Problem 63 - Algebra and Trigonometry

Question

\(59-64=\) 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. $$ P(x)=x^{6}-64 $$

Step-by-Step Solution

Answer

(a) \((x - 2)(x^2 + 2x + 4)(x + 2)(x^2 - 2x + 4)\); (b) \((x - 2)(x + 2)(x + 1 - i\sqrt{3})(x + 1 + i\sqrt{3})(x - 1 - i\sqrt{3})(x - 1 + i\sqrt{3})\).

Step 1: Recognize the Structure of the Polynomial

The polynomial given is \( P(x) = x^6 - 64 \). Notice that it is a difference of two terms: a squared term \((x^3)^2\) and a constant, 64, which is a square, \(8^2\). This suggests an application of the difference of squares method.

Step 2: Factor as a Difference of Squares

The expression can be rewritten and factored as a difference of squares: \[ x^6 - 64 = (x^3)^2 - 8^2 = (x^3 - 8)(x^3 + 8) \] Thus, we can decompose \(P(x)\) into two cubic factors: \(x^3 - 8\) and \(x^3 + 8\).

Step 3: Factor Each Cubic Expression as Sum and Difference of Cubes

For the two cubic factors, we apply the formulas for the sum and difference of cubes: - The difference of cubes formula is \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \). - The sum of cubes formula is \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \). Apply these:1. \(x^3 - 8 = (x - 2)(x^2 + 2x + 4)\)2. \(x^3 + 8 = (x + 2)(x^2 - 2x + 4)\)

Step 4: Identify Linear and Irreducible Quadratic Factors with Real Coefficients

Thus, we have factored \(P(x)\) using real coefficients: \( (x - 2)(x^2 + 2x + 4)(x + 2)(x^2 - 2x + 4) \). The linear factors are \(x - 2\) and \(x + 2\). The quadratic factors \(x^2 + 2x + 4\) and \(x^2 - 2x + 4\) are irreducible over the real numbers.

Step 5: Further Factor into Linear Factors with Complex Coefficients

To factor completely into complex linear factors, solve for roots of the irreducible quadratic expressions using the quadratic formula: The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). 1. For \(x^2 + 2x + 4\), \(a = 1, b = 2, c = 4\): - Discriminant \(b^2 - 4ac = 4 - 16 = -12\) leads to complex roots \(x = -1 \pm i\sqrt{3}\). 2. For \(x^2 - 2x + 4\), \(a = 1, b = -2, c = 4\): - Discriminant \(b^2 - 4ac = 4 - 16 = -12\) leads to complex roots \(x = 1 \pm i\sqrt{3}\). With these roots, the complete factorization is: \( (x-2)(x+2)(x+1-i\sqrt{3})(x+1+i\sqrt{3})(x-1-i\sqrt{3})(x-1+i\sqrt{3}) \).

Key Concepts

Difference of Squares

When dealing with polynomials, the difference of squares formula is a valuable tool for simplification. The formula is: \[ a^2 - b^2 = (a - b)(a + b) \]This means that we can express the difference between two squares as a product of two binomials. In our given polynomial, we have \( x^6 - 64 \), which can be structured as \((x^3)^2 - 8^2\).
This fits neatly into the formula, allowing us to express it as \((x^3 - 8)(x^3 + 8)\).
Using this technique reduces higher-degree polynomials into more manageable ones, making them easier to work with in subsequent steps.

Sum and Difference of Cubes

The sum and difference of cubes are special factoring techniques used when dealing with cubic terms. The formulas are:

• Difference of cubes: \(a^3 - b^3 = (a - b)(a^2 + ab + b^2) \)
• Sum of cubes: \(a^3 + b^3 = (a + b)(a^2 - ab + b^2) \)

These formulas help in breaking down cubic expressions into a linear factor and a quadratic factor. For the polynomial \(P(x) = x^6 - 64\), once factored into \((x^3 - 8)(x^3 + 8)\), we can apply these formulas. Thus, \(x^3 - 8\) becomes \((x - 2)(x^2 + 2x + 4)\), and \(x^3 + 8\) becomes \((x + 2)(x^2 - 2x + 4)\).
This step simplifies the cubic expressions into factors more suitable for further analysis or solutions.

Quadratic Formula

The quadratic formula is a universal method used to find the roots of any quadratic equation in the form \(ax^2 + bx + c = 0\). The formula is expressed as:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula is particularly useful when the quadratic cannot easily be factored by other methods. In the solution of \(x^2 + 2x + 4\) and \(x^2 - 2x + 4\), we use the quadratic formula to determine the roots. For these quadratics, the discriminant \(b^2 - 4ac\) is negative, resulting in complex roots. These roots are calculated as \(x = -1 \pm i\sqrt{3}\) for the first, and \(x = 1 \pm i\sqrt{3}\) for the second.
Applying the quadratic formula helps reveal deeper structures and solutions within polynomial equations.

Complex Roots

When a polynomial has a negative discriminant in its quadratic expression, it means the solutions are not real numbers. Instead, they are complex numbers. A complex root takes the form \(a \pm bi\), where \(i\) is the imaginary unit, defined by \(i^2 = -1\).
This is evident in the way the quadratics \(x^2 + 2x + 4\) and \(x^2 - 2x + 4\) were solved using the quadratic formula. The roots \(-1 \pm i\sqrt{3}\) and \(1 \pm i\sqrt{3}\) respectively highlight that complex roots come in conjugate pairs. These pairs are crucial for completely factoring a polynomial into linear factors.
Understanding complex roots expands the scope of solutions, ensuring comprehensiveness in finding all roots of polynomials.