Problem 28 - Algebra and Trigonometry

Question

Factor the polynomial completely, and find all its zeros. State the multiplicity of each zero. \(P(x)=x^{6}-729\)

Step-by-Step Solution

Answer

Factors: \((x - 3)(x + 3)(x^2 + 3x + 9)(x^2 - 3x + 9)\). Real zeros: \(x = 3\), \(x = -3\); complex zeros have multiplicity 1.

Step 1: Identify the Structure

The polynomial given is \(P(x) = x^6 - 729\). Notice that it is a difference of two powers: \(x^6\) and \(9^6\). This is a difference of squares structure: \((a^3)^2 - (b^3)^2\).

Step 2: Apply the Difference of Squares Formula

The formula for factoring a difference of squares is \((A^2 - B^2) = (A - B)(A + B)\). Here, \(A = x^3\) and \(B = 9\), so applying the formula gives: \((x^3 - 9)(x^3 + 9)\).

Step 3: Factor Each Term Further

Each of the terms \(x^3 - 9\) and \(x^3 + 9\) can be factored using the sum and difference of cubes formulas. First, factor \(x^3 - 9 = (x - 3)(x^2 + 3x + 9)\). For \(x^3 + 9 = (x + 3)(x^2 - 3x + 9)\).

Step 4: Verify Complete Factoring

Combine the factors obtained: \((x - 3)(x^2 + 3x + 9)(x + 3)(x^2 - 3x + 9)\) is the complete factorization of \(P(x)\). Each quadratic term is not factorable further over the real numbers.

Step 5: Solve for Real and Complex Zeros

Solve each factor for zeros. \(x - 3 = 0\) gives root \(x = 3\); \(x + 3 = 0\) gives root \(x = -3\). For \(x^2 + 3x + 9\) and \(x^2 - 3x + 9\), use the quadratic formula: For \(x^2 + 3x + 9\), calculate \(x = \frac{-3 \pm \sqrt{3^2 - 4 \times 1 \times 9}}{2 \times 1}\), and for \(x^2 - 3x + 9\), \(x = \frac{3 \pm \sqrt{-27}}{2}\). These give complex roots.

Step 6: State the Zeros and Their Multiplicities

Zeros are \(x = 3\) and \(x = -3\), each with multiplicity 1, and the complex zeros extracted earlier from \((x^2 + 3x + 9)\) and \((x^2 - 3x + 9)\) also have multiplicity 1.

Key Concepts

Difference of Squares

The difference of squares is a powerful algebraic tool used to factorize expressions of the form \(A^2 - B^2\). This formula allows us to break down large expressions into simpler components. The key idea is that any expression which can be written as a square minus another square can be factored into

• \((A - B)(A + B)\)

In our original problem, \(P(x) = x^6 - 729\), we identified this as a difference of squares by recognizing \(x^6\) as

• \((x^3)^2\)

and \(729\) is

• \((9)^2\)

Thus, it fit perfectly into the difference of squares form, setting \(A = x^3\) and \(B = 9\). This gave us the factorization

• \((x^3 - 9)(x^3 + 9)\)

ensuring the polynomial is expressed as a product of smaller terms which can further be simplified.

Difference of Cubes

Factorization involving cubes can often be tricky, but is manageable with the right approach. For expressions like \(A^3 - B^3\) or \(A^3 + B^3\), we use:

• Difference of cubes: \(A^3 - B^3 = (A - B)(A^2 + AB + B^2)\)
• Sum of cubes: \(A^3 + B^3 = (A + B)(A^2 - AB + B^2)\)

Returning to our polynomial, once factored as \((x^3 - 9)(x^3 + 9)\), each term is still a cube expression. For \(x^3 - 9\), use the difference of cubes formula:

• \(A = x\), \(B = 3\)

This results in \((x - 3)(x^2 + 3x + 9)\). Similarly, \(x^3 + 9\) is a sum of cubes with:

• \((x + 3)(x^2 - 3x + 9)\)

These steps yield polynomials factored down to the cubic levels.

Complex Roots

Identifying complex roots is crucial for polynomials that do not factorize fully over the real numbers. In our case, after breaking down the polynomial, some quadratic factors remain, specifically

• \(x^2 + 3x + 9\)
• \(x^2 - 3x + 9\)

These do not have real roots and require solving to find complex solutions.
For each quadratic equation, you solve using the quadratic formula,

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Here, the discriminants ( \(b^2 - 4ac\)) are negative, leading to imaginary terms under the square root. Thus, the solutions will include complex numbers,
which are expressed in the form of

• \(a \pm bi\)

where \(i\) is the imaginary unit \(\sqrt{-1}\). This indicates that these quadratic equations contribute complex roots to the polynomial's complete set of solutions.

Quadratic Formula

The quadratic formula is a versatile tool for finding roots of any quadratic equation \(ax^2 + bx + c = 0\). It states:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

This formula helps us find both real and complex roots by calculating the discriminant \(b^2 - 4ac\).

• If positive, the roots are real and distinct.
• If zero, the roots are real and identical (a double root).
• If negative, the roots are complex.

In our polynomial's factors, \(x^2 + 3x + 9\) and \(x^2 - 3x + 9\) each need this formula to find their roots. Applying it reveals that both have complex solutions, as their discriminants are negative. Understanding this method ensures we don't overlook complex solutions when real ones aren't present.