Problem 28 - Algebra 2 and Trigonometry

Question

In \(19-28 :\) a. Find \(\mathrm{f}(a)\) for each given function. b. Is \(a\) a root of the function? $$ \mathrm{f}(x)=-5 x^{3}+5 x^{2}+2 x+3 \text { and } a=\frac{3}{2} i $$

Step-by-Step Solution

Answer

\( f\left(\frac{3}{2} i \right) \neq 0 \), so \( \frac{3}{2} i \) is not a root.

Step 1: Finding the value of f(a)

We start by substituting the value of \( a = \frac{3}{2} i \) into the function \( f(x) = -5x^3 + 5x^2 + 2x + 3 \). This means we need to evaluate \( f\left(\frac{3}{2} i\right) = -5 \left(\frac{3}{2} i\right)^3 + 5 \left(\frac{3}{2} i\right)^2 + 2 \left(\frac{3}{2} i\right) + 3 \).

Step 2: Calculating each term

First, calculate \( \left(\frac{3}{2} i\right)^2 = \left(\frac{9}{4}\right) i^2 = -\frac{9}{4} \) because \( i^2 = -1 \). Then calculate \( \left(\frac{3}{2} i\right)^3 = \frac{3}{2} i \cdot \left(-\frac{9}{4}\right) = -\frac{27}{8} i \).

Step 3: Evaluating the polynomial terms

Substitute the calculated values back into the function. Compute the terms: 1. \( -5 \left(-\frac{27}{8}i\right) = \frac{135}{8} i \)2. \( 5 \left(-\frac{9}{4}\right) = -\frac{45}{4} \)3. \( 2 \left(\frac{3}{2}i\right) = 3i \)4. The constant term 3 is already given as 3.

Step 4: Summing up all terms

Combine all the terms to get:\[\frac{135}{8} i - \frac{45}{4} + 3i + 3\]First, convert all terms to have a common denominator:\[\frac{135}{8} i + \frac{24}{8} i - \frac{90}{8} + \frac{24}{8}\]Combine the like terms:\[\left(\frac{159}{8} i\right) + \left(-\frac{66}{8}\right)\]

Step 5: Final computation and conclusion

The final result after combining all terms gives us \( \frac{159}{8}i - \frac{66}{8} \). Since this is not zero, we conclude that \( f(a) eq 0 \).

Step 6: Conclusion on whether a is a root

Since \( f(a) eq 0 \), \( a = \frac{3}{2} i \) is not a root of the function \( f(x) = -5x^3 + 5x^2 + 2x + 3 \).

Key Concepts

Polynomial Evaluation

Evaluating a polynomial is all about substituting a given number, which could be real or complex, into the polynomial expression and simplifying it. In this exercise, we evaluated the polynomial function \( f(x) = -5x^3 + 5x^2 + 2x + 3 \) by replacing \( x \) with \( a = \frac{3}{2} i \). This means we're plugging the complex number \( \frac{3}{2} i \) where \( x \) is, and solving the expression.

• First, substitute the complex number into each term of the polynomial.
• Calculate each term separately, especially when dealing with powers, as these will involve multiplying the complex number by itself.
• Use the property of the imaginary unit \( i \), where \( i^2 = -1 \), to simplify calculations.

Each substituted term requires careful computation, especially because complex numbers can make calculations tricky if not handled properly. Once all terms are calculated, add them together for the final result. The calculated value helps in determining if the given \( a \) is a root of the polynomial.

Roots of Polynomials

The roots of a polynomial are the values of \( x \) that make the polynomial equal to zero. If substituting \( a \) into \( f(x) \) results in zero, then \( a \) is indeed a root of the polynomial \( f(x) \).

• Substitute \( a \) into the polynomial.
• Calculate the expression to find \( f(a) \).
• If \( f(a) = 0 \), then \( a \) is a root; otherwise, it is not.

In this case, after calculating \( f\left(\frac{3}{2} i\right) \), the polynomial did not equal zero. This meant \( \frac{3}{2} i \) was not a root. Identifying roots is crucial for understanding the behavior of polynomials, especially their graphs and solutions.

Imaginary Unit

The imaginary unit, denoted as \( i \), is a fundamental concept in complex numbers. It is defined as the square root of negative one: \( i^2 = -1 \).

• When using \( i \) in calculations, remember this squaring property.
• In expressions, whenever you find \( i^2 \), replace it with \(-1\) to simplify the computation.

In the step-by-step solution, knowing that \( i^2 = -1 \) was essential in correctly simplifying the powers of the imaginary part. Handling \( i \) correctly ensures accurate results in evaluations and confirms or refutes whether a given complex number is a root of the polynomial. The imaginary unit is not just a mathematical tool but also a gateway to more advanced topics in mathematics, involving complex analysis and signal processing.