w^{3} - 96w^{2}\) is \(2w^{2}(w - 16)(w + 3)\).

Step 1: Identify the Greatest Common Factor (GCF)

The GCF of \(2w^{4}\), \(-26w^{3}\), and \(-96w^{2}\) is \(2w^{2}\). So we factor out \(2w^{2}\) and rewrite the polynomial as a product.

Step 2: Rewrite the Polynomial as a Product

After factoring out \(2w^{2}\), write the polynomial as a product: \(2w^{2} (w^{2} - 13w - 48)\).

Step 3: Factor the Quadratic Polynomial

The quadratic polynomial inside the parenthesis, \(w^2 - 13w - 48\), can be factored using the formula for factoring quadratic polynomials, which states that \(x^{2}-a x+b\) can be factored as \((x-r)(x-s)\), where \(r\) and \(s\) are roots of the equation \(x^{2}-a x + b = 0\). In this case, the roots \(r\) and \(s\) are 16 and -3, so the quadratic polynomial could be factored into \((w - 16)(w + 3)\).

Key Concepts

Greatest Common Factor

When working with polynomials, finding the greatest common factor (GCF) is often the first step in factoring. The GCF is the largest polynomial that divides all terms in the expression. It simplifies the polynomial and makes further factoring easier.

• Identifying the GCF: Look at each term in the polynomial. For example, in the polynomial \(2w^{4} - 26w^{3} - 96w^{2}\), identify the common factors in both the coefficients and the variable parts.
• The GCF for the coefficients 2, -26, and -96 is 2. For the variables, \(w^4, w^3, w^2\), the highest power of \(w\) common to all is \(w^2\).
• Therefore, the GCF of the polynomial is \(2w^{2}\).

This is crucial because it allows us to express the polynomial as \(2w^{2}(w^{2} - 13w - 48)\), simplifying further manipulation.

Quadratic Polynomial

A quadratic polynomial is a polynomial of degree 2, which generally takes the form \(ax^2 + bx + c\). To factor a quadratic polynomial, we often look for two numbers that multiply to make 'c' (the constant term) and add to make 'b' (the coefficient of the linear term).

• In this exercise, the polynomial we factor is \(w^2 - 13w - 48\).
• You're searching for two numbers whose product is -48 and sum is -13.
• These numbers are 16 and -3, as \(16 \times (-3) = -48\) and \(16 + (-3) = -13\).

Thus, the quadratic polynomial \(w^2 - 13w - 48\) factors into \((w - 16)(w + 3)\). Recognizing and factoring quadratics is a key skill in algebra. Once you factor these correctly, the expression simplifies significantly.

Roots of Equations

Finding the roots of an equation is about identifying which values of the variable make the equation true. It's an important skill, especially when dissecting quadratic equations.

• When you factor a quadratic, such as \((w - 16)(w + 3)\), you essentially determine its roots.
• The roots or solutions will make each factor zero, so solve \(w - 16 = 0\) and \(w + 3 = 0\).
• From \(w - 16 = 0\), solve for \(w = 16\).
• From \(w + 3 = 0\), solve for \(w = -3\).

These solutions, 16 and -3, are the points where the polynomial equals zero, meaning they are the roots of the quadratic equation \(w^2 - 13w - 48 = 0\). Understanding this helps bridge the gap between the algebraic manipulation of polynomials and their graphical representation as parabolas intersecting the x-axis.