Problem 64 - 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. $$P(x)=x^{3}-2 x-4$$

Step-by-Step Solution

Answer

(a) \((x - 2)(x^2 + 2x + 2)\); (b) \((x - 2)(x + 1 - i)(x + 1 + i)\).

Step 1: Find Real Roots

Use the Rational Root Theorem to test possible rational roots of the polynomial \(P(x) = x^3 - 2x - 4\). The possible rational roots are the factors of the constant term (-4), which are \(\pm 1, \pm 2, \pm 4\). After testing these, you'll find that \(P(2) = 0\). Thus, \(x = 2\) is a real root.

Step 2: Perform Polynomial Division

Divide \(P(x)\) by \(x - 2\) to find the quotient, which will be a quadratic polynomial. Using synthetic or long division, divide: \[P(x) \div (x - 2) = x^2 + 2x + 2\]So, \(P(x) = (x - 2)(x^2 + 2x + 2)\).

Step 3: Factor Quadratic Completely

Next, factor the quadratic \(x^2 + 2x + 2\) using the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]where \(a = 1\), \(b = 2\), and \(c = 2\). Plug these into the formula to get: \[x = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} = \frac{-2 \pm \sqrt{-4}}{2} = -1 \pm i\]Thus, the factors are \((x + 1 - i)(x + 1 + i)\).

Step 4: Linear and Irreducible Quadratic Factors

For part (a), express \(P(x)\) in terms of linear and irreducible quadratic factors with real coefficients. The already obtained factorization is:\[P(x) = (x - 2)(x^2 + 2x + 2)\].This representation uses one linear and one irreducible quadratic factor.

Step 5: Completely Factor into Linear Factors

For part (b), the polynomial is to be completely factored into linear factors with complex coefficients. Using the factors obtained:\[P(x) = (x - 2)(x + 1 - i)(x + 1 + i)\].These are all linear factors over the complex numbers.

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a valuable tool when factoring polynomials, especially if you'd like to determine any rational roots they might have. It states that for a polynomial with integer coefficients, any rational solution, or root, would be in the form \( \frac{p}{q} \), where:

• \( p \) is a factor of the constant term.
• \( q \) is a factor of the leading coefficient.

In the polynomial \( P(x) = x^3 - 2x - 4 \), the constant term is \(-4\) and the leading coefficient is \(1\). Thus, the possible rational roots are the factors of \(-4\), which are \( \pm 1, \pm 2, \pm 4 \). We test these values by substituting them into the polynomial. If substituting the value results in zero, it is a root. For \( P(x) \), it turns out \( x = 2 \) is a root, helping us simplify further steps.

Polynomial Division

Once a root is found using the Rational Root Theorem, the next step is often to divide the polynomial by that root to simplify it further. This can be done using polynomial division. There are two main methods: synthetic and long division. Both yield the same result but have different processes.In our problem, after identifying that \( x = 2 \) is a root, we can divide \( P(x) = x^3 - 2x - 4 \) by \( x - 2 \). This division simplifies \( P(x) \) into a more manageable polynomial, \( x^2 + 2x + 2 \). Successfully dividing through this process is crucial as it sets up the next steps in factoring the polynomial further, either into more rational, integer, or even complex factors.

Complex Roots

Complex roots are essential for completely factoring certain polynomials. These roots come into play especially when dealing with quadratics that cannot be factored using real numbers alone. The quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) helps in finding these roots.In the quadratic \( x^2 + 2x + 2 \), using the formula, we get roots \( x = -1 \pm i \). These roots are not real, as indicated by the negative value under the square root (the discriminant). Instead, they are complex. When factoring the polynomial completely, these roots are represented as \((x + 1 - i)(x + 1 + i)\). Complex roots occur in conjugate pairs (e.g., \(-1 + i\) and \(-1 - i\)), which is a property of polynomials with real coefficients. Understanding and identifying complex roots is necessary for achieving a complete factorization that includes every type of root the polynomial might have.