Problem 6 - Algebra 2

Question

Find all of the zeros of each function. \(f(x)=x^{3}-4 x^{2}+6 x-4\)

Step-by-Step Solution

Answer

The zeros of the function are \(x = 2, 1+i,\) and \(1-i\).

Step 1: Understanding the problem

The function given is a cubic polynomial, and we need to find all its zeros. Zeros of a function are the values of x that make the function equal to zero: \(f(x) = 0\).

Step 2: Trying possible rational roots

According to the Rational Root Theorem, any rational root of the polynomial \(x^3 - 4x^2 + 6x - 4\) will be a factor of the constant term (-4) over the leading coefficient (1). The possible rational roots are \(\pm 1, \pm 2, \pm 4\).

Step 3: Substituting possible rational roots

Evaluate the polynomial \(f(x)\) at possible rational roots: - \(f(1) = 1^3 - 4 \cdot 1^2 + 6 \cdot 1 - 4 = -1\)- \(f(-1) = (-1)^3 - 4(-1)^2 + 6(-1) - 4 = -15\)- \(f(2) = 2^3 - 4 \cdot 2^2 + 6 \cdot 2 - 4 = 0\)- \(f(-2) = (-2)^3 - 4(-2)^2 + 6(-2) - 4 = -28\)- \(f(4) = 4^3 - 4 \cdot 4^2 + 6 \cdot 4 - 4 = 12\)The value \(x = 2\) is a root of the polynomial.

Step 4: Polynomial division to find other roots

Since \(x=2\) is a root, we can factor \((x-2)\) out of \(f(x) = x^3 - 4x^2 + 6x - 4\). Perform synthetic division:\[\begin{array}{r|rrr}2 & 1 & -4 & 6 & -4 \ & & 2 & -4 & 4 \\hline & 1 & -2 & 2 & 0 \\end{array}\]The quotient is \(x^2 - 2x + 2\).

Step 5: Solving the quadratic equation

The quotient \(x^2 - 2x + 2 = 0\) is a quadratic equation. Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=1\), \(b=-2\), \(c=2\):\[x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4\cdot1\cdot2}}{2\cdot1}\]\[x = \frac{2 \pm \sqrt{4 - 8}}{2}\]\[x = \frac{2 \pm \sqrt{-4}}{2}\]\[x = \frac{2 \pm 2i}{2}\]\[x = 1 \pm i\].

Step 6: Identifying all zeros

The zeros of the function \(f(x) = x^3 - 4x^2 + 6x - 4\) are \(x = 2, 1+i,\) and \(1-i\). These include one real zero and two complex conjugate zeros.

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a useful tool when dealing with polynomial equations, especially when you need to find rational roots quickly and effectively. In our problem, the polynomial is given as \( f(x) = x^3 - 4x^2 + 6x - 4 \). The theorem states that if there are any rational roots (or zeros) to this polynomial, they will be a fraction \( \frac{p}{q} \), where \( p \) divides the constant term, and \( q \) divides the leading coefficient.

For \( f(x) \), the constant term is \(-4\), and the leading coefficient is \(1\). Thus, the possible rational roots that we can test are \( \pm 1, \pm 2, \pm 4 \). By trying these values as potential roots, you can quickly identify any rational solutions without needing to do more complex algebra.

• Set the polynomial equation as \( f(x) = 0 \) and substitute each rational root to verify which value results in zero.
• In our problem, attempting \( x = 2 \), we find that \( f(2) = 0 \), therefore, \( x = 2 \) is a rational root.

This step allows us to understand where to start simplifying the polynomial by factoring out \( (x - 2) \). Using this tool saves time and narrows down the potential root candidates significantly.

Synthetic Division

Once a rational root such as \( x = 2 \) is identified, you can use synthetic division to simplify the polynomial. Synthetic division is a streamlined version of polynomial division. It helps to divide a polynomial by \( (x - c) \), where \( c \) is a known root, to find what remains after factoring out \( (x - c) \).

For \( f(x) = x^3 - 4x^2 + 6x - 4 \), we perform synthetic division by \( x - 2 \):

• Arrange the coefficients of the polynomial: \( 1, -4, 6, -4 \).
• Write \( 2 \) (the known root) to the left and carry out synthetic division.
• Drop down the leading coefficient (1), multiply by \( 2 \) and add to the next coefficient, repeating these steps.

This gives us a quotient of \( x^2 - 2x + 2 \). The output signifies a quadratic polynomial, revealing simpler components of the original cubic polynomial. We need to solve it next using the quadratic formula.

Quadratic Formula

The quadratic formula is a robust method for solving any quadratic equation of the form \( ax^2 + bx + c = 0 \). It is written as:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]For the quadratic \( x^2 - 2x + 2 = 0 \) obtained from the synthetic division of \( f(x) \), we substitute \( a = 1 \), \( b = -2 \), and \( c = 2 \) into the formula.

• First, calculate the discriminant: \( b^2 - 4ac = (-2)^2 - 4 \cdot 1 \cdot 2 = 4 - 8 = -4 \).
• The negative discriminant indicates the roots will be complex, which usually appear in conjugate pairs.
• Solve for \( x \): \( x = \frac{2 \pm \sqrt{-4}}{2} = 1 \pm i \).

Thus, the quadratic formula reveals that the remaining zeros are complex numbers \( 1 + i \) and \( 1 - i \). Along with the real root \( x = 2 \), they complete the solution for all zeros of the cubic polynomial \( f(x) \).