Factor To Solve $x^4 - 5x^2 + 4 = 0$.

Introduction

Quartic equations are a type of polynomial equation of degree four, which can be challenging to solve. The equation $x^4 - 5x^2 + 4 = 0$ is a classic example of a quartic equation that can be factored using various techniques. In this article, we will explore the different methods to factor this equation and provide a step-by-step solution.

Understanding the Equation

The given equation is a quartic equation in the form of $ax^4 + bx^3 + cx^2 + dx + e = 0$. In this case, the equation is $x^4 - 5x^2 + 4 = 0$, where $a = 1$, $b = 0$, $c = -5$, $d = 0$, and $e = 4$. The equation has no linear terms, making it a quadratic equation in disguise.

Factoring the Equation

To factor the equation, we can use the method of substitution. Let's substitute $y = x^2$ into the equation. This will transform the equation into a quadratic equation in terms of $y$. The new equation becomes $y^2 - 5y + 4 = 0$.

Solving the Quadratic Equation

The quadratic equation $y^2 - 5y + 4 = 0$ can be factored using the method of factorization. We can write the equation as $(y - 4)(y - 1) = 0$. This gives us two possible solutions for $y$: $y = 4$ and $y = 1$.

Substituting Back

Now that we have the solutions for $y$, we can substitute back to find the solutions for $x$. Since $y = x^2$, we can write $x^2 = 4$ and $x^2 = 1$. Taking the square root of both sides, we get $x = \pm 2$ and $x = \pm 1$.

Alternative Method: Using the Rational Root Theorem

Another method to factor the equation is to use the rational root theorem. This theorem states that if a rational number $p/q$ is a root of the polynomial equation $a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 = 0$, then $p$ must be a factor of $a_0$ and $q$ must be a factor of $a_n$. In this case, the factors of $a_0 = 4$ are $\pm 1, \pm 2, \pm 4$ and the factors of $a_n = 1$ are $\pm 1$.

Finding the Rational Roots

Using the rational root theorem, we can find the possible rational roots of the equation. The possible rational roots are $\pm 1, \pm 2, \pm 4$. We can test these values by substituting them into the equation to see if any of them satisfy the equation.

Testing the Rational Roots

Let's test the rational roots by substituting them into the equation. We can start by substituting $x = 1$ into the equation. This gives us $1^4 - 5(1)^2 + 4 = 0$, which simplifies to $0 = 0$. This means that $x = 1$ is a root of the equation.

Factoring the Equation

Now that we have found one of the roots, we can factor the equation using polynomial division. We can divide the original equation by $(x - 1)$ to get the other factor. This gives us $(x - 1)(x^3 + x^2 - 4x - 4) = 0$.

Solving the Cubic Equation

The cubic equation $x^3 + x^2 - 4x - 4 = 0$ can be solved using various methods, including the method of substitution and the rational root theorem. However, in this case, we can use the method of factorization to solve the equation.

Factoring the Cubic Equation

We can factor the cubic equation $x^3 + x^2 - 4x - 4 = 0$ as $(x + 2)(x^2 - 2x - 2) = 0$. This gives us two possible solutions for $x$: $x = -2$ and $x^2 - 2x - 2 = 0$.

Solving the Quadratic Equation

The quadratic equation $x^2 - 2x - 2 = 0$ can be solved using the method of factorization. We can write the equation as $(x - 1 - \sqrt{3})(x - 1 + \sqrt{3}) = 0$. This gives us two possible solutions for $x$: $x = 1 + \sqrt{3}$ and $x = 1 - \sqrt{3}$.

Conclusion

In this article, we have explored the different methods to factor the quartic equation $x^4 - 5x^2 + 4 = 0$. We have used the method of substitution, the rational root theorem, and polynomial division to solve the equation. The solutions to the equation are $x = \pm 2$, $x = \pm 1$, and $x = 1 + \sqrt{3}$, $x = 1 - \sqrt{3}$.

Introduction

The quartic equation $x^4 - 5x^2 + 4 = 0$ is a classic example of a polynomial equation of degree four. In our previous article, we explored the different methods to factor this equation and provided a step-by-step solution. In this article, we will answer some of the frequently asked questions about the quartic equation and provide additional information to help you better understand this topic.

Q: What is a quartic equation?

A: A quartic equation is a polynomial equation of degree four, which means that the highest power of the variable is four. The general form of a quartic equation is $ax^4 + bx^3 + cx^2 + dx + e = 0$, where $a$, $b$, $c$, $d$, and $e$ are constants.

Q: How do I solve a quartic equation?

A: There are several methods to solve a quartic equation, including the method of substitution, the rational root theorem, and polynomial division. In our previous article, we used these methods to solve the equation $x^4 - 5x^2 + 4 = 0$.

Q: What is the rational root theorem?

A: The rational root theorem is a theorem that states that if a rational number $p/q$ is a root of the polynomial equation $a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 = 0$, then $p$ must be a factor of $a_0$ and $q$ must be a factor of $a_n$.

Q: How do I use the rational root theorem to solve a quartic equation?

A: To use the rational root theorem to solve a quartic equation, you need to find the possible rational roots of the equation. You can do this by listing the factors of the constant term $a_0$ and the leading coefficient $a_n$. Then, you can test these values by substituting them into the equation to see if any of them satisfy the equation.

Q: What is polynomial division?

A: Polynomial division is a method of dividing one polynomial by another to find the quotient and remainder. In the context of solving a quartic equation, polynomial division can be used to divide the original equation by a factor to find the other factor.

Q: How do I use polynomial division to solve a quartic equation?

A: To use polynomial division to solve a quartic equation, you need to divide the original equation by a factor to find the other factor. You can do this by using long division or synthetic division.

Q: What are the solutions to the quartic equation $x^4 - 5x^2 + 4 = 0$?

A: The solutions to the quartic equation $x^4 - 5x^2 + 4 = 0$ are $x = \pm 2$, $x = \pm 1$, and $x = 1 + \sqrt{3}$, $x = 1 - \sqrt{3}$.

Q: How do I verify the solutions to a quartic equation?

A: To verify the solutions to a quartic equation, you need to substitute each solution into the original equation to see if it satisfies the equation. If the solution satisfies the equation, then it is a valid solution.

Conclusion

In this article, we have answered some of the frequently asked questions about the quartic equation and provided additional information to help you better understand this topic. We hope that this article has been helpful in clarifying any confusion you may have had about the quartic equation.

Additional Resources

If you are interested in learning more about the quartic equation, we recommend the following resources:

• Textbooks: "Algebra" by Michael Artin, "Calculus" by Michael Spivak
• Online Resources: Khan Academy, MIT OpenCourseWare, Wolfram Alpha
• Software: Mathematica, Maple, MATLAB

We hope that this article has been helpful in providing you with a better understanding of the quartic equation. If you have any further questions or need additional help, please don't hesitate to ask.