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Magazine Chapter 3: Problem 2
Find all real and imaginary solutions to each equation. Check your answers. $$x^{3}-x^{2}-5 x+5=0$$
Short Answer
Expert verified
x = 1, x = \( \sqrt{5} \), and x = \( -\sqrt{5} \). All roots are real.
Step by step solution
01
- Finding Initial Rational Root
To find any rational roots, use the Rational Root Theorem. The possible rational roots of the equation are 卤1, 卤5.
02
- Testing Possible Rational Roots
Test these values in the polynomial equation: 1) For x=1: \(1^3 - 1^2 - 5 \times 1 + 5 = 0\) This is true, so 1 is a root.
03
- Polynomial Division
Since x = 1 is a root, use polynomial division or synthetic division to divide the polynomial \(x^3 - x^2 - 5x + 5\) by \(x - 1\).
04
- Polynomial Division Continues
Perform the polynomial division: The result is \(x^2 - 5\).
05
- Solve the Quadratic Equation
Solve the quadratic equation \(x^2 - 5 = 0\) by setting it equal to zero: \(x^2 - 5 = 0\). Thus, \(x^2 = 5\).
06
- Finding the Roots of the Quadratic Equation
Solve for x to find: \(x = \pm\sqrt{5}\).
07
- Listing All Solutions
Combine the solutions found: x = 1, x = \( \sqrt{5} \) and \( x = - \sqrt{5} \). There are no imaginary solutions.
08
- Verification
Check each solution by substituting them back into the original equation to verify they satisfy it.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rational Root Theorem
The Rational Root Theorem is a useful tool for finding rational roots of polynomial equations. It states that if a polynomial has a rational root, it must be a factor of the constant term divided by a factor of the leading coefficient. In our example, the polynomial is \(x^3 - x^2 - 5x + 5\). Here, the constant term is 5, and the leading coefficient is 1. The possible rational roots are the factors of 5 divided by the factors of 1, which gives us 卤1 and 卤5.
To use this theorem, we test these possible values in the polynomial equation to see if any of them make the equation equal zero. In our equation, we find that x = 1 is a root.
To use this theorem, we test these possible values in the polynomial equation to see if any of them make the equation equal zero. In our equation, we find that x = 1 is a root.
Polynomial Division
Once a root is identified using the Rational Root Theorem, the next step is to simplify the polynomial using polynomial division. In our example, we discovered that x = 1 is a root. Therefore, we divide the polynomial \(x^3 - x^2 - 5x + 5\) by \(x - 1\).
Polynomial division helps break down a polynomial into simpler components. You can use either long division or synthetic division. Both methods will arrive at the same result. When dividing \(x^3 - x^2 - 5x + 5\) by \(x - 1\), you will result in a quotient of \(x^2 - 5\). This reduced polynomial is much simpler to work with in subsequent steps.
Polynomial division helps break down a polynomial into simpler components. You can use either long division or synthetic division. Both methods will arrive at the same result. When dividing \(x^3 - x^2 - 5x + 5\) by \(x - 1\), you will result in a quotient of \(x^2 - 5\). This reduced polynomial is much simpler to work with in subsequent steps.
Quadratic Equations
When you reduce a polynomial using polynomial division, you often end up with a quadratic equation. In our case, the reduced polynomial is \(x^2 - 5\). Quadratic equations usually take the form \(ax^2 + bx + c = 0\). These equations can be solved by factoring, completing the square, or using the quadratic formula.
For our specific quadratic equation \(x^2 - 5 = 0\), we move the constant term to the other side to get \(x^2 = 5\). To find the values of x, we take the square root of both sides, resulting in \(x = \pm\sqrt{5}\). These are the solutions to our quadratic equation.
For our specific quadratic equation \(x^2 - 5 = 0\), we move the constant term to the other side to get \(x^2 = 5\). To find the values of x, we take the square root of both sides, resulting in \(x = \pm\sqrt{5}\). These are the solutions to our quadratic equation.
Real and Imaginary Roots
Polynomials can have both real and imaginary roots. Real roots are the values of x that make the polynomial equal to zero and are real numbers. Imaginary roots involve the imaginary unit \(i\), which is the square root of -1. In our example equation \(x^3 - x^2 - 5x + 5 = 0\), we found three real roots: x = 1, x = \( \sqrt{5} \), and x = \( -\sqrt{5} \).
There are no imaginary roots in this particular problem. Imaginary roots typically appear in pairs and can be identified when the quadratic formula results in the square root of a negative number. Understanding both real and imaginary roots is crucial because they give a complete picture of the solutions to polynomial equations.
There are no imaginary roots in this particular problem. Imaginary roots typically appear in pairs and can be identified when the quadratic formula results in the square root of a negative number. Understanding both real and imaginary roots is crucial because they give a complete picture of the solutions to polynomial equations.
