Chapter 4: Problem 17
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. }
Short Answer
Expert verified
The zeros of the polynomial are \(3i, -3i, \frac{1+i\sqrt{3}}{2}, \frac{1-i\sqrt{3}}{2}\).
Step by step solution
01
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 \(-3i\).
02
Apply the Factor Theorem
Since \(3i\) and \(-3i\) are zeros, \((x-3i)(x+3i) = x^2+9\) is a factor of the polynomial \(P(x)\).
03
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 \(x^2-x+1\).
04
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 \(\frac{1-i\sqrt{3}}{2}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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.
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.
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:
- \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\)
- 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 \(\frac{1-i\sqrt{3}}{2}\).