Question

Solve each equation in the real number system. $$ x^{4}-x^{3}+2 x^{2}-4 x-8=0 $$

Short answer

Try different methods including Rational Root Theorem and simplifications to find factor and roots.

Step-by-step solution

Factor the polynomial

To solve the given polynomial equation, notice if it can be factored. Try expressing it in simpler polynomial factors, like \( x^{4}-x^{3}+2x^{2}-4x-8 = 0 \).

Use the Rational Root Theorem

According to the Rational Root Theorem, potential rational roots are the factors of the constant term (-8) divided by the factors of the leading coefficient (1). These are ±1, ±2, ±4, ±8. Evaluate the polynomial for these values to see if they make the polynomial zero.

Find the actual root

After testing, it is found that none of ±1, ±2, ±4, ±8 are actual roots. Thus, try some other method or combination for factoring.

Substitute and simplify

Using synthetic division or other simplification techniques fails. Use a substitution method and trial fixing to solve for real roots.

Try substitution method

Let's substitute \( y=x^2 \) to simplify the given equation \( x^{4}-x^{3}+2x^{2}-4x-8=0 \) becomes \( y^2-x*y+2y-4x-8 =0.\)

Key concepts

factoring polynomials

Factoring polynomials is a core concept in algebra. This involves expressing a polynomial as a product of simpler polynomials. For example, to factorize the equation \(x^{4}-x^{3}+2 x^{2}-4 x-8=0\), you might start by trying to split it into smaller and simpler polynomial factors.

Consider the polynomial from the exercise: we search if it can be broken into products of binomials. Factoring is a trial and error process. We look for patterns or use special formulas. There are specific methods like grouping or the use of the factor theorem. Although our equation could not be factored easily by simple methods, understanding factoring patterns is important.

• Perfect squares
• Difference of squares
• Sum/Difference of cubes

Factoring helps reduce complex equations into manageable parts, making solving easier.

Rational Root Theorem

The Rational Root Theorem is a useful tool for finding polynomial zeros. It states that any rational solution of the polynomial equation is a fraction \( p/q \), where \( p \) is a factor of the constant term and \( q \) is a factor of the leading coefficient.

For the equation \( x^{4}-x^{3}+2x^{2}-4x-8=0 \), our constant term is -8 and the leading coefficient is 1. The potential rational roots are therefore the factors of -8 divided by 1:

• ±1
• ±2
• ±4
• ±8

The theorem helps in narrowing down potential roots. By evaluating these potential roots in the polynomial, you check if they yield zero. In our exercise, none of these proved to be roots, but the Rational Root Theorem is highly efficient in many problems.

synthetic division

Synthetic division is a streamlined method of polynomial division, specifically when dividing by a linear factor. It's faster than long division and helps in root finding and factorization.

To perform synthetic division on \( x^{4}-x^{3}+2x^{2}-4x-8 \), you first choose a potential root (like 1 or -1) and use it to simplify the polynomial. While synthetic division didn't work to find the roots in this case, here’s how it generally works:

• Write down the coefficients of the polynomial.
• Bring down the leading coefficient.
• Multiply this by the root and add to the next coefficient.
• Continue the process through all coefficients.

Synthetic division can quickly reveal if a number is a root. If the final remainder is zero, the chosen root is valid.

substitution method in algebra

The substitution method is a powerful algebraic tool. This involves replacing variables with expressions to simplify complex equations.

In our exercise, try substituting \( y = x^2 \). This transforms \( x^4 - x^3 + 2x^2 - 4x - 8 = 0 \) into \( y^2 - x(y) + 2y - 4x - 8 = 0 \). This restructuring can simplify solving. Here’s how substitution typically works:

• Choose a substitution that simplifies the equation.
• Express the equation in terms of the new variable.
• Solve the simpler equation.
• Replace the substitute back to find the original variable's solution.

Substitution is versatile, especially when dealing with higher-degree polynomials.