/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 67 A polynomial P is given. (a) Fac... [FREE SOLUTION] | 91Ó°ÊÓ

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A polynomial P is given. (a) Factor P into linear and irreducible quadratic factors with real coefficients. (b) Factor P completely into linear factors with complex coefficients. \(P(x)=x^{6}-64\)

Short Answer

Expert verified
Factor as \((x-2)(x^2+2x+4)(x+2)(x^2-2x+4)\) using real coefficients, and \((x-2)(x+2)(x+1+i\sqrt{3})(x+1-i\sqrt{3})(x-1+i\sqrt{3})(x-1-i\sqrt{3})\) using complex coefficients.

Step by step solution

01

Identify the Polynomial Type

The given polynomial is of the form \( x^6 - 64 \). It is a difference of squares since \( 64 = 8^2 \), and can be expressed as \((x^3)^2 - 8^2\). This is recognizable as a difference of squares formula: \( a^2 - b^2 = (a - b)(a + b) \).
02

Factor Using Difference of Squares

Apply the difference of squares formula to factor \( (x^3)^2 - 8^2 \) as \((x^3 - 8)(x^3 + 8)\). Now the polynomial is factored into two cubic expressions.
03

Factor the Cubic Expressions

Each cubic expression \( x^3 - 8 \) and \( x^3 + 8 \) can be factored further. Recognize these as \( a^3 - b^3 \) and \( a^3 + b^3 \), where \( b = 2 \). Use the formulas:- \( a^3 - b^3 = (a-b)(a^2 + ab + b^2) \)- \( a^3 + b^3 = (a+b)(a^2 - ab + b^2) \)For \( x^3 - 8 \):\[ x^3 - 8 = (x - 2)(x^2 + 2x + 4) \]For \( x^3 + 8 \):\[ x^3 + 8 = (x + 2)(x^2 - 2x + 4) \]
04

Combine Linear and Irreducible Quadratic Factors

The polynomial \( P(x) \) is now factored into linear and irreducible quadratic factors with real coefficients as follows:\[ P(x) = (x - 2)(x^2 + 2x + 4)(x + 2)(x^2 - 2x + 4) \]
05

Factor Completely Over Complex Numbers

To factor the quadratic terms \( x^2 + 2x + 4 \) and \( x^2 - 2x + 4 \) completely into linear factors with complex coefficients, use the quadratic formula:\( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \).For \( x^2 + 2x + 4 \): \( a = 1, b = 2, c = 4 \).The roots are:\[ x = \frac{-2 \pm \sqrt{2^2 - 4\cdot1\cdot4}}{2\cdot1} = \frac{-2 \pm \sqrt{-12}}{2} = -1 \pm i\sqrt{3} \]For \( x^2 - 2x + 4 \): \( a = 1, b = -2, c = 4 \).The roots are:\[ x = \frac{2 \pm \sqrt{(-2)^2 - 4\cdot1\cdot4}}{2\cdot1} = \frac{2 \pm \sqrt{-12}}{2} = 1 \pm i\sqrt{3} \]Thus, \( P(x) \) is fully factored as:\[ P(x) = (x - 2)(x + 2)(x + 1 + i\sqrt{3})(x + 1 - i\sqrt{3})(x - 1 + i\sqrt{3})(x - 1 - i\sqrt{3}) \]

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Key Concepts

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

Difference of Squares
The difference of squares is a special polynomial pattern. It is expressed as \( a^2 - b^2 \), and it factors beautifully into two terms: \((a - b)(a + b)\). This pattern relies on the fact that squaring a term removes any negative signs, so the subtraction (difference) results in conjugate pairs that cancel out intermediate terms when expanded. This is particularly useful when dealing with polynomials like the given \( x^6 - 64 \), where \( x^6 \) and \( 64 \) are both perfect squares. Recognizing this allows us to rewrite the expression and apply the difference of squares formula. This first step simplifies complex polynomials, enabling you to break them into smaller, more manageable parts.
Cubic Expressions
When a polynomial is factored down using the difference of squares, it often results in cubic expressions, such as \( x^3 - 8 \) and \( x^3 + 8 \) in our problem. These can be further factored using formulas for sums and differences of cubes. For a cubic difference \( a^3 - b^3 \), the formula is \( (a-b)(a^2 + ab + b^2) \). For a cubic sum \( a^3 + b^3 \), it’s \( (a+b)(a^2 - ab + b^2) \). Breaking these down further helps us find even simpler linear factors and quadratic terms and reveals more about the polynomial's roots and structure.
Quadratic Formula
The quadratic formula is a core tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). It's derived from completing the square and provides a way to find roots when factoring isn't straightforward. The formula is:
  • \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \)
For quadratic expressions like \( x^2 + 2x + 4 \), the formula allows us to find complex roots. This finding is crucial, especially when factoring over complex numbers, because it gives us a means to split the quadratic terms into linear factors that may not be immediately obvious by inspection.
Complex Numbers
Complex numbers expand our ability to solve many equations that have no real solutions. They are in the form of \( a + bi \), where \( i \) is the imaginary unit with the property that \( i^2 = -1 \). In the context of the polynomial \( P(x) \), they allow us to delve deeper into factoring quadratic expressions like \( x^2 + 2x + 4 \), where the discriminant \( b^2 - 4ac \) is negative. This Standard form encompasses all possible numbers in a two-dimensional plane, making complex numbers versatile tools for thoroughly solving and factoring equations.
Linear and Quadratic Factors
Factoring a polynomial into linear and quadratic factors helps in understanding the polynomial's behavior, such as its roots and graph intersections. Linear factors are expressions like \( (x-a) \) where \( a \) is a real or complex root of the polynomial. Quadratic factors appear when further factoring isn't possible over the real numbers, as they are irreducible in that domain. This concept is crucial when transitioning between factoring over real coefficients and complex coefficients, illustrating that even higher-degree polynomials can be broken down into simpler components that are easier to analyze or solve.

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