Problem 19 - 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.) $$f(x)=x^{2}-12 x+26$$

Step-by-Step Solution

Answer

The zeros of the function \( f(x)=x^{2}-12 x+26 \) are \( x = 6 \pm \sqrt{68}i \). The polynomial as a product of linear factors is written as \( (x - 6 - \sqrt{68}i)(x - 6 + \sqrt{68}i) \).

Step 1: Set the function to zero

First off, we need to set the given function equal to zero: \(x^{2}-12 x+26 = 0\).

Step 2: Apply the quadratic formula

Using the quadratic formula, which states that if you have a quadratic equation in the form \( ax^{2} + bx + c = 0 \), then the solutions for \( x \) will be \(\frac{-b \pm \sqrt{b^{2}-4ac}}{2a} \). Here \( a = 1, b = -12, c = 26 \). Substitute these values into the formula to get \(x = \frac{12 \pm \sqrt{(-12)^{2}-4(1)(26)}}{2(1)} = 6 \pm \sqrt{36 - 104},\)which simplifies to \(6 \pm \sqrt{-68}.\)

Step 3: Simplify the square root

Here we notice that the square root has a negative argument, implying that we have complex solutions. We can simplify it by separating the negative sign and treating it as \( \sqrt{-1} = i \). Therefore, the solutions to the equation are \( x = 6 \pm \sqrt{68}i.\)

Step 4: Product of linear factors

The polynomial as a product of linear factors can be written as \( x^{2}-12 x+26 = (x - 6 - \sqrt{68}i)(x - 6 + \sqrt{68}i) \).

Step 5: Check graphically

You can plot the function using a graphing utility to verify the results. However, keep in mind that the x-intercepts of the graph won't be visible, as the solutions are complex.

Key Concepts

Quadratic Formula

Understanding the quadratic formula is paramount when dealing with polynomial functions, especially when finding zeros. The general form of a quadratic equation is \( ax^2 + bx + c = 0 \), and the quadratic formula is the solution to this equation: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

When applying the quadratic formula, it is key to identify the coefficients \( a \), \( b\), and \( c \) from the equation and then to calculate the discriminant, \( b^2 - 4ac \). The discriminant can tell you the nature of the roots – if it's positive, there are two distinct real roots; if zero, there is one repeated real root; and if negative, the solutions are not real but complex. In our example, \( f(x) = x^2-12x+26 \), by setting \(a=1\), \(b=-12\), and \(c=26\), we were able to find the complex roots of the equation using the quadratic formula.

Complex Solutions

When the discriminant of a quadratic equation is negative, the solutions are complex. Complex solutions always come in pairs, known as complex conjugates. For the equation \( ax^2+bx+c=0 \), if the discriminant \( b^2-4ac < 0 \) indicates a negative value under the square root, then the solutions take the form \( a \pm bi \) where \( i \) is the imaginary unit representing \( \sqrt{-1}\).

In the given example with \( f(x) = x^{2} - 12x + 26 = 0 \), the simplification under the square root led to \( \sqrt{-68} \), which we rewrite using \( i \) as \( 6 \pm \sqrt{68}i \). These complex solutions may not correspond to x-intercepts on a graph, and they are crucial for understanding the behavior of polynomial functions that do not cross the x-axis.

Linear Factors of Polynomials

A polynomial function can be expressed as a product of linear factors, which are polynomials of the first degree. To find the linear factors of a polynomial, you need to find the roots or zeros of the polynomial function. Once these zeros are known, each zero, say \( r \), can be transformed into a linear factor of the form \( x - r \).

For real zeros, the linear factors correspond to the crossing points on the x-axis of the graph of the polynomial function. However, for complex zeros, such as the ones we found in our example \( x = 6 \pm \sqrt{68}i \), the linear factors \( (x - 6 - \sqrt{68}i) \) and \( (x - 6 + \sqrt{68}i) \) do not represent x-intercepts, as they do not correspond to points on the real number line. Instead, they reflect the fact that the polynomial has factors that when multiplied together, will give the original polynomial, indicating its roots in the complex plane.