Problem 25 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find all the zeros of the function and write the polynomial as a product of linear factors. Use a graphing utility to verify your results graphically. (If possible, use the graphing utility to verify the imaginary zeros.) $$f(z)=z^{2}-z+56$$

Step-by-Step Solution

Answer

The zeroes are \(x = \frac{1 + i\sqrt{223}}{2}\) and \(x = \frac{1 - i\sqrt{223}}{2}\). The polynomial as a product of linear factors is \(f(z)=(z-\frac{1 + i\sqrt{223}}{2}) * (z-\frac{1 - i\sqrt{223}}{2})\).

Step 1 - Finding the zeros

Use the quadratic formula to find the roots. The quadratic formula is \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) where \(a\), \(b\), and \(c\) are the coefficients from our equation \(ax^{2} + bx + c\). Here \(a=1, b=-1, c=56\). So the roots are \(x = \frac{1\pm \sqrt{(-1)^{2} - 4*1*56}}{2*1}\) which simplifies to \(x = \frac{1 \pm \sqrt{1 - 224}}{2}\).

Step 2 - Simplifying root expression

Now simplify the above expression to find the actual roots. Because the value under the square root is negative, we know that the roots will be complex. The simplification of the root expression leads to: \(x = \frac{1\pm \sqrt{-223}}{2}\), which can be written as \(x = \frac{1\pm i\sqrt{223}}{2}\) where \(i\) is the imaginary unit. So, zeroes of \(f(z)\) are \( x = \frac{1 + i\sqrt{223}}{2}\) and \( x = \frac{1 - i\sqrt{223}}{2}\).

Step 3 - Representing Polynomial as a Product of Linear Factors

We have two roots from the step above, so the corresponding factors will be \((z-\frac{1 + i\sqrt{223}}{2})\) and \((z-\frac{1 - i\sqrt{223}}{2})\). The polynomial \(f(z)\) as a product of these factors will be: \( f(z) = (z-\frac{1 + i\sqrt{223}}{2}) * (z-\frac{1 - i\sqrt{223}}{2})\).

Step 4 - Verifying Results Graphically

Graph the function \(y=f(z)\) on a graphing utility. Due to imaginary roots, they cannot be visually represented on a real number plane graph. However, when graphing this function, there would normally be a parabolic curve if the roots were real numbers, and it would cross the x-axis at the roots of the equation.

Key Concepts

Complex Numbers

Complex numbers are numbers that consist of a real part and an imaginary part. They are usually expressed in the form of \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit. The imaginary unit is defined by the property that \(i^2 = -1\). This allows us to extend the real number system to include solutions to equations that do not have solutions in the real numbers alone. For example, the equation \(x^2 + 1 = 0\) has no real solution, because the square of any real number is non-negative. But using complex numbers, we find that \(x = i\) or \(x = -i\) are solutions.

• Complex numbers are useful in various mathematical contexts, including solving polynomials that do not have real number roots.
• In the context of quadratic equations, complex roots occur when the discriminant (the expression under the square root in the quadratic formula) is negative.
• The two complex roots in the form \(a \pm bi\) are conjugates, meaning they are mirror images along the real axis in the complex plane.

Understanding complex numbers is crucial in handling the complete set of solutions for polynomial equations, thus broadening the scope of problem-solving.

Roots of Polynomial

The roots of a polynomial are the values of the variable that make the polynomial equal to zero. For a quadratic polynomial like \(f(z) = z^2 - z + 56\), the roots are found by setting the polynomial equal to zero and solving for \(z\). These roots can be real or complex depending on the discriminant of the quadratic equation.

• If the discriminant \(b^2 - 4ac\) is positive, the quadratic has two distinct real roots.
• If it is zero, there is exactly one real root (a repeated root).
• If the discriminant is negative, the quadratic has two complex conjugate roots.

In this exercise, the discriminant was negative \(-223\), indicating that the roots are complex. They appear as conjugates \(\frac{1 \pm i\sqrt{223}}{2}\). Ultimately, understanding the roots helps in expressing the original polynomial as a product of linear factors. This transformation allows further simplification or analysis through expansion and other algebraic methods.

Quadratic Formula

The quadratic formula is a key tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\). It is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula provides the solution for any quadratic equation, granting insights into the nature of its roots. No matter the coefficients, this formula helps determine whether the roots are real or complex.

• The term \(b^2 - 4ac\) is known as the discriminant. It reveals information about the nature of the roots;
• The \(\pm\) symbol in the formula indicates that there are generally two solutions or roots.
• The quadratic formula simplifies the process of solving quadratic equations, especially ones that are not easily factorable.

By using the quadratic formula, we determined in the original exercise that the roots of \(f(z) = z^2 - z + 56\) are complex, facilitating the decomposition of the polynomial into linear factors, even when the roots are not observable on a traditional graphing plane. The quadratic formula remains an indispensable tool in algebra and calculus, ensuring the ability to solve any quadratic equation efficiently.