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Solve each equation by factoring. [Hint for Exer cises 19-22: First factor out a fractional power.] x^{3}+2 x^{4}-3 x^{3}=0

Short Answer

Expert verified
The solutions are x = 0 and x = 1.

Step by step solution

01

Simplify the Equation

First, we simplify the equation by combining like terms. Notice that there are two terms involving \( x^3 \): \( x^3 \) and \(-3x^3\). We combine them to transform the equation into: \[ 2x^4 - 2x^3 = 0 \].
02

Factor Out the Common Term

Next, factor out the common term in the expression. Both terms share \(2x^3\) as a factor. Factoring \(2x^3\) out gives us: \[ 2x^3(x - 1) = 0 \].
03

Solve Each Factor Separately

For the product of factors to equal zero, each individual factor must equal zero. We have:1. \(2x^3 = 0\)2. \(x - 1 = 0\).
04

Solve for x from the First Factor

Solving \(2x^3 = 0\) involves dividing both sides by 2 which simplifies to \(x^3 = 0\). Taking the cube root of both sides of the equation gives \(x = 0\).
05

Solve for x from the Second Factor

Solving \(x - 1 = 0\) by adding 1 to both sides results in \(x = 1\).
06

Combine the Solutions

The solutions to the equation are \(x = 0\) and \(x = 1\).

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Key Concepts

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

Simplifying Equations
Simplifying equations is an essential first step in solving polynomial equations. It involves reducing the equation as much as possible to make the next steps easier to manage. Think of it as trying to untangle a knot; you start by loosening sections that are easiest to manage.
In our example equation, we had terms involving the power of 3. Specifically, \(x^3\) and \(-3x^3\). By simplifying, we combine these like terms to streamline the equation, leaving us with \(2x^4 - 2x^3 = 0\).
This simplification is crucial as it sets the stage for identifying common factors, which is the next vital step in the solving process. Remember, the fewer terms you have to work with, the easier the equation will be to factor.
Factoring Out Common Terms
Once an equation is simplified, the next step is to factor out any common terms. Factoring is essentially pulling out any shared component from each part of the equation, a bit like finding a common thread in different sections of a rope.
In our simplified equation, \(2x^4 - 2x^3 = 0\), we noticed that both terms included \(2x^3\). By factoring out \(2x^3\), we re-wrote the expression as \(2x^3(x - 1) = 0\).
Finding a common factor and re-expressing the equation is like laying the groundwork for solving. This makes the entire equation much easier to handle since it breaks down one complex problem into smaller, more manageable parts.
Solving Polynomial Equations
Now that the equation is factored, each factor can be solved separately, paving the way for finding the solution to the original equation. When an equation is expressed as a product of factors set to zero, you can apply the zero product property which states that if a product equals zero, at least one of the factors must also be zero.
In our instance, \(2x^3(x - 1) = 0\) gives us two separate equations to solve:
  • \(2x^3 = 0\)
  • \(x - 1 = 0\)
Solving \(2x^3 = 0\) involves dividing by 2 to simplify to \(x^3 = 0\), and then taking the cube root to find \(x = 0\).
For \(x - 1 = 0\), you simply add 1 to both sides, resulting in \(x = 1\).
These individual solutions, \(x = 0\) and \(x = 1\), come together as the solutions to the original polynomial equation. Being able to break down a polynomial equation in this way makes seemingly complex problems much easier to solve.

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Most popular questions from this chapter

BUSINESS: Straight-Line Depreciation Straight-line depreciation is a method for estimating the value of an asset (such as a piece of machinery) as it loses value ( \({ }^{\prime \prime}\) depreciates" \()\) through use. Given the original price of an asset, its useful lifetime, and its scrap value (its value at the end of its useful lifetime), the value of the asset after \(t\) years is given by the formula: $$ \begin{aligned} \text { Value }=(\text { Price })-&\left(\frac{(\text { Price })-(\text { Scrap value })}{(\text { Useful lifetime })}\right) \cdot t \\ & \text { for } 0 \leq t \leq(\text { Useful lifetime }) \end{aligned} $$ a. A newspaper buys a printing press for $$\$ 800,000$$ and estimates its useful life to be 20 years, after which its scrap value will be $$\$ 60,000$$. Use the formula above Exercise 63 to find a formula for the value \(V\) of the press after \(t\) years, for \(0 \leq t \leq 20\) b. Use your formula to find the value of the press after 10 years. c. Graph the function found in part (a) on a graphing calculator on the window \([0,20]\) by \([0,800,000] .\) [Hint: Use \(x\) instead of \(t\).]

SOCIAL SCIENCE: Age at First Marriage Americans are marrying later and later. Based on data for the years 2000 to 2007 , the median age at first marriage for men is \(y_{1}=0.12 x+26.8\), and for women it is \(y_{2}=0.12 x+25\), where \(x\) is the number of years since 2000 . a. Graph these lines on the window \([0,30]\) by \([0,35] .\) b. Use these lines to predict the median marriage ages for men and women in the year 2020 . [Hint: Which \(x\) -value corresponds to 2020 ? Then use TRACE, EVALUATE, or TABLE.] c. Predict the median marriage ages for men and women in the year 2030 .

BUSINESS: Research Expenditures An electronics company's research budget is \(R(p)=3 p^{0.25}\), where \(p\) is the company's profit, and the profit is predicted to be \(p(t)=55+4 t\), where \(t\) is the number of years from now. (Both \(R\) and \(p\) are in millions of dollars.) Express the research expenditure \(R\) as a function of \(t\), and evaluate the function at \(t=5\).

GENERAL: Impact Velocity If a marble is dropped from a height of \(x\) feet, it will hit the ground with velocity \(v(x)=\frac{60}{11} \sqrt{x}\) miles per hour (neglecting air resistance). Use this formula to find the velocity with which a marble will strike the ground if it dropped from the top of the tallest building in the United States, the 1451 -foot Willis Tower in Chicago.

$$ \text { If } f(x)=x+a \text { , then } f(f(x))=\text { ? } $$

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