Chapter 6: Problem 6
Find all of the zeros of each function. \(f(x)=x^{3}-4 x^{2}+6 x-4\)
Short Answer
Expert verified
The zeros of the function are \(x = 2, 1+i,\) and \(1-i\).
Step by step solution
01
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: \(f(x) = 0\).
02
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 \(\pm 1, \pm 2, \pm 4\).
03
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.
04
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 \(x^2 - 2x + 2\).
05
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 \(a=1\), \(b=-2\), \(c=2\):\[x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4\cdot1\cdot2}}{2\cdot1}\]\[x = \frac{2 \pm \sqrt{4 - 8}}{2}\]\[x = \frac{2 \pm \sqrt{-4}}{2}\]\[x = \frac{2 \pm 2i}{2}\]\[x = 1 \pm i\].
06
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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.
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 \( x - 2 \):
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.
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 \( x \): \( x = \frac{2 \pm \sqrt{-4}}{2} = 1 \pm i \).