Problem 79
Question
Simplify. See Example \(6 .\) $$2 x^{2}+4\left(3 x-x^{2}\right)+3 x$$
Step-by-Step Solution
Verified Answer
The expression simplifies to \(-2x^2 + 15x\).
1Step 1: Distribute the 4
First, distribute the 4 into the terms inside the parentheses: \[ 4(3x - x^2) = 4 imes 3x - 4 imes x^2 = 12x - 4x^2 \]This simplifies the expression to:\[ 2x^2 + 12x - 4x^2 + 3x \]
2Step 2: Combine Like Terms
Now, combine the like terms by grouping the expressions with the same variable part:1. Combine the \( x^2 \) terms: \[ 2x^2 - 4x^2 = -2x^2 \]2. Combine the x terms: \[ 12x + 3x = 15x \]This results in the simplified expression:\[ -2x^2 + 15x \]
Key Concepts
Distributive PropertyCombining Like TermsQuadratic Expression
Distributive Property
The distributive property is a key mathematical principle that makes simplifying expressions much easier. Put simply, it tells us how to multiply a single term by each term within a set of parentheses. For example, in our exercise, we need to distribute the 4 across both terms inside the parentheses \((3x - x^2)\).
Here's how it works:- Multiply 4 by \(3x\), which gives you \(12x\)- Multiply 4 by \(-x^2\), resulting in \(-4x^2\)After distributing, you can take a deep breath because now our original expression \(2x^2 + 4(3x - x^2) + 3x\) finds itself simplified to \(2x^2 + 12x - 4x^2 + 3x\).
This clever property helps break down and untangle complex-looking problems into simpler forms that we can more easily manage.
Here's how it works:- Multiply 4 by \(3x\), which gives you \(12x\)- Multiply 4 by \(-x^2\), resulting in \(-4x^2\)After distributing, you can take a deep breath because now our original expression \(2x^2 + 4(3x - x^2) + 3x\) finds itself simplified to \(2x^2 + 12x - 4x^2 + 3x\).
This clever property helps break down and untangle complex-looking problems into simpler forms that we can more easily manage.
Combining Like Terms
Combining like terms is a fundamental step in simplifying a polynomial. This means putting together terms that have the same variable raised to the same power. Finding these groups can help make the expression a lot more compact.
In our exercise, there are two groups of like terms:
This is how combining like terms works its magic: it streamlines and declutters a polynomial.
In our exercise, there are two groups of like terms:
- The \(x^2\) terms: \(2x^2\) and \(-4x^2\)
- The x terms: \(12x\) and \(3x\)
This is how combining like terms works its magic: it streamlines and declutters a polynomial.
Quadratic Expression
A quadratic expression is a polynomial that involves the square of a variable, usually represented in the form \(ax^2 + bx + c\). Our final simplified expression from the exercise, \(-2x^2 + 15x\), is a perfect example of a quadratic expression.
Here's why this is important:- The term \(-2x^2\) is the quadratic term since it features \(x^2\).- The term \(15x\) is the linear term because it involves just \(x\).- These types of expressions are pivotal because they show up in various real-world phenomena, like calculating areas, modeling paths of objects, and finding maximum and minimum values.
Quadratic expressions can be more than just numbers and letters; they form equations and functions that describe real, everyday situations.
Here's why this is important:- The term \(-2x^2\) is the quadratic term since it features \(x^2\).- The term \(15x\) is the linear term because it involves just \(x\).- These types of expressions are pivotal because they show up in various real-world phenomena, like calculating areas, modeling paths of objects, and finding maximum and minimum values.
Quadratic expressions can be more than just numbers and letters; they form equations and functions that describe real, everyday situations.
Other exercises in this chapter
Problem 78
Evaluate each expression. See Example \(9 .\) $$ (-2)^{3}\left(\frac{-6}{-2}\right)(-1) $$
View solution Problem 78
Find the value of each expression. $$ -|1 . \overline{27}| $$
View solution Problem 79
Solve each equation. $$ -x+12=-17 $$
View solution Problem 79
Solve for the specified variable. $$ E=I R+I r \quad \text { for } R $$
View solution