Problem 71 Factor each polynomial. The vari... [FREE SOLUTION]

Chapter 5: Problem 71

Factor each polynomial. The variables used as exponents represent positive integers. \(x^{8}-x^{4}-6\)

Short Answer

Expert verified

(x^4 - 3)(x^4 + 2)

Step by step solution

- Recognize Quadratic Form

Identify that the polynomial can be treated as a quadratic equation in terms of a variable substitution. Let鈥檚 use substitution: let \(y = x^4\), then we rewrite the polynomial as \(y^2 - y - 6\).

- Factor the Quadratic Equation

Factor the quadratic equation \(y^2 - y - 6\). We need to find two numbers that multiply to -6 and add up to -1. These numbers are -3 and 2. Thus, we can factor the polynomial as \((y - 3)(y + 2)\).

- Substitute Back

Replace \(y\) with \(x^4\) in the factored form \( (y - 3)(y + 2) \). This gives \((x^4 - 3)(x^4 + 2)\).

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Key Concepts

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

Quadratic Form

Understanding the quadratic form is crucial in polynomial factoring. A polynomial is said to be in quadratic form if it can be represented as ax^2 + bx + cgiven some substitution of variables. In our exercise, we recognized that the polynomial x^8 - x^4 - 6resembles a quadratic form in terms of a different variable, making it easier to factor. Substituting y = x^4transformed it to a familiar quadratic equation:y^2 - y - 6This substitution allowed us to apply our knowledge of quadratic equations to solve the problem effectively.

Variable Substitution

Variable substitution is a powerful technique in algebra. It simplifies complex polynomials by transforming them into more familiar forms. In our exercise, substituting y = x^4was key to simplifying x^8 - x^4 - 6. After substitution, the polynomial becomes y^2 - y - 6which is much easier to factor. This transformation helps because most students are familiar with factoring simple quadratics. By substituting back y = x^4after factoring, we return to the original variable, providing the factored form of the original polynomial.

Factoring Quadratics

Factoring quadratic equations is a fundamental technique. To factor y^2 - y - 6,we look for two numbers that multiply to -6 and add up to -1. These numbers are -3 and 2. Thus, the equation factors into:(y - 3)(y + 2).Then, we substitute back using our earlier variable substitution y = x^4.Replacing ywith x^4.in our factors gives us the final factored form of the original polynomial:(x^4 - 3)(x^4 + 2).Understanding the process of identifying factors based on multiplication and addition properties is essential. It simplifies polynomials and solves equations efficiently.

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91影视

Factor each polynomial. The variables used as exponents represent positive integers. \(x^{8}-x^{4}-6\)

Short Answer

Step by step solution

- Recognize Quadratic Form

- Factor the Quadratic Equation

- Substitute Back

Key Concepts

Quadratic Form

Variable Substitution

Factoring Quadratics

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