Problem 76 - A Graphical Approach to Precalculus with Limits

Question

Use the given zero to completely factor \(P(x)\) into linear factors. Zero: \(2-i ; \quad P(x)=x^{4}-4 x^{3}+9 x^{2}-16 x+20\)

Step-by-Step Solution

Answer

Factor: \((x - 2)^3 (x^2 - 4x + 5)\).

Step 1: Identify Complex Roots

Given the zero, which is a complex number in the form \(2-i\), its conjugate \(2+i\) is also a zero because polynomials with real coefficients have conjugate pairs as roots.

Step 2: Form Quadratic Factor from Complex Roots

Using the complex zeros \(2-i\) and \(2+i\), we can create a quadratic factor: \[(x-(2-i))(x-(2+i)) = (x-2+i)(x-2-i) = (x-2)^2 - (i)^2 = (x-2)^2 + 1.\]This simplifies to \((x-2)^2 + 1 = x^2 - 4x + 5\).

Step 3: Perform Polynomial Division

Divide the original polynomial \(P(x) = x^4 - 4x^3 + 9x^2 - 16x + 20\) by the quadratic factor \(x^2 - 4x + 5\) to find another factor. Use synthetic or long division to simplify the polynomial.The result of this division will be a quadratic polynomial factor.

Step 4: Find Remaining Roots

The result of the division is \(x^2 - 4x + 4\).Notice this is a perfect square trinomial that can be factored further: \((x - 2)(x - 2) = (x - 2)^2\).

Step 5: Combine All Factors

Combine all linear factors from the steps above:The complete factorization of \(P(x)\) is \[(x - (2-i))(x - (2+i))(x - 2)(x - 2),\]which simplifies to \((x - 2)^3 (x^2 -4x + 5)\), confirming all factors are linear with complex roots included.

Key Concepts

Complex Numbers

Complex numbers are numbers that consist of two parts: a real part and an imaginary part. They are written in the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit, defined by \(i^2 = -1\). These numbers are essential in algebra as they allow for the solutions of equations that have no real-number solutions.
They appear in many algebraic situations, especially when working with polynomials. For example, if you encounter a polynomial with real coefficients, finding zeros could lead you to complex numbers. In our exercise, the complex number provided is \(2-i\), which is one zero of the polynomial \(P(x)\).

Conjugate Roots

Conjugate roots are pairs of complex numbers that are interconnected. If a polynomial has a complex root \(a + bi\), its conjugate \(a - bi\) is also a root. This comes from the fact that polynomials with real coefficients must have real or conjugate pairs of complex roots.
Understanding conjugate roots is vital because they ensure all roots of a polynomial with real coefficients occur in pairs. In our example, the provided zero is \(2 - i\), which means its conjugate \(2 + i\) is also a zero. This is crucial in forming a quadratic factor as we'll see next.

Quadratic Factors

Quadratic factors are polynomial expressions of degree two, such as \(x^2 - 4x + 5\). They serve as building blocks for dividing higher-degree polynomials.
When a complex number and its conjugate are roots, they combine to create a quadratic factor with real coefficients. Using the zeros \(2-i\) and \(2+i\), we create the quadratic factor:

• Find the product \((x - (2-i))(x - (2+i))\)
• This simplifies to \((x - 2)^2 + 1 = x^2 - 4x + 5\)

This quadratic factor is essential for further dividing the polynomial to find other factors.

Polynomial Division

Polynomial division allows you to break down a large polynomial into smaller, more manageable factors. There are techniques like synthetic or long division to achieve this. In the exercise, given the polynomial \(P(x)\) is divided by the quadratic factor \(x^2 - 4x + 5\) to extract another factor.
By carefully performing the division:

• The goal is simplifying \(P(x) = x^4 - 4x^3 + 9x^2 - 16x + 20\).
• The division results in \(x^2 - 4x + 4\), a quadratic itself.
• This further factors into \((x-2)^2\), revealing remaining linear factors.

Polynomial division aids in pinpointing remaining roots and fully factorizing the polynomial into linear terms.