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Chapter 5: Problem 64

Solve each equation. $$5 x^{3}=125 x$$

Short Answer

Expert verified
x = 0, 5, -5

Step by step solution

01

Simplify the Equation

Start by dividing both sides of the equation by 5: divide by 5: \[ \frac{5x^3}{5} = \frac{125x}{5} \]\[ x^3 = 25x \]
02

Set the Equation to Zero

Rearrange the equation by subtracting 25x from both sides to set it to zero: \[ x^3 - 25x = 0 \]
03

Factorize the Equation

Take out the common factor, which is x: \[ x(x^2 - 25) = 0 \]
04

Solve for x

Set each factor to zero and solve: \[ x = 0 \] For \(x^2 - 25 = 0\): Solve for x: \[ x^2 = 25 \] Take the square root of both sides: \[ x = \pm 5 \]
05

Compile the Solutions

Combine all solutions found: \( x = 0, 5, -5 \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Factoring Polynomials
Factoring polynomials is a key concept in algebra that simplifies equations and helps solve them. A polynomial like the one in the problem, x鲁 - 25x, can be factored to simplify solving. Let's break it down.
Begin by identifying the greatest common factor (GCF). For x鲁 - 25x, both terms share 'x' as a factor.
This allows us to rewrite the polynomial:
  • x鲁 - 25x = x(x虏 - 25)
Factoring can break down a polynomial into simpler terms. By breaking it down, we turn hard equations into easier-to-solve parts. Here, x is pulled out, leaving x虏 - 25 which is easier to handle.
Setting Equations to Zero
Another key mathematical strategy is setting the equation to zero. This step is essential for using several algebraic techniques, including factoring and solving quadratic equations.
In our problem, the equation was:

  • x鲁 = 25x
By subtracting 25x from both sides, we set the equation to zero:
  • x鲁 - 25x = 0
When we set an equation to zero, we prepare it for different methods like factoring, making it more manageable. Equations set to zero indicate potential solutions where the expression equals zero, revealing critical information about possible x-values.
Solving Quadratic Equations
The final step involves solving quadratic equations. Once a polynomial is factored and set to zero, solving involves finding values of 'x' that make each factor equal to zero.
Consider our factored equation:
  • x(x虏 - 25) = 0
We solve for x by setting each factor to zero:
  • x = 0
  • x虏 - 25 = 0
For x虏 - 25 = 0, we rearrange and solve like a standard quadratic equation:
  • x虏 = 25
  • x = 卤5
This yields our solutions: x = 0, 5, -5. Solving quadratic equations nets the values of x, which satisfy the conditions of the equation.

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Most popular questions from this chapter

Factor each polynomial by grouping. $$ 2 q h-h+8 q-4 $$

Work with a group to examine the following steps in the solution to \(5 x^{2}-5=0\) $$\begin{array}{r}5\left(x^{2}-1\right)=0 \\\5(x-1)(x+1)=0 \\\x-1=0 \quad \text { or } \quad x+1=0 \\\x=1 \quad \text { or } \quad x=-1\end{array}$$ What happened to the 5? Explain.

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