Problem 67 - Precalculus Mathematics for Calculus

Question

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) linear factors: \((x+2)(x-2)(x-(1+i\sqrt{3}))(x-(1-i\sqrt{3}))(x-(-1+i\sqrt{3}))(x-(-1-i\sqrt{3}))\)."

Step 1: Recognize the Structure of the Polynomial

Notice that the polynomial \(P(x) = x^6 - 64\) is a difference of squares because it can be rewritten as \((x^3)^2 - 8^2\). This suggests that we can factor it using the identity for the difference of squares.

Step 2: Apply the Difference of Squares Identity

Apply the difference of squares factoring identity: \(a^2 - b^2 = (a+b)(a-b)\). Here, \(a = x^3\) and \(b = 8\). Therefore,\[P(x) = (x^3 + 8)(x^3 - 8).\]

Step 3: Factor Each Cubic as Sum/Difference of Cubes

Notice that each factor is a sum or difference of cubes. 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 \(x^3 + 8\), where \(a = x\), \(b = 2\):\[x^3 + 8 = (x+2)(x^2 - 2x + 4).\]For \(x^3 - 8\), where \(a = x\), \(b = 2\):\[x^3 - 8 = (x-2)(x^2 + 2x + 4).\]

Step 4: Combine Factors for (a)

Recombine the linear and irreducible quadratic factors obtained from Step 3.Thus, the factorization of \(P(x) = x^6 - 64\) over the reals is:\[(x+2)(x^2 - 2x + 4)(x-2)(x^2 + 2x + 4).\]

Step 5: Factor Irreducible Quadratics Over Complex Numbers

Factor each irreducible quadratic by finding its complex roots, if any.For \(x^2 - 2x + 4 = 0\): The roots are \(x = 1 \pm i\sqrt{3}\).This can be factored as \((x - (1 + i\sqrt{3}))(x - (1 - i\sqrt{3}))\).For \(x^2 + 2x + 4 = 0\): The roots are \(x = -1 \pm i\sqrt{3}\).This can be factored as \((x - (-1 + i\sqrt{3}))(x - (-1 - i\sqrt{3}))\).

Step 6: Combine Factors for (b)

Combine all linear factors over the complex numbers. The complete factorization of \(P(x)\) over the complex numbers 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

Polynomials can often be simplified using special formulas. The difference of squares is one such formula. When you see something like \(a^2 - b^2\), it can always be turned into \((a+b)(a-b)\). This is because the squares make equal terms cancel when you expand.In our problem, the polynomial \(P(x) = x^6 - 64\) fits this form. Think of it as \((x^3)^2 - 8^2\). Here, \(x^3\) is like \(a\), and 8 is like \(b\). You can split it into \((x^3 + 8)(x^3 - 8)\). This is the backbone for finding polynomial factors.Why is this important? Factoring polynomials can simplify calculations and reveal roots or solutions. Recognizing the difference of squares is a powerful tool every math student should know.

Sum and Difference of Cubes

Once you've split a polynomial using the difference of squares, you might find cubes waiting to be factored. Cubes have their own special formulas. When you have something like \(a^3 + b^3\), it can be rewritten as \((a+b)(a^2-ab+b^2)\). Similarly, \(a^3 - b^3\) becomes \((a-b)(a^2+ab+b^2)\).For \(x^3 + 8\), think about it as \(x^3 + 2^3\). It's a sum of cubes with \(x\) as \(a\) and 2 as \(b\). The factorization is

• \((x+2)\)
• \((x^2 - 2x + 4)\)

For \(x^3 - 8\), consider it as \(x^3 - 2^3\). It's a difference of cubes with similar rules:

• \((x-2)\)
• \((x^2 + 2x + 4)\)

Understanding how to handle cubes helps break down complex polynomials into simpler parts, making calculations more manageable.

Complex Numbers

Complex numbers might seem intimidating at first, but they are just an extension of regular numbers. Regular numbers, or "real numbers," lie on a line – the number line. Complex numbers add a new dimension using \(i\), the imaginary unit, where \(i^2 = -1\).In our exercise, once we've factored out real components, we're left with irreducible quadratics like \(x^2 - 2x + 4\). To factor these further into linear components, we find their complex roots. Solving the equation gives us roots involving \(i\):

• For \(x^2 - 2x + 4 = 0\), roots are \(x = 1 \pm i\sqrt{3}\).

These roots allow us to write:

• \((x - (1 + i\sqrt{3}))(x - (1 - i\sqrt{3}))\)

By utilizing complex numbers, you can achieve a complete factorization of polynomials, offering insights into all possible solutions.

Irreducible Quadratic

Irreducible quadratics are polynomial expressions that cannot be factored further using real numbers. They typically appear in the form \(ax^2 + bx + c\) where no real numbers satisfy the equation when set to zero.In our exercise, quadratics like \(x^2 - 2x + 4\) and \(x^2 + 2x + 4\) remain irreducible over real numbers due to their complex roots. However, their factorization can be achieved using complex numbers by finding their two complex roots.For example, solving \(x^2 - 2x + 4 = 0\) using the quadratic formula yields complex solutions:

• \(x = 1 \pm i\sqrt{3}\)

These roots indicate that the quadratic can be written as two linear complex factors. Understanding irreducible quadratics is crucial for accurately factoring polynomials and gaining insights beyond real number limitations.