Chapter 7: Problem 29
Factor completely, if possible. Check your answer. $$w^{2}+4 w-5$$
Short Answer
Expert verified
The quadratic expression \(w^2 + 4w - 5\) can be factored completely as \((w + 5)(w - 1)\).
Step by step solution
01
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 \(c = -5\).
02
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 \(5 + -1 = 4\).
03
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: \(w^2 + 5w - 1w - 5\).
04
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:
\(w(w + 5) - 1(w + 5)\)
05
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.
06
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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.
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.
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.
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:
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 \(-1(w + 5)\).