Question
Solve the polynomial equation by factoring and check the solutions graphically.
x cubed minus 4 x equals 0
Question content area bottom
Part 1
The solutions are
enter your response here.
(Simplify your answer. Use a comma to separate answers as needed.)
x cubed minus 4 x equals 0
Question content area bottom
Part 1
The solutions are
enter your response here.
(Simplify your answer. Use a comma to separate answers as needed.)
Answers
Answered by
GPT-4o mini
To solve the polynomial equation \(x^3 - 4x = 0\) by factoring, we can start by factoring out the common term.
1. The equation can be rewritten as:
\[
x^3 - 4x = 0
\]
Factoring out \(x\), we have:
\[
x(x^2 - 4) = 0
\]
2. Next, we further factor \(x^2 - 4\) as it is a difference of squares:
\[
x(x - 2)(x + 2) = 0
\]
3. Setting each factor to zero gives us the solutions:
- \(x = 0\)
- \(x - 2 = 0 \Rightarrow x = 2\)
- \(x + 2 = 0 \Rightarrow x = -2\)
Thus, the solutions to the equation \(x^3 - 4x = 0\) are:
\[
x = 0, 2, -2
\]
**Part 1 Answer:**
The solutions are \(0, 2, -2\).
### Graphical Check
To check the solutions graphically, you can plot the function \(y = x^3 - 4x\) and observe where it intersects the x-axis. The intersections with the x-axis correspond to the solutions of the equation.
When plotted, the graph of the function will cross the x-axis at \(x = -2\), \(x = 0\), and \(x = 2\), confirming our solutions.
1. The equation can be rewritten as:
\[
x^3 - 4x = 0
\]
Factoring out \(x\), we have:
\[
x(x^2 - 4) = 0
\]
2. Next, we further factor \(x^2 - 4\) as it is a difference of squares:
\[
x(x - 2)(x + 2) = 0
\]
3. Setting each factor to zero gives us the solutions:
- \(x = 0\)
- \(x - 2 = 0 \Rightarrow x = 2\)
- \(x + 2 = 0 \Rightarrow x = -2\)
Thus, the solutions to the equation \(x^3 - 4x = 0\) are:
\[
x = 0, 2, -2
\]
**Part 1 Answer:**
The solutions are \(0, 2, -2\).
### Graphical Check
To check the solutions graphically, you can plot the function \(y = x^3 - 4x\) and observe where it intersects the x-axis. The intersections with the x-axis correspond to the solutions of the equation.
When plotted, the graph of the function will cross the x-axis at \(x = -2\), \(x = 0\), and \(x = 2\), confirming our solutions.