Problem 20
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.) $$f(x)=x^{2}+6 x-2$$
Step-by-Step Solution
Verified Answer
The zeros of the function \(f(x) = x^2 + 6x - 2\) are \(x = -3 + sqrt(11)\) and \(x = -3 - sqrt(11)\). The polynomial written as a product of linear factors is \((x + 3 - sqrt(11))(x + 3 + sqrt(11))\).
1Step 1: Use the Quadratic Formula to Find the Roots
The quadratic formula is given by: \(x = [-b ± sqrt(b^2 - 4ac)] / 2a \) where a, b, c are coefficients of the quadratic equation. For \( f(x) = x^2 + 6x - 2 \), a = 1, b = 6, and c = -2. Substituting these values into the quadratic formula, the roots will be obtained as: \(x = [-6 ± sqrt((6^2) - 4*1*(-2))] / 2*1\)
2Step 2: Simplify the expression for the roots
Simplifying, \(x = [-6 ± sqrt(36 + 8)] / 2 = [-6 ± sqrt(44)] / 2 = [-6 ± 2 sqrt(11)] / 2 = -3 ± sqrt(11)\). Therefore, the roots of the equation \(x^2 + 6x - 2\) are \(x = -3 + sqrt(11) \) and \(x = -3 - sqrt(11)\)
3Step 3: Write the polynomial as a product of linear factors
A quadratic polynomial \( ax^2 + bx + c \) can be factored into \( a(x - r)(x - s) \), where r and s are roots of the equation. From step 2, the roots are: \( x_1 = -3 + sqrt(11) \) and \( x_2 = -3 - sqrt(11)\). Therefore, the polynomial \( x^2 + 6x - 2 \) can be written as \( (x - (-3 + sqrt(11)))(x - (-3 - sqrt(11))) \) or \( (x + 3 - sqrt(11))(x + 3 + sqrt(11))\).
4Step 4: Verification using graphing utility
In this step, graphing utility software or a calculator can be used for plotting the function \(f(x) = x^2 + 6x - 2\). By setting the function equal to zero and solving for x, the zeros of the function should match with the ones obtained from the quadratic formula
Key Concepts
Quadratic FormulaFactoring PolynomialsGraphing Utility VerificationComplex Roots
Quadratic Formula
Understanding how the quadratic formula is a crucial tool for finding the roots of polynomial equations is essential, especially for equations of the second degree. It provides a methodical approach to obtaining the zeroes of a quadratic function, which are the values for which the polynomial equals zero.
The general form of the quadratic equation is given by: \( ax^2 + bx + c = 0 \) where 'a', 'b', and 'c' are coefficient values that determine the shape and position of the parabola represented by the equation. The quadratic formula itself is: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where the symbol '±' indicates that there will be two solutions, one involving addition and the other subtraction. These solutions represent the x-intercepts of the parabola. For the function \( f(x) = x^2 + 6x - 2 \), we use '1' for a, '6' for b, and '-2' for c. Substituting these into the formula results in the roots \( -3 + \sqrt{11} \) and \( -3 - \sqrt{11} \) which are essential for understanding the polynomial's behavior.
The general form of the quadratic equation is given by: \( ax^2 + bx + c = 0 \) where 'a', 'b', and 'c' are coefficient values that determine the shape and position of the parabola represented by the equation. The quadratic formula itself is: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where the symbol '±' indicates that there will be two solutions, one involving addition and the other subtraction. These solutions represent the x-intercepts of the parabola. For the function \( f(x) = x^2 + 6x - 2 \), we use '1' for a, '6' for b, and '-2' for c. Substituting these into the formula results in the roots \( -3 + \sqrt{11} \) and \( -3 - \sqrt{11} \) which are essential for understanding the polynomial's behavior.
Factoring Polynomials
Factoring polynomials can be likened to breaking down a complex structure into simpler components. This mathematical process involves expressing the polynomial as a product of its factors, which are polynomials of lower degrees. Specifically, for quadratic polynomials, factoring aims to write them as a product of two linear factors.
To factor the function \( f(x) = x^2 + 6x - 2 \), we use the roots obtained from the Quadratic Formula. The factored form is constructed using the formula \( a(x - r)(x - s) \), where 'r' and 's' are the roots. In this case, ‘a’ is 1, and our roots are \( x_1 = -3 + \sqrt{11} \) and \( x_2 = -3 - \sqrt{11} \). Therefore, the factored form of our polynomial is \( (x - (-3 + \sqrt{11}))(x - (-3 - \sqrt{11})) \) or simplified to \( (x + 3 - \sqrt{11})(x + 3 + \sqrt{11}) \). Factoring can greatly simplify problem-solving and further demonstrate the relationships between algebraic expressions.
To factor the function \( f(x) = x^2 + 6x - 2 \), we use the roots obtained from the Quadratic Formula. The factored form is constructed using the formula \( a(x - r)(x - s) \), where 'r' and 's' are the roots. In this case, ‘a’ is 1, and our roots are \( x_1 = -3 + \sqrt{11} \) and \( x_2 = -3 - \sqrt{11} \). Therefore, the factored form of our polynomial is \( (x - (-3 + \sqrt{11}))(x - (-3 - \sqrt{11})) \) or simplified to \( (x + 3 - \sqrt{11})(x + 3 + \sqrt{11}) \). Factoring can greatly simplify problem-solving and further demonstrate the relationships between algebraic expressions.
Graphing Utility Verification
After solving equations analytically, it's advantageous to verify results with a graphing utility. This verification process gives a visual representation of polynomial functions and helps confirm the accuracy of calculated roots.
By plotting the function \( f(x) = x^2 + 6x - 2 \) using a graphing utility, students can see where the graph intersects the x-axis. These intersections represent the function's zeroes. In our case, the graph should intersect at the points corresponding to the calculated roots, \( -3 + \sqrt{11} \) and \( -3 - \sqrt{11} \). In addition to this, for functions yielding complex roots, certain graphing utilities are capable of providing illustration, although the visualization may differ as complex numbers cannot be represented on the standard Cartesian plane. When the graph is plotted and the intercepts match our analytical roots, we achieve graphical verification of our results.
By plotting the function \( f(x) = x^2 + 6x - 2 \) using a graphing utility, students can see where the graph intersects the x-axis. These intersections represent the function's zeroes. In our case, the graph should intersect at the points corresponding to the calculated roots, \( -3 + \sqrt{11} \) and \( -3 - \sqrt{11} \). In addition to this, for functions yielding complex roots, certain graphing utilities are capable of providing illustration, although the visualization may differ as complex numbers cannot be represented on the standard Cartesian plane. When the graph is plotted and the intercepts match our analytical roots, we achieve graphical verification of our results.
Complex Roots
Complex roots occur when the discriminant (the part under the square root sign) in the quadratic formula is negative, leading to square roots of negative numbers. These are not graphable on a standard xy-coordinate plane because they do not have a real-number value. Complex roots always come in conjugate pairs in the form of \( a ± bi \) where \( i \) is the imaginary unit, the square root of -1.
In cases where a polynomial has complex roots, such as when working with coefficients that produce a negative discriminant, the roots can be expressed as \( r ± si \) where \( r \) and \( s \) are real numbers and \( si \) represents the imaginary component. Although complex roots were not part of the roots for our function \( f(x) = x^2 + 6x - 2 \), it’s important to understand that they are a fundamental aspect of polynomial equations, extending the real number system to ensure that every nonconstant polynomial equation has a root.
In cases where a polynomial has complex roots, such as when working with coefficients that produce a negative discriminant, the roots can be expressed as \( r ± si \) where \( r \) and \( s \) are real numbers and \( si \) represents the imaginary component. Although complex roots were not part of the roots for our function \( f(x) = x^2 + 6x - 2 \), it’s important to understand that they are a fundamental aspect of polynomial equations, extending the real number system to ensure that every nonconstant polynomial equation has a root.
Other exercises in this chapter
Problem 20
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