Solve for x x^6-2x^4-4x^2+8=0

Solution:

Step 1: Factorization

Begin by factoring the equation.

Step 1.1: Group and Factor Common Terms

Group the terms as follows: $(x^6 - 2x^4) - (4x^2 - 8) = 0$

Step 1.1.1: Factor Out GCF from Each Group

Extract the greatest common factor from each group: $x^4(x^2 - 2) - 4(x^2 - 2) = 0$

Step 1.2: Factor Out the Common Binomial

Factor out the common binomial $x^2 - 2$: $(x^2 - 2)(x^4 - 4) = 0$

Step 1.3: Express $x^4$ as a Square

Rewrite $x^4$ as $(x^2)^2$: $(x^2 - 2)((x^2)^2 - 4) = 0$

Step 1.4: Express 4 as a Square

Express 4 as $2^2$: $(x^2 - 2)((x^2)^2 - 2^2) = 0$

Step 1.5: Apply Difference of Squares

Apply the difference of squares formula: $(x^2 - 2)(x^2 + 2)(x^2 - 2) = 0$

Step 1.6: Combine Like Factors

Combine the like factors: $((x^2 - 2)^2)(x^2 + 2) = 0$

Step 2: Apply Zero Product Property

Use the zero product property to set each factor equal to zero: $(x^2 - 2)^2 = 0$ and $x^2 + 2 = 0$

Step 3: Solve the First Factor

Solve $(x^2 - 2)^2 = 0$ for $x$:

Step 3.1: Take Square Root

Take the square root of both sides: $x^2 - 2 = 0$

Step 3.2: Solve for $x$

Add 2 to both sides: $x^2 = 2$

Step 3.3: Find $x$

Take the square root: $x = \pm\sqrt{2}$

Step 4: Solve the Second Factor

Solve $x^2 + 2 = 0$ for $x$:

Step 4.1: Isolate $x^2$

Subtract 2 from both sides: $x^2 = -2$

Step 4.2: Solve for $x$

Take the square root, introducing $i$: $x = \pm i\sqrt{2}$

Step 5: Combine Solutions

Combine all solutions: $x = \sqrt{2}, -\sqrt{2}, i\sqrt{2}, -i\sqrt{2}$

Knowledge Notes:

To solve the given polynomial equation $x^6 - 2x^4 - 4x^2 + 8 = 0$, we use the following concepts:

1. Factorization: The process of breaking down a complex expression into simpler factors that, when multiplied together, give the original expression.

2. Greatest Common Factor (GCF): The largest factor that divides two or more numbers. When factoring polynomials, the GCF is the highest degree of the variable that is common to all terms or the largest coefficient that is common.

3. Difference of Squares: A binomial of the form $a^2 - b^2$ can be factored into $(a + b)(a - b)$.

4. Zero Product Property: If the product of several factors is zero, then at least one of the factors must be zero. This property allows us to set each factor of a product equal to zero to find the roots of the equation.

5. Solving Quadratic Equations: When we have an equation of the form $x^2 = a$, we can solve for $x$ by taking the square root of both sides, yielding $x = \pm\sqrt{a}$.

6. Complex Numbers: When we encounter a negative number under a square root, we introduce the imaginary unit $i$, where $i^2 = -1$. The square root of a negative number can be expressed as a multiple of $i$. For example, $\sqrt{-2} = i\sqrt{2}$.

By applying these concepts, we can factor the given polynomial and solve for the variable $x$ to find all possible solutions.