Problem 43 - Algebra and Trigonometry

Question

\(41-58=\) Find all zeros of the polynomial. $$ P(x)=x^{3}-2 x^{2}+2 x-1 $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Polynomial

We are given the polynomial \(P(x) = x^3 - 2x^2 + 2x - 1\). Our goal is to find all its zeros. This means finding the values of \(x\) such that \(P(x) = 0\).

Step 2: Apply Rational Root Theorem

To find possible rational roots, we apply the Rational Root Theorem. This suggests that any rational root, \(p/q\), is a factor of the constant term (-1) divided by a factor of the leading coefficient (1). The possible rational roots are \(\pm 1\).

Step 3: Test Possible Roots

We test the possible roots \(x = 1\) and \(x = -1\).1. For \(x = 1\): \[P(1) = 1^3 - 2(1^2) + 2(1) - 1 = 1 - 2 + 2 - 1 = 0\] Since \(P(1) = 0\), \(x = 1\) is a root.2. For \(x = -1\): \[P(-1) = (-1)^3 - 2(-1)^2 + 2(-1) - 1 = -1 - 2 - 2 - 1 = -6\] Since \(P(-1) eq 0\), \(x = -1\) is not a root.

Step 4: Perform Synthetic Division

Since \(x = 1\) is a root, we use synthetic division on \(P(x)\) divided by \(x - 1\) to find a quadratic factor.Performing synthetic division with 1 yields a quotient of \(x^2 - x + 1\) and a remainder of 0.

Step 5: Solve the Quadratic Equation

We now solve the quadratic equation \(x^2 - x + 1 = 0\) for its roots using the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]With \(a = 1\), \(b = -1\), \(c = 1\), we have:\[x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1 \pm \sqrt{-3}}{2}\]The solutions are \(x = \frac{1}{2} + \frac{i \sqrt{3}}{2}\) and \(x = \frac{1}{2} - \frac{i \sqrt{3}}{2}\).

Step 6: Conclude the Roots

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

Key Concepts

Rational Root Theorem

When solving polynomials, it's important to consider methods that help find potential rational roots. One such method is the Rational Root Theorem. This theorem suggests that any rational solution, or root, of a polynomial can be expressed as \( \frac{p}{q} \) where \( p \) is a factor of the constant term, and \( q \) is a factor of the leading coefficient.

For example, if you have a polynomial like \( P(x) = x^3 - 2x^2 + 2x - 1 \), the constant term is -1 and the leading coefficient is 1. The factors of -1 are ±1, thus the possible rational roots to consider are \( x = 1 \) and \( x = -1 \).

• First, you test these values in the polynomial equation to check which make the equation true, or equal to zero.
• In the given polynomial, testing \( x = 1 \) results in zero, proving it as a root.
• On the other hand, testing \( x = -1 \) does not satisfy the polynomial, thus eliminating it as a root.

After identifying potential rational roots using this theorem, further methods like synthetic division can simplify finding the remaining roots.

synthetic division

Once a rational root of a polynomial is found, synthetic division becomes a valuable tool for simplifying the polynomial. Synthetic division allows you to divide the polynomial by a binomial of the form \( x - r \) quickly, where \( r \) is the root already discovered.

For the polynomial \( P(x) = x^3 - 2x^2 + 2x - 1 \), having found that \( x = 1 \) is a root, we perform synthetic division with \( x - 1 \).

• The process condenses a polynomial to a lower degree: a cubic polynomial becomes quadratic.
• In our example, synthetic division of \( P(x) \) by \( x - 1 \) results in the quadratic \( x^2 - x + 1 \).
• This quotient represents the factor of the original polynomial after dividing out the known root.

Remember, synthetic division offers an efficient way to zero in on simpler equations, making it easier to find additional roots or solve for them using other methods, like the quadratic formula.

quadratic formula

The quadratic formula is a classic, essential tool when it comes to solving quadratic equations, particularly those that do not factor easily. When faced with a quadratic such as \( x^2 - x + 1 = 0 \), the formula allows us to find any potential real or complex roots.

The formula is expressed as:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, \( a \), \( b \), and \( c \) are coefficients from the quadratic equation \( ax^2 + bx + c = 0 \). Applying this with \( a = 1 \), \( b = -1 \), and \( c = 1 \), you get:\[x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{1 \pm \sqrt{-3}}{2}\]

• The term under the square root, known as the discriminant, is negative (\(-3\)), indicating complex roots.
• These complex solutions involve the imaginary unit \( i \), where \( i = \sqrt{-1} \).

The roots are \( x = \frac{1}{2} + \frac{i \sqrt{3}}{2} \) and \( x = \frac{1}{2} - \frac{i \sqrt{3}}{2} \). Comprehensive use of the quadratic formula ensures all roots, whether real or complex, are systematically found, rounding out the complete solution to the polynomial.