Problem 47 - Algebra and Trigonometry

Question

Find all zeros of the polynomial. \(P(x)=x^{3}-2 x^{2}+2 x-1\)

Step-by-Step Solution

Answer

The zeros are \(x = 1\), \(x = \frac{1 + \sqrt{3}i}{2}\), and \(x = \frac{1 - \sqrt{3}i}{2}\).

Step 1: Recognize the Polynomial Type

The polynomial given is a cubic polynomial: \(P(x) = x^3 - 2x^2 + 2x - 1\). Cubic polynomials can have up to three real roots. Our task is to find all real and possibly complex roots.

Step 2: Use the Rational Root Theorem

The Rational Root Theorem suggests that any rational solution (or root) of the polynomial equation is a factor of the constant term (-1) divided by a factor of the leading coefficient (1). Hence, the possible rational roots are \(\pm 1\).

Step 3: Test Possible Rational Roots

Let's test \(x = 1\) and \(x = -1\):For \(x = 1\): \(P(1) = 1^3 - 2(1)^2 + 2(1) - 1 = 1 - 2 + 2 - 1 = 0\).Thus, \(x = 1\) is a root.

Step 4: Perform Polynomial Division

Since \(x = 1\) is a root, we can divide \(P(x)\) by \(x-1\) using synthetic or long division. This will help us find the other roots. Perform synthetic division:```1 | 1 -2 2 -1 | 1 -1 1 ---------------- 1 -1 1 0```We obtain the quotient \(x^2 - x + 1\).

Step 5: Solve the Quadratic Equation

We now solve the quadratic equation \(x^2 - x + 1 = 0\) using the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]where \(a = 1\), \(b = -1\), and \(c = 1\). This gives:\[x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{1 \pm \sqrt{-3}}{2}\]The discriminant \(b^2 - 4ac = -3\) is negative, indicating two complex conjugate roots: \(x = \frac{1 + \sqrt{3}i}{2}\) and \(x = \frac{1 - \sqrt{3}i}{2}\).

Step 6: Compile All Roots

Based on our calculations, the zeros of the polynomial \(P(x) = x^3 - 2x^2 + 2x - 1\) are:1. \(x = 1\)2. \(x = \frac{1 + \sqrt{3}i}{2}\)3. \(x = \frac{1 - \sqrt{3}i}{2}\).

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a useful tool when dealing with polynomial equations. It provides a systematic way to find all possible rational roots of a polynomial equation. For a polynomial of the form \(a_nx^n + a_{n-1}x^{n-1} + ... + a_0\), the possible rational roots are the ratios \(\frac{p}{q}\), where \(p\) is a factor of the constant term \(a_0\) and \(q\) is a factor of the leading coefficient \(a_n\). In simpler terms:

• List all factors of the constant term.
• List all factors of the leading term's coefficient.
• Form all possible ratios \(\frac{p}{q}\).

For example, when using the Rational Root Theorem on the polynomial \(P(x) = x^3 - 2x^2 + 2x - 1\), with a constant term \(-1\) and a leading coefficient of 1, the possible rational roots are \(\pm 1\). By systematically testing these, you can begin to find the actual roots of the polynomial.

Polynomial Division

Once a root of a polynomial is identified, like in the case where \(x = 1\) is a root of \(P(x) = x^3 - 2x^2 + 2x - 1\), polynomial division provides a method to find the remaining roots. Dividing the original polynomial by the factor associated with the known root reduces the polynomial's degree, simplifying the process of finding other roots.
There are two primary methods of polynomial division:

• Synthetic Division: A streamlined process ideal for dividing polynomials when the divisor is of the form \(x - c\).
• Long Division: Similar to the division of numbers, it is a more general approach but involves more steps when dividing unlike synthetic division.

For the given polynomial, synthetic division can be used as it quickly reduces the problem to the quotient \(x^2 - x + 1\), making it easier to find additional zeros of the polynomial.

Complex Roots

Complex roots often emerge when dealing with polynomials, especially when the discriminant of a quadratic equation is negative. Complex numbers include an imaginary unit \(i\), where \(i\) is the square root of \(-1\).
In our example, after identifying one real root \(x = 1\) and simplifying the polynomial to \(x^2 - x + 1\) using polynomial division, the quadratic formula reveals the other two roots are complex. The equation is solved as follows:

• Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
• Calculate the discriminant \(b^2 - 4ac\); if negative, it indicates complex roots.
• The roots obtained are \(x = \frac{1 + \sqrt{-3}i}{2}\) and \(x = \frac{1 - \sqrt{-3}i}{2}\).

These complex roots appear as conjugates, typical in polynomial equations with real coefficients, ensuring all roots are found for the cubic polynomial.