Problem 47 - Precalculus Mathematics for Calculus

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

Step 1: Identify Possible Rational Roots

Using the Rational Root Theorem, we determine the possible rational roots of the polynomial \( P(x) = x^3 - 2x^2 + 2x - 1 \). These are the divisors of the constant term (-1) divided by the divisors of the leading coefficient (1). Thus, the possible rational roots are \( \pm 1 \).

Step 2: Test Possible Roots

Test each of the possible rational roots by substituting them into the polynomial. 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.

Step 3: Perform Polynomial Division

Now that we've found a root, use polynomial division to divide \( P(x) \) by \( x - 1 \).\[ (x^3 - 2x^2 + 2x - 1) \div (x - 1) = x^2 - x + 1 \]This gives us the quotient \( x^2 - x + 1 \).

Step 4: Find Roots of the Quadratic Factor

Solve the quadratic \( x^2 - x + 1 = 0 \) using the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Substitute \( a = 1, b = -1, c = 1 \): \[x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1 \pm \sqrt{-3}}{2}\]Thus, the roots are \( x = \frac{1 + i \sqrt{3}}{2} \) and \( x = \frac{1 - i \sqrt{3}}{2} \).

Step 5: List All Roots

The zeros of the polynomial \( P(x) = x^3 - 2x^2 + 2x - 1 \) are \( x = 1 \), \( x = \frac{1 + i \sqrt{3}}{2} \), and \( x = \frac{1 - i \sqrt{3}}{2} \). This includes one real root and two complex conjugate roots.

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a useful tool in algebra for identifying potential rational roots of a polynomial equation. It comes in handy, especially when dealing with polynomials that have integer coefficients. The theorem states that any possible rational root of the polynomial \( P(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 \) (where all coefficients \( a_i \) are integers) is of the form \( \frac{p}{q} \) where:

• \( p \) is a factor of the constant term \( a_0 \)
• \( q \) is a factor of the leading coefficient \( a_n \)

For example, in the polynomial \( P(x) = x^3 - 2x^2 + 2x - 1 \), the constant term is \(-1\) and the leading coefficient is \(1\). Therefore, the possible rational roots are \( \pm 1 \), derived from the factors of \(-1\) over the factors of \(1\). Once identified, these potential roots can be tested by plugging them into the polynomial to see if they result in zero, confirming that they are indeed roots.

Polynomial Division

When you find a root of a polynomial, the next step often involves polynomial division. This method divides the original polynomial by a binomial of the form \( x - r \), where \( r \) is a root. The process helps simplify the polynomial by reducing its degree, making further analysis easier.In our example, we identified \( x = 1 \) as a root. Thus, we perform polynomial division of \( P(x) = x^3 - 2x^2 + 2x - 1 \) by \( x - 1 \). By dividing, we get a quotient of \( x^2 - x + 1 \). This quotient represents the remaining polynomial that holds the other roots.Polynomial division is similar to traditional long division. You divide the leading term of the dividend by the leading term of the divisor, multiply, and subtract, repeating until completion. This allows for streamlining the polynomial, facilitating the use of simpler methods, like factoring or applying formulas, to discover additional roots.

Quadratic Formula

When you have a quadratic polynomial and need to find its roots, the quadratic formula is a go-to tool. The formula is given by:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This equation gives solutions to any quadratic equation of the form \( ax^2 + bx + c = 0 \). Here:

• \( a \) is the coefficient of \( x^2 \)
• \( b \) is the coefficient of \( x \)
• \( c \) is the constant term

Applying the quadratic formula to the quotient \( x^2 - x + 1 \) from our previous step, we substitute \( a = 1, b = -1, c = 1 \) into the formula. The calculation results in:\[ x = \frac{1 \pm \sqrt{-3}}{2} \]Hence, the roots are complex and given by \( x = \frac{1 + i\sqrt{3}}{2} \) and \( x = \frac{1 - i\sqrt{3}}{2} \). These are non-real roots (specifically complex conjugates), often appearing together due to the nature of quadratic equations that lack real solutions. Understanding this formula is essential, as it systematically provides the solutions for any solvable quadratic equation.