Problem 22
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.) $$f(x)=x^{2}+36$$
Step-by-Step Solution
Verified Answer
-- The zeroes of the function are \(6i\) and \(-6i\); the polynomial expressed as a product of linear factors is \( (x - 6i) (x + 6i).
1Step 1: Solve for zeroes
To solve for zeroes in the polynomial function, it's necessary to solve the equation for 'x' when \(f(x) = 0\). That gives us the equation: \(x^{2} + 36 = 0\).
2Step 2: Simplify and Solve
By rearranging terms, \(x^{2} = -36\). Since it's not possible to square root a negative number in real numbers, we need to use the concept of imaginary numbers. So, \(x = \sqrt{-36}\) gives us \( x = i\sqrt{36} and x = -i\sqrt{36}\) which simplifies to \(x = 6i and x = -6i\). These are the zeroes.
3Step 3: Factorize the polynomial
We can now write the polynomial as a product of linear factors using the zeroes that we found: \(f(x) = (x - 6i)(x + 6i)\).
4Step 4: Verify graphically
If possible, one should verify the solution graphically. Plot the function \(f(x) = x^{2} + 36\) and observe for intersection points with the x-axis (the real zeroes). However, in this case, the zeroes are imaginary, so they cannot be displayed graphically on the real number plane. But with an appropriate software, it's possible to display them on a complex plane.
Key Concepts
Complex NumbersImaginary ZerosFactorizationQuadratic Functions
Complex Numbers
Complex numbers are an extension of the real numbers that can represent quantities beyond just those we encounter in everyday arithmetic. They include a real part and an imaginary part. Writing these numbers takes the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit, defined by the property \(i^2 = -1\).
In the polynomial \(x^2 + 36\), we encountered a scenario where we took the square root of a negative number, leading to an imaginary result. This operation would be impossible within the realm of real numbers.
In the polynomial \(x^2 + 36\), we encountered a scenario where we took the square root of a negative number, leading to an imaginary result. This operation would be impossible within the realm of real numbers.
- The imaginary unit \(i\) allows us to extend our number system to solve equations that have no real solutions.
- Introducing complex numbers lets us solve equations like \(x^2 = -36\) by taking the square root of both sides, resulting in solutions of \(x = 6i\) and \(x = -6i\).
Imaginary Zeros
In the world of polynomials, zeros are the values of \(x\) where the polynomial equals zero. Usually, we find these zeros by setting the polynomial function equal to zero and solving for \(x\).
In our exercise, the zeros of \(f(x) = x^2 + 36\) are not real numbers but imaginary due to the negative inside the square root. Imaginary zeros occur when we have a negative number under the square root sign, leading us to use complex numbers to express the solutions.
Imaginary zeros reveal the points where the graph of the polynomial would intersect the x-axis in a plane where the y-axis represents purely imaginary values.
In our exercise, the zeros of \(f(x) = x^2 + 36\) are not real numbers but imaginary due to the negative inside the square root. Imaginary zeros occur when we have a negative number under the square root sign, leading us to use complex numbers to express the solutions.
Imaginary zeros reveal the points where the graph of the polynomial would intersect the x-axis in a plane where the y-axis represents purely imaginary values.
- For \(f(x) = x^2 + 36\), the imaginary zeros are \(6i\) and \(-6i\).
- Such zeros indicate that the polynomial doesn’t intersect the real x-axis, making them unobservable in a standard graph.
Factorization
Factorization is a process where we express a polynomial as a product of simpler polynomials called factors. It’s a crucial algebraic technique, particularly when solving polynomial equations. In the case of quadratic polynomials, like our \(f(x) = x^2 + 36\), factorization often revolves around determining its zeros.
Given the imaginary zeros \(6i\) and \(-6i\), we can write the polynomial as a product of its linear factors:
\(f(x) = (x - 6i)(x + 6i)\).
Given the imaginary zeros \(6i\) and \(-6i\), we can write the polynomial as a product of its linear factors:
\(f(x) = (x - 6i)(x + 6i)\).
- Each factor in this context is a binomial of the form \(x - ext{zero}\).
- These factors arise from solving the equation \(f(x) = 0\), revealing where the polynomial outputs zero.
Quadratic Functions
Quadratic functions are a family of equations that form a parabola when graphed on the Cartesian plane. The general form is \(ax^2 + bx + c\), where \(a, b,\) and \(c\) are constants. These functions hold great importance in algebra due to their ability to model a wide range of phenomena.
In our example, the quadratic \(x^2 + 36\) simplifies to reveal only imaginary solutions, demonstrating that not all quadratics have real x-intercepts.
Key characteristics of quadratic functions include:
In our example, the quadratic \(x^2 + 36\) simplifies to reveal only imaginary solutions, demonstrating that not all quadratics have real x-intercepts.
Key characteristics of quadratic functions include:
- The vertex, which is the highest or lowest point of the parabola.
- The axis of symmetry, a vertical line through the vertex dividing the parabola into mirrored halves.
- The discriminant, derived from the quadratic formula, determines the nature of the roots (real or imaginary).
Other exercises in this chapter
Problem 22
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