Chapter 3: Problem 108
Find all complex zeros of each polynomial function. Give exact values. List multiple zeros as necessary. $$f(x)=4 x^{4}-65 x^{2}+16$$
Short Answer
Expert verified
The complex zeros are \[ x = 4, -4, \frac{1}{2}, -\frac{1}{2} \].
Step by step solution
01
- Identify the Form of the Polynomial
Observe that the polynomial is given by \(f(x)=4x^{4}-65x^{2}+16\). Notice that the polynomial is in quadratic form with respect to \(x^2\).
02
- Substitute to Simplify the Polynomial
To make solving easier, let \(u = x^2\), so the polynomial becomes \(4u^2 - 65u + 16\).
03
- Solve the Quadratic Equation
Use the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 4\), \(b = -65\), and \(c = 16\). Plug in these values to get: \[ u= \frac{65 \pm \sqrt{(-65)^2 - 4 \cdot 4 \cdot 16}}{2 \cdot 4} = \frac{65 \pm \sqrt{4225 - 256}}{8} = \frac{65 \pm \sqrt{3969}}{8} = \frac{65 \pm 63}{8} \] This simplifies to two solutions: \(u = \frac{128}{8} = 16\) and \(u = \frac{2}{8} = \frac{1}{4}\).
04
- Substitute Back for x and Solve
Recall that \(u = x^2\). Therefore, we solve for \(x\) in \(x^2 = 16\) and \(x^2 = \frac{1}{4}\): \[ x^2 = 16 \Rightarrow x = \pm 4 \] \(x^2 = \frac{1}{4} \Rightarrow x = \pm \frac{1}{2}\).
05
- Write the Final Answer
Collect all the solutions: \[ x = 4, -4, \frac{1}{2}, -\frac{1}{2} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
quadratic form
When dealing with polynomial equations, a quadratic form is one where we can rewrite the equation in the format of a quadratic equation. In our example, the original polynomial given was: \[ f(x) = 4x^4 - 65x^2 + 16 \] To notice the quadratic form, we observe that the powers of x are multiples of 2. Here, we rewrite the equation in terms of \(x^2\), making it easier to solve using quadratic methods. Think of it like this: substituting \(u = x^2\) transforms the polynomial into a simpler equation: \[ 4u^2 - 65u + 16 \] Recognizing quadratic forms allows us to apply familiar techniques, such as the quadratic formula, which simplifies the problem-solving process significantly.
substitution method
The substitution method is a vital tool for simplifying complex polynomial equations. By substituting \(u = x^2\), we reduce the complexity of the polynomial. Here's a step-by-step breakdown:
- First, we identify a simpler variable to replace a more complex expression. In our case, we let \(u = x^2\).
- This transforms our polynomial from \(4x^4 - 65x^2 + 16\) to \(4u^2 - 65u + 16\).
- Now, solving the quadratic equation in \(u\) form becomes straightforward.
quadratic formula
The quadratic formula is a powerful tool specifically designed to solve quadratic equations of the form \(au^2 + bu + c = 0\). The formula is: \[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our exercise, the quadratic equation derived after substitution was \(4u^2 - 65u + 16\). We identified the coefficients as:
In our exercise, the quadratic equation derived after substitution was \(4u^2 - 65u + 16\). We identified the coefficients as:
- a = 4
- b = -65
- c = 16
solving polynomial equations
Solving polynomial equations involves finding the values of the variable that satisfy the equation. In our case, here’s the comprehensive process:
- First, identify the polynomial and determine if it can be transformed (or simplified). Our example was in quadratic form, so we substituted \(u = x^2\), reducing it to \(4u^2 - 65u + 16\).
- Next, use the quadratic formula to find solutions for \(u\). We found \(u = 16\) and \(u = \frac{1}{4}\).
- Then, revert the substitution by solving \(x^2 = 16\), giving \(x = \pm 4\). For \(x^2 = \frac{1}{4}\), the solutions are \(x = \pm \frac{1}{2}\).
- Combine all roots for the final answer: \(x = 4, -4, \frac{1}{2}, -\frac{1}{2}\).