/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 Factor the expressions in exerci... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 28

Factor the expressions in exercises. $$ n^{2}-n-30 $$

Short Answer

Expert verified
Answer: The factored form of the quadratic expression is $$(n-6)(n+5)$$.

Step by step solution

01

Identify the quadratic expression

The given quadratic expression is $$n^2 - n - 30$$.
02

Find two numbers with a product of -30 and a sum of -1

The two numbers that satisfy the condition are 5 and -6. They have a product of -30 and a sum of -1 (5 + (-6) = -1).
03

Rewrite the middle term using the two numbers found in step 2

Rewrite the quadratic expression: $$n^2 - n - 30 = n^2 + 5n - 6n - 30$$.
04

Factor by grouping

Group the terms in pairs and apply the distributive property (factoring out any common factors): $$(n^2 + 5n) + (-6n - 30) = n(n+5) - 6(n+5)$$.
05

Factor out the common binomial term

Notice that both terms in the previous step have a common binomial term of $$(n+5)$$. Factor out this term: $$n(n+5) - 6(n+5) = (n-6)(n+5)$$. The factored expression is $$(n-6)(n+5)$$.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Expressions
Quadratic expressions are algebraic expressions of the form \(ax^2 + bx + c\). The highest exponent of the variable is 2, which makes it a quadratic. The expression "\(n^2 - n - 30\)" follows this format with coefficients: \(a = 1\), \(b = -1\), and \(c = -30\). Quadratic expressions can often be factored into the product of two binomials. This is done to simplify the expression or to find the roots of the related quadratic equation. Knowing the format helps to quickly identify the expression as quadratic when working on problems.
Factoring by Grouping
Factoring by grouping is a method used to factor expressions by rearranging terms and grouping them into pairs. In the exercise, the expression \(n^2 - n - 30\) is rewritten by using two numbers that have a specified product and sum. These numbers help split the middle term, \(-n\), into two parts. For this example, numbers 5 and -6 are used, giving us the expression \(n^2 + 5n - 6n - 30\).
  • Group the terms: \((n^2+5n) + (-6n-30)\).
  • Factor each group individually: \(n(n+5)\) and \(-6(n+5)\).
Notice how grouping the terms allows us to factor each part separately, revealing the common factor \((n+5)\). This results in a factored expression that can be simplified further.
Algebraic Equations
Algebraic equations are mathematical statements that express the equality of two expressions. They can be solved to find the values of variables. When dealing with quadratic expressions like \(n^2 - n - 30\), factoring plays a central role. By transforming the original quadratic expression into a product of binomials \((n-6)(n+5) = 0\), we can set each factor equal to zero:
  • \(n-6=0\) gives solution \(n=6\).
  • \(n+5=0\) gives solution \(n=-5\).
These solutions represent the values where the original equation "\(n^2 - n - 30 = 0\)" holds true. Factoring quadratics into simpler expressions is a key skill in solving these types of algebraic equations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In electronics, when two resistors, with resistances \(A\) and \(B,\) are connected in a parallel circuit, the total resistance is $$ \frac{1}{1 / A+1 / B} $$ Rewrite this so there are no fractions in the numerator or denominator.

Are the two expressions equivalent? \(a b(a+b+1) \quad\) and \(\quad a^{2} b+a b^{2}+a b\)

Rewrite the expression by taking out the common factors. \(5 x+100\)

Quabbin Reservoir in Massachusetts provides much of Boston's water. At the start of 2009 the reservoir contained 412 billion gallons of water. (a) If in January 2009 the amount (in billions of gallons) flowing into the reservoir is \(A\) and the amount flowing out is \(B\), write an expression for the amount of water in the reservoir at the end of January \(2009 .\) (b) If in February 2009 the amount flowing in is 20 billion gallons less than the previous month and the amount flowing out is 12 billion gallons more than the previous month, write an expression for the amount of water in the reservoir at the end of February \(2009 .\) (c) Combine like terms to simplify your answer to part (b).

Is the fraction \(\frac{x+3}{x}\) equivalent to \(1+\frac{3}{x}\) ?
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation

Calculus

Read Explanation

Probability and Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.