Problem 74 - A Graphical Approach to Precalculus with Limits

Question

Use the given zero to completely factor \(P(x)\) into linear factors. Zero: \(5 i ; \quad P(x)=x^{4}-x^{3}+23 x^{2}-25 x-50\)

Step-by-Step Solution

Answer

P(x) = (x - 5i)(x + 5i)(x - 2)(x + 1)

Step 1: Identify the Given Zero and Use the Conjugate Pair

Since the given zero is a complex number, its conjugate, defining that the zero of a function expressed as a complex number in the form `a+bi`, will have conjugate as zero in the form of `a-bi`, will also be a zero of the polynomial. Therefore, if the given zero is `5i`, the zero `-5i` will also be a zero of the polynomial.

Step 2: Form the Quadratic Factor from Complex Zeros

Since we have the zeroes `5i` and `-5i`, we can form a quadratic factor using them: `(x - 5i)(x + 5i) = x^2 + 25`.

Step 3: Polynomial Division to Find Remaining Factors

Using synthetic or polynomial long division, divide the original polynomial `P(x) = x^4 - x^3 + 23x^2 - 25x - 50` by the quadratic factor `x^2 + 25` to find the other factors. The division gives the result `x^2 - x - 2` with no remainder, implying that `(x^2 + 25)` is indeed a factor of `P(x)`.

Step 4: Factor the Quadratic Result

Next, we need to factor the quadratic `x^2 - x - 2` in linear terms. By factoring, we get `(x - 2)(x + 1)`.

Step 5: Express P(x) in Linear Factors

The complete factorisation of polynomial `P(x)` becomes: `P(x) = (x - 5i)(x + 5i)(x - 2)(x + 1)`.

Key Concepts

Complex Numbers

Complex numbers are fascinating and important in mathematics for representing numbers that are not on the ordinary number line. These numbers are in the form of

• \( a + bi \) where \( a \) is the real part, and \( b \) is the imaginary part of the complex number.
• The imaginary unit \( i \) is defined with the unique property that \( i^2 = -1 \).

When dealing with polynomials, if a complex number is a zero, its conjugate is invariably also a zero. For example, for the zero \( 5i \), its conjugate \( -5i \) also plays a key role.
Understanding complex numbers and their conjugates allows you to better handle polynomial equations, especially when factoring, which is crucial for solving these equations.

Quadratic Factors

When dealing with complex numbers in polynomials, constructing quadratic factors is a crucial step. Given a complex zero, its conjugate forms another zero as mentioned. From these zeros

• For \( 5i \) and \( -5i \), the quadratic factor is \((x - 5i)(x + 5i)\).
• Expanding this results in the real-valued quadratic \(x^2 + 25\), because the imaginary parts cancel each other out.

These quadratic factors are significant because they reduce the potential complexity of polynomials by sticking to real numbers, which are easier for many students to work with. They also make polynomial division simpler, keeping calculations manageable and open to further factoring if needed. Remembering how to construct and multiply these factors is invaluable for unraveling more complicated polynomials.

Polynomial Division

Polynomial division is a powerful tool used to simplify polynomials and extract their factors. When dividing a given polynomial by another, you must use either synthetic division or polynomial long division.
This technique is similar to numerical long division but applied to algebraic expressions.

• For example, to find the factors of our specific polynomial \(P(x) = x^4 - x^3 + 23x^2 - 25x - 50\), we divide it by the quadratic \(x^2 + 25\).
• The division results in \(x^2 - x - 2\) with no remainder, confirming that \(x^2 + 25\) is indeed a factor.

Once this division step is done, focus on further factorizing the resulting polynomial, if possible. In our exercise, this subsequent polynomial divides further into linear factors \((x - 2)\) and \((x + 1)\). Polynomial division, therefore, not only simplifies the expression but also reveals additional factorable components.