Problem 29 - Beginning and Intermediate Algebra

Question

Factor completely, if possible. Check your answer. $$w^{2}+4 w-5$$

Step-by-Step Solution

Answer

The quadratic expression \(w^2 + 4w - 5\) can be factored completely as \((w + 5)(w - 1)\).

Step 1: Identify the structure of the given quadratic expression

The quadratic expression is given as \(w^2 + 4w - 5\), and its general form is \(aw^2 + bw + c\), where \(a = 1\), \(b = 4\), and \(c = -5\).

Step 2: Find two numbers whose product is ac and sum is b

The product of \(a = 1\) and \(c = -5\) is -5, and the sum is \(b = 4\). We need to find two numbers that multiply to -5 and add up to 4. These numbers are 5 and -1 because \(5 \times -1 = -5\) and \(5 + -1 = 4\).

Step 3: Rewrite the quadratic expression using the two numbers found

Rewrite the middle term of the expression using the two numbers 5 and -1. The expression becomes: \(w^2 + 5w - 1w - 5\).

Step 4: Factor by grouping

Group the terms in pairs and factor out the greatest common factor (GCF) in each pair: \((w^2 + 5w) + (-1w - 5)\) Factor out the GCF in both pairs: \(w(w + 5) - 1(w + 5)\)

Step 5: Factor out the common binomial

The binomial \((w + 5)\) is common in both terms, so factor it out: \((w + 5)(w - 1)\) We have now factored the quadratic expression completely.

Step 6: Check the answer

To check our answer, we'll use FOIL (First, Outer, Inner, Last) to multiply the binomials back together: \((w + 5)(w - 1) = w^2 - w + 5w -5\) Combine like terms: \((w + 5)(w - 1) = w^2 + 4w - 5\) The final check confirms that our factoring is correct, as we have the original quadratic expression.

Key Concepts

Quadratic Expressions

A quadratic expression is a polynomial of degree two, typically written in the form \(ax^2 + bx + c\). Here, \(a\), \(b\), and \(c\) are constants, and \(x\) represents the variable. The expression for this exercise is \(w^2 + 4w - 5\), where \(a = 1\), \(b = 4\), and \(c = -5\).
Quadratic expressions are foundational in algebra, appearing in many contexts such as physics, engineering, and economics. Recognizing the structure \(ax^2 + bx + c\) is crucial for tackling various problems.

• The term \(ax^2\): Represents the quadratic term.
• The term \(bx\): Represents the linear component.
• The term \(c\): Acts as the constant part.

Factoring quadratics is a method of rewriting the expression as a product of simpler expressions, which is a key step in solving quadratic equations.

Polynomial Factorization

Polynomial factorization involves breaking down a polynomial into a product of its factors. In the case of quadratic expressions, we aim to express \(ax^2 + bx + c\) as \((dx + e)(fx + g)\).
To factor, we look for two numbers that add up to \(b\) and multiply to \(ac\). For the expression \(w^2 + 4w - 5\), we found 5 and -1 to be the numbers needed.

• Product of two numbers \(= ac\): Here, \(a = 1\) and \(c = -5\), so the product is -5.
• Sum of two numbers \(= b\): We need a sum of 4.

Polynomials like quadratics can often be broken into binomials (expressions with two terms), making them easier to manipulate or solve. Successfully factoring requires finding common factors and rearranging terms strategically.

Algebraic Manipulation

Algebraic manipulation is the process of using algebraic techniques to simplify or solve expressions. In the exercise, after identifying the numbers 5 and -1 as suitable for factoring, we rewrite the quadratic from \(w^2 + 4w - 5\) into \(w^2 + 5w - 1w - 5\).
This manipulation allows us to group terms, creating pairs such as \((w^2 + 5w) + (-1w - 5)\). We factor out common elements in each pair:

• First pair: Factor \(w\) from \(w^2 + 5w\) to get \(w(w + 5)\).
• Second pair: Factor \(-1\) from \(-1w - 5\) to get \(-1(w + 5)\).

The common binomial \((w + 5)\) can then be factored out, resulting in \((w + 5)(w - 1)\). This kind of manipulation is a critical skill for solving equations and understanding algebra deeply. Practicing this ensures that you can tackle more complex algebraic problems.