Solve for x 4x^7-4x^3=0

The question is asking to find all the values of the variable x that satisfy the equation 4x^7 - 4x^3 = 0. This requires factoring the equation and using the Zero Product Property, which states that if the product of several factors is equal to zero, at least one of those factors must be zero. The solver would need to identify and solve each factor for x to find all possible solutions for the equation.

$4 x^{7} - 4 x^{3} = 0$

Answer

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Solution:

Step:1

Extract common factors from the equation.

Step:1.1

Extract $4 x^{3}$ from $4 x^{7} - 4 x^{3}$ .

Step:1.1.1

Pull out $4 x^{3}$ from $4 x^{7}$ : $4 x^{3} (x^{4}) - 4 x^{3} = 0$

Step:1.1.2

Pull out $4 x^{3}$ from $- 4 x^{3}$ : $4 x^{3} (x^{4}) + 4 x^{3} (- 1) = 0$

Step:1.1.3

Combine the factored terms: $4 x^{3} (x^{4} - 1) = 0$

Step:1.2

Express $x^{4}$ as $(x^{2})^{2}$ : $4 x^{3} ((x^{2})^{2} - 1) = 0$

Step:1.3

Represent $1$ as $1^{2}$ : $4 x^{3} ((x^{2})^{2} - 1^{2}) = 0$

Step:1.4

Apply the difference of squares formula $a^{2} - b^{2} = (a + b) (a - b)$ , where $a = x^{2}$ and $b = 1$ : $4 x^{3} (x^{2} + 1) (x^{2} - 1) = 0$

Step:1.5

Proceed to factor further.

Step:1.5.1

Simplify the expression.

Step:1.5.1.1

Rewrite $1$ as $1^{2}$ : $4 x^{3} (x^{2} + 1) (x^{2} - 1^{2}) = 0$

Step:1.5.1.2

Continue factoring.

Step:1.5.1.2.1

Apply the difference of squares formula again, where $a = x$ and $b = 1$ : $4 x^{3} (x^{2} + 1) (x + 1) (x - 1) = 0$

Step:1.5.1.2.2

Eliminate superfluous parentheses: $4 x^{3} (x^{2} + 1) (x + 1) (x - 1) = 0$

Step:1.5.2

Remove any unnecessary parentheses: $4 x^{3} (x^{2} + 1) (x + 1) (x - 1) = 0$

Step:2

Set each factor equal to $0$ to find the roots: $x^{3} = 0$ , $x^{2} + 1 = 0$ , $x + 1 = 0$ , $x - 1 = 0$

Step:3

Solve for $x$ when $x^{3} = 0$ .

Step:3.1

Equating $x^{3}$ to $0$ : $x^{3} = 0$

Step:3.2

Find $x$ by solving $x^{3} = 0$ .

Step:3.2.1

Take the cube root of both sides: $x = \sqrt[3]{0}$

Step:3.2.2

Simplify the cube root of $0$ .

Step:3.2.2.1

Express $0$ as $0^{3}$ : $x = \sqrt[3]{0^{3}}$

Step:3.2.2.2

Extract terms from under the radical: $x = 0$

Step:4

Solve for $x$ when $x^{2} + 1 = 0$ .

Step:4.1

Set $x^{2} + 1$ equal to $0$ : $x^{2} + 1 = 0$

Step:4.2

Find $x$ by solving $x^{2} + 1 = 0$ .

Step:4.2.1

Subtract $1$ from both sides: $x^{2} = - 1$

Step:4.2.2

Take the square root of both sides: $x = \pm \sqrt{- 1}$

Step:4.2.3

Represent $\sqrt{- 1}$ as $i$ : $x = \pm i$

Step:4.2.4

Combine both positive and negative solutions.

Step:4.2.4.1

Use the positive value from $\pm$ : $x = i$

Step:4.2.4.2

Use the negative value from $\pm$ : $x = - i$

Step:4.2.4.3

The complete solution includes both positive and negative values: $x = i, - i$

Step:5

Solve for $x$ when $x + 1 = 0$ .

Step:5.1

Set $x + 1$ equal to $0$ : $x + 1 = 0$

Step:5.2

Subtract $1$ from both sides: $x = - 1$

Step:6

Solve for $x$ when $x - 1 = 0$ .

Step:6.1

Set $x - 1$ equal to $0$ : $x - 1 = 0$

Step:6.2

Add $1$ to both sides: $x = 1$

Step:7

Combine all values of $x$ that satisfy the equation: $x = 0, i, - i, - 1, 1$

Knowledge Notes:

The problem-solving process involves several mathematical concepts:

Factoring: The process of breaking down an expression into a product of simpler expressions. In this case, we factored out $4 x^{3}$ from the original polynomial.
Difference of Squares: A specific factoring technique that applies when an expression can be written as $a^{2} - b^{2}$ , which factors into $(a + b) (a - b)$ . This was used twice in the solution.
Complex Numbers: Numbers of the form $a + b i$ , where $a$ and $b$ are real numbers and $i$ is the imaginary unit, satisfying $i^{2} = - 1$ . In this problem, we encounter complex solutions $i$ and $- i$ .
Roots of an Equation: The solutions to an equation where the expression equals zero. Each factor of the equation can be set to zero to find the roots.
Cube Root: The number that, when multiplied by itself three times, gives the original number. For example, $\sqrt[3]{0} = 0$ because $0 \times 0 \times 0 = 0$ .
Square Root: The number that, when multiplied by itself, gives the original number. The square root of a negative number introduces the imaginary unit $i$ .