Problem 17 - A Graphical Approach to Precalculus with Limits

Question

One or more zeros are given for each polynomial. Find all remaining zeros. P(x)=x^{4}-x^{3}+10 x^{2}-9 x+9 ; \quad 3 i \text { is a zero. }

Step-by-Step Solution

Answer

The zeros of the polynomial are \(3i, -3i, \frac{1+i\sqrt{3}}{2}, \frac{1-i\sqrt{3}}{2}\).

Step 1: Use the Complex Conjugate Root Theorem

Since the polynomial has real coefficients and a complex zero \(3i\), its conjugate \(-3i\) is also a zero. This gives us zeros at \(3i\) and \(-3i\).

Step 2: Apply the Factor Theorem

Since \(3i\) and \(-3i\) are zeros, \((x-3i)(x+3i) = x^2+9\) is a factor of the polynomial \(P(x)\).

Step 3: Polynomial Division

Divide the polynomial \(P(x)\) by the factor \(x^2+9\) to find the quotient polynomial. Performing the polynomial division gives:\[P(x) \div (x^2+9) = x^2-x+1\]which means the quotient is \(x^2-x+1\).

Step 4: Find Remaining Zeros

To find the remaining zeros, solve the equation \(x^2-x+1=0\). Use the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\]Substitute \(a=1\), \(b=-1\), and \(c=1\) to get:\[x = \frac{-(-1) \pm \sqrt{(-1)^2-4 \times 1 \times 1}}{2 \times 1} = \frac{1 \pm \sqrt{-3}}{2}\]This simplifies to:\[x = \frac{1 \pm i\sqrt{3}}{2}\]So, the remaining zeros are \(\frac{1+i\sqrt{3}}{2}\) and \(\frac{1-i\sqrt{3}}{2}\).

Key Concepts

Complex Conjugate Root Theorem

The Complex Conjugate Root Theorem is a helpful tool when working with polynomials that have real coefficients and complex roots. If a polynomial has a complex root, such as \(3i\), this theorem tells us that the conjugate of that root must also be a root. For example:

• If \(3i\) is a root, then \(-3i\) must also be a root because complex roots occur in conjugate pairs for polynomials with real coefficients.
• This means when one zero of a polynomial is a complex number, you automatically know another zero without additional calculations.

This symmetry in roots simplifies our work by helping quickly determine the structure and solutions of the polynomial equations.

Factor Theorem

The Factor Theorem is a key concept in polynomial algebra that aids in breaking down polynomials into simpler, more manageable pieces. If you know a zero of the polynomial, the Factor Theorem tells us:

• If \(c\) is a zero of the polynomial \(P(x)\), then \((x-c)\) is a factor of \(P(x)\).
• With given roots \(3i\) and \(-3i\), we can construct a quadratic factor: \((x - 3i)(x + 3i) = x^2 + 9\).
• This helps transform a higher degree polynomial into smaller parts, making further analysis or division possible.

Understanding how each zero corresponds to a factor provides a way to tackle increasingly complex polynomial equations with confidence.

Polynomial Division

Polynomial division is a fundamental process similar to long division with numbers, but instead, it involves polynomials. The goal here is to divide the polynomial \(P(x)\) by a factor, such as \(x^2 + 9\), to determine the quotient polynomial. Here's how it generally works:

• Write the polynomials in descending order of their degrees.
• Divide the first term of the dividend by the first term of the divisor, applying it to the entire divisor.
• Subtract the result from the original polynomial.
• Repeat the process with the new polynomial result until you reach a degree lower than the divisor.

In our exercise, dividing \(P(x)\) by \(x^2 + 9\) gave us the quotient \(x^2-x+1\), turning it into a more straightforward form to solve for additional zeros.

Quadratic Formula

To solve quadratic polynomials like \(x^2-x+1=0\), the Quadratic Formula is a powerful tool. It provides the roots of any quadratic equation \(ax^2 + bx + c = 0\) as:

• \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\)

Applying it to find the remaining zeros:

• Set \(a=1\), \(b=-1\), and \(c=1\).
• The discriminant here \(b^2 - 4ac\) equals \(-3\), which is negative, indicating complex roots.
• From the formula, \(x = \frac{1 \pm i\sqrt{3}}{2}\), resulting in roots \(\frac{1+i\sqrt{3}}{2}\) and \(\frac{1-i\sqrt{3}}{2}\).

This formula makes it straightforward to find roots, including complex numbers, for any quadratic equation, concluding the solution to our original polynomial problem.