Problem 65 - 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. $$P(x)=x^{4}+8 x^{2}-9$$

Step-by-Step Solution

Answer

(a) Over the reals: \((x^2 + 9)(x - 1)(x + 1)\), (b) Over the complexes: \((x - 1)(x + 1)(x + 3i)(x - 3i)\).

Step 1: Rewrite Polynomial and Identify Terms

First, we rewrite the given polynomial to identify potential patterns or strategies for factoring. The polynomial is \(P(x) = x^4 + 8x^2 - 9\). We notice that the polynomial is quadratic in form by letting \(u = x^2\), so it becomes \(u^2 + 8u - 9\).

Step 2: Factor Using Substitution

With \(u = x^2\), we now factor the quadratic \(u^2 + 8u - 9\). We look for two numbers that multiply to \(-9\) and add to \(8\). These numbers are \(9\) and \(-1\), allowing us to factor as \((u + 9)(u - 1)\). Substituting back, we get \((x^2 + 9)(x^2 - 1)\).

Step 3: Factor Completely for Real Coefficients

Next, we factor these expressions further. \(x^2 - 1\) can be factored using the difference of squares formula \((x - 1)(x + 1)\). Thus, \(P(x) = (x^2 + 9)(x - 1)(x + 1)\). The factor \(x^2 + 9\) is an irreducible quadratic over the reals.

Step 4: Factor Completely for Complex Coefficients

To factor completely in terms of linear factors over the complex numbers, we need to also factor \(x^2 + 9\). Using the identity \(x^2 + a^2 = (x + ai)(x - ai)\), we find \(x^2 + 9 = (x + 3i)(x - 3i)\). Therefore, the polynomial can be written as \((x - 1)(x + 1)(x + 3i)(x - 3i)\).

Key Concepts

Quadratic in Form

Sometimes, a higher-degree polynomial like the one we have here can be disguised as a quadratic equation. This is where we describe the polynomial in a "quadratic in form" way. To approach such polynomials, we use substitution to simplify our task. In the exercise, we see the polynomial: \[ P(x) = x^4 + 8x^2 - 9 \]This looks complex at first, but we can cleverly substitute to make it look like a simple quadratic equation. We let \(u = x^2\). This means \(x^4\) can be rewritten as \(u^2\). So, our polynomial now becomes:\[ u^2 + 8u - 9 \]This substitution helps us identify the terms and factor it like a regular quadratic, making complex problems more approachable.

Irreducible Quadratic

When we solve a polynomial, sometimes we encounter quadratics that cannot be factored further using real numbers. These are called irreducible quadratics. In this exercise, we factored part of the polynomial into an irreducible quadratic. After substituting and factoring, we reached:\[ (x^2 + 9)(x^2 - 1) \]The quadratic \(x^2 + 9\) is irreducible over the real numbers because there are no two numbers that multiply to 9 and sum up to zero. Its roots are complex, hence, we leave it as it is when factoring for real coefficients. This is an important concept to understand when dealing with real number factors.

Difference of Squares

Identifying and factoring differences of squares is a powerful technique in solving polynomials. The expression \(x^2 - 1\) in our problem perfectly showcases this method. A difference of squares is any expression that can be written in the form \(a^2 - b^2\). This formula can be factored as:\[ a^2 - b^2 = (a - b)(a + b) \]For \(x^2 - 1\), this follows directly, since:- \(a = x\)- \(b = 1\)This simplifies as:\[ x^2 - 1 = (x - 1)(x + 1) \]Recognizing and applying the difference of squares formula allows us to quickly factor parts of a polynomial, simplifying it to find the other factors.

Complex Coefficients

Factoring polynomials over the complex numbers requires acknowledging roots that include imaginary numbers. The polynomial factor we identified, \(x^2 + 9\), is irreducible over the reals but can be factored into linear terms using complex coefficients.The formula to factor \(x^2 + a^2\) using complex numbers is:\[ x^2 + a^2 = (x + ai)(x - ai) \]Here, for our example, we have:- \(a^2 = 9\), so \(a = 3\)And thus:\[ x^2 + 9 = (x + 3i)(x - 3i) \]Incorporating complex coefficients allows us to express any polynomial in terms of linear factors, improving our understanding of its complete factorization. This is particularly useful in more advanced mathematical contexts.