Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 5: Problem 27
Solve each equation. $$ x^{3}-x=2 x^{2}-2 $$
Short Answer
Expert verified
The solutions are \(x = 1\), \(x = \frac{1 + \sqrt{5}}{2}\), and \(x = \frac{1 - \sqrt{5}}{2}\).
Step by step solution
01
- Bring All Terms to One Side
Start by moving all terms to one side of the equation to set it equal to zero. Subtract \(2x^2\) and \(2\) from both sides:\[ x^3 - x - 2x^2 + 2 = 0 \]This equation simplifies to:\[ x^3 - 2x^2 - x + 2 = 0 \]
02
- Factor the Cubic Polynomial
To factor \(x^3 - 2x^2 - x + 2\), use the synthetic division or trial/error method to find roots. Checking possible rational roots, we can test \(x = 1\):Plug into polynomial: \(1^3 - 2(1)^2 - 1 + 2 = 0\). Therefore, \(x = 1\) is a root.
03
- Perform Synthetic Division
Perform synthetic division using \(x = 1\):``` 1 | 1 -2 -1 2 | 1 -1 0---------------- 1 -1 -1 0```The quotient is \(x^2 - x - 1\). Thus, the polynomial factors as:\[ (x - 1)(x^2 - x - 1) = 0 \]
04
- Solve the Quadratic Equation
Now solve \(x^2 - x - 1 = 0\) using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \(a = 1\), \(b = -1\), and \(c = -1\):\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)} = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} \]Thus, the solutions for the quadratic are \(x = \frac{1 + \sqrt{5}}{2}\) and \(x = \frac{1 - \sqrt{5}}{2}\).
05
- Combine All Solutions
Combine the root from the synthetic division and the solutions from the quadratic formula. The solutions to the equation are:\[ x = 1, \, x = \frac{1 + \sqrt{5}}{2}, \, x = \frac{1 - \sqrt{5}}{2} \]
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.
Synthetic Division
Synthetic division is a streamlined way to divide a polynomial by a binomial of the form \(x - c\). It's particularly useful for finding roots or simplifying higher degree polynomials. To use synthetic division, you follow these steps:
- Write down the coefficients of the polynomial you want to divide.
- Put the value of \(c\) to the left, which is the value for which \(x - c = 0\).
- Bring down the first coefficient as is.
- Multiply \(c\) by this coefficient and place it under the next coefficient.
- Add this result to the next coefficient, and repeat until all coefficients are processed.
Factoring Cubic Polynomials
Factoring cubic polynomials can initially seem daunting, but it's approachable with the right methods. For polynomials like \(x^3 - 2x^2 - x + 2\), we aim to express them as a product of simpler polynomials.
- First, find a root using methods like the Rational Root Theorem, which involves testing possible rational roots.
- Once a root is found, use synthetic division to reduce the polynomial's degree.
- The result gives you a factorization into \((x - \, \text{root})\) times a simpler polynomial.
Quadratic Formula
The quadratic formula is a fail-safe tool for solving any quadratic equation of the form \(ax^2 + bx + c = 0\). It is expressed as:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\
\]Here's how to use it:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\
\]Here's how to use it:
- Identify the coefficients \(a\), \(b\), and \(c\) in your equation.
- Plug these values into the formula.
- Simplify under the square root and complete the calculation to find the roots \(x\).
Rational Root Theorem
The Rational Root Theorem is a handy technique to help find possible rational roots of a polynomial equation. It states that for a polynomial \(ax^n + bx^{n-1} + \ldots + k = 0\), any potential rational root \(\frac{p}{q}\) is such that:
In the exercise, this theorem narrows down the rational roots of \(x^3 - 2x^2 - x + 2\), which hints at \(x = 1\) being a root, verified by substituting back into the equation. This root discovery simplifies the next steps, further showcasing the theorem's utility.
- \(p\) is a factor of the constant term \(k\).
- \(q\) is a factor of the leading coefficient \(a\).
In the exercise, this theorem narrows down the rational roots of \(x^3 - 2x^2 - x + 2\), which hints at \(x = 1\) being a root, verified by substituting back into the equation. This root discovery simplifies the next steps, further showcasing the theorem's utility.
