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

Solution:

Step:1

Extract common factors from the equation.

Step:1.1

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

Step:1.1.1

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

Step:1.1.2

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

Step:1.1.3

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

Step:1.2

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

Step:1.3

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

Step:1.4

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

Step:1.5

Proceed to factor further.

Step:1.5.1

Simplify the expression.

Step:1.5.1.1

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

Step:1.5.1.2

Continue factoring.

Step:1.5.1.2.1

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

Step:1.5.1.2.2

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

Step:1.5.2

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

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:

1. Factoring: The process of breaking down an expression into a product of simpler expressions. In this case, we factored out $4x^{3}$ from the original polynomial.

2. 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.

3. Complex Numbers: Numbers of the form $a + bi$, 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 $-i$.

4. 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.

5. Cube Root: The number that, when multiplied by itself three times, gives the original number. For example, $\sqrt[3]{0} = 0$ because $0 \times 0 \times 0 = 0$.

6. Square Root: The number that, when multiplied by itself, gives the original number. The square root of a negative number introduces the imaginary unit $i$.