One Of The Solutions To The Equation ${ X^3 + X^2 - 16x - 16 = 0 }$is { -4$}$.Test The Solution To Find The Full Solution Set. Which Of The Following Statements Correctly Describes The Solution Set?A. The Equation Has One Real

Introduction

In mathematics, solving cubic equations can be a challenging task. However, with the right approach, we can find the full solution set of a given cubic equation. In this article, we will explore one of the solutions to the cubic equation $x^3 + x^2 - 16x - 16 = 0$ and test it to find the full solution set.

The Given Cubic Equation

The given cubic equation is $x^3 + x^2 - 16x - 16 = 0$. We are told that one of the solutions to this equation is $x = -4$. Our goal is to test this solution and find the full solution set of the equation.

Testing the Solution

To test the solution, we substitute $x = -4$ into the given equation:

$(-4)^3 + (-4)^2 - 16(-4) - 16 = 0$

Expanding the terms, we get:

$-64 + 16 + 64 - 16 = 0$

Simplifying the expression, we get:

$0 = 0$

Since the equation holds true for $x = -4$, we can conclude that $x = -4$ is indeed a solution to the given cubic equation.

Finding the Full Solution Set

Now that we have confirmed that $x = -4$ is a solution to the equation, we need to find the full solution set. To do this, we can use the fact that if $x = r$ is a solution to a polynomial equation, then $(x - r)$ is a factor of the polynomial.

In this case, we know that $x = -4$ is a solution to the equation, so $(x + 4)$ is a factor of the polynomial. We can use polynomial division to divide the given polynomial by $(x + 4)$ and find the other factor.

Polynomial Division

To divide the given polynomial by $(x + 4)$, we can use polynomial long division or synthetic division. Let's use polynomial long division:

$x^3 + x^2 - 16x - 16 \div (x + 4)$

Performing the division, we get:

$x^2 - 4x - 4$

So, we have found that the other factor of the polynomial is $x^2 - 4x - 4$.

Factoring the Quadratic Factor

Now that we have found the other factor, we can try to factor it further. The quadratic factor $x^2 - 4x - 4$ can be factored as:

$(x - 2)^2 - 4$

Expanding the expression, we get:

$x^2 - 4x + 4 - 4$

Simplifying the expression, we get:

$x^2 - 4x$

So, we have factored the quadratic factor as $(x - 2)^2$.

Finding the Full Solution Set

Now that we have factored the polynomial, we can find the full solution set. We know that $(x + 4)$ is a factor of the polynomial, so we can set it equal to zero to find one solution:

$x + 4 = 0$

Solving for $x$, we get:

$x = -4$

We also know that $(x - 2)^2$ is a factor of the polynomial, so we can set it equal to zero to find the other solutions:

$(x - 2)^2 = 0$

Solving for $x$, we get:

$x = 2$

Since $(x - 2)^2$ is a squared factor, we know that $x = 2$ is a repeated root.

Conclusion

In conclusion, we have found the full solution set of the given cubic equation $x^3 + x^2 - 16x - 16 = 0$. The solution set consists of three solutions: $x = -4$, $x = 2$, and $x = 2$. The equation has one real solution, $x = -4$, and two complex solutions, $x = 2$ and $x = 2$.

The Solution Set

The solution set of the equation is:

$x = -4, x = 2, x = 2$

The Correct Statement

The correct statement that describes the solution set is:

A. The equation has one real solution.

Final Thoughts

In this article, we have explored one of the solutions to the cubic equation $x^3 + x^2 - 16x - 16 = 0$ and tested it to find the full solution set. We have used polynomial division and factoring to find the other factor of the polynomial and have found that the equation has one real solution and two complex solutions.

Q&A: Frequently Asked Questions

In this section, we will answer some of the frequently asked questions related to the cubic equation $x^3 + x^2 - 16x - 16 = 0$ and its solution set.

Q: What is the given cubic equation?

A: The given cubic equation is $x^3 + x^2 - 16x - 16 = 0$.

Q: What is one of the solutions to the equation?

A: One of the solutions to the equation is $x = -4$.

Q: How did you test the solution?

A: We tested the solution by substituting $x = -4$ into the given equation and simplifying the expression. We found that the equation holds true for $x = -4$, confirming that it is indeed a solution to the equation.

Q: How did you find the full solution set?

A: We found the full solution set by using polynomial division to divide the given polynomial by $(x + 4)$ and finding the other factor. We then factored the quadratic factor and found the other solutions.

Q: What is the full solution set of the equation?

A: The full solution set of the equation is $x = -4, x = 2, x = 2$.

Q: How many real solutions does the equation have?

A: The equation has one real solution, $x = -4$.

Q: How many complex solutions does the equation have?

A: The equation has two complex solutions, $x = 2$ and $x = 2$.

Q: What is the significance of the repeated root $x = 2$?

A: The repeated root $x = 2$ indicates that the equation has a multiplicity of 2 at $x = 2$. This means that the graph of the equation touches the x-axis at $x = 2$ but does not cross it.

Q: Can you provide a visual representation of the solution set?

A: Yes, we can provide a visual representation of the solution set. The graph of the equation $x^3 + x^2 - 16x - 16 = 0$ has a single real root at $x = -4$ and two complex roots at $x = 2$ and $x = 2$. The graph touches the x-axis at $x = 2$ but does not cross it.

Q: What is the next step in solving the equation?

A: The next step in solving the equation would be to use the solution set to find the corresponding values of the function. This would involve substituting the values of $x$ into the original equation and simplifying the expression.

Q: Can you provide a summary of the solution set?

A: Yes, we can provide a summary of the solution set. The solution set of the equation $x^3 + x^2 - 16x - 16 = 0$ consists of three solutions: $x = -4, x = 2, x = 2$. The equation has one real solution, $x = -4$, and two complex solutions, $x = 2$ and $x = 2$.

Conclusion

In this article, we have explored one of the solutions to the cubic equation $x^3 + x^2 - 16x - 16 = 0$ and tested it to find the full solution set. We have used polynomial division and factoring to find the other factor of the polynomial and have found that the equation has one real solution and two complex solutions. We have also answered some of the frequently asked questions related to the equation and its solution set.