StudySmarter - The all-in-one study app.

4.8 • +11k Ratings

More than 3 Million Downloads

Free

Suggested languages for you:

Americas

Europe

Roots of Polynomials

- Calculus
- Absolute Maxima and Minima
- Absolute and Conditional Convergence
- Accumulation Function
- Accumulation Problems
- Algebraic Functions
- Alternating Series
- Antiderivatives
- Application of Derivatives
- Approximating Areas
- Arc Length of a Curve
- Area Between Two Curves
- Arithmetic Series
- Average Value of a Function
- Calculus of Parametric Curves
- Candidate Test
- Combining Differentiation Rules
- Combining Functions
- Continuity
- Continuity Over an Interval
- Convergence Tests
- Cost and Revenue
- Density and Center of Mass
- Derivative Functions
- Derivative of Exponential Function
- Derivative of Inverse Function
- Derivative of Logarithmic Functions
- Derivative of Trigonometric Functions
- Derivatives
- Derivatives and Continuity
- Derivatives and the Shape of a Graph
- Derivatives of Inverse Trigonometric Functions
- Derivatives of Polar Functions
- Derivatives of Sec, Csc and Cot
- Derivatives of Sin, Cos and Tan
- Determining Volumes by Slicing
- Direction Fields
- Disk Method
- Divergence Test
- Eliminating the Parameter
- Euler's Method
- Evaluating a Definite Integral
- Evaluation Theorem
- Exponential Functions
- Finding Limits
- Finding Limits of Specific Functions
- First Derivative Test
- Function Transformations
- General Solution of Differential Equation
- Geometric Series
- Growth Rate of Functions
- Higher-Order Derivatives
- Hydrostatic Pressure
- Hyperbolic Functions
- Implicit Differentiation Tangent Line
- Implicit Relations
- Improper Integrals
- Indefinite Integral
- Indeterminate Forms
- Initial Value Problem Differential Equations
- Integral Test
- Integrals of Exponential Functions
- Integrals of Motion
- Integrating Even and Odd Functions
- Integration Formula
- Integration Tables
- Integration Using Long Division
- Integration of Logarithmic Functions
- Integration using Inverse Trigonometric Functions
- Intermediate Value Theorem
- Inverse Trigonometric Functions
- Jump Discontinuity
- Lagrange Error Bound
- Limit Laws
- Limit of Vector Valued Function
- Limit of a Sequence
- Limits
- Limits at Infinity
- Limits at Infinity and Asymptotes
- Limits of a Function
- Linear Approximations and Differentials
- Linear Differential Equation
- Linear Functions
- Logarithmic Differentiation
- Logarithmic Functions
- Logistic Differential Equation
- Maclaurin Series
- Manipulating Functions
- Maxima and Minima
- Maxima and Minima Problems
- Mean Value Theorem for Integrals
- Models for Population Growth
- Motion Along a Line
- Motion in Space
- Natural Logarithmic Function
- Net Change Theorem
- Newton's Method
- Nonhomogeneous Differential Equation
- One-Sided Limits
- Optimization Problems
- P Series
- Particle Model Motion
- Particular Solutions to Differential Equations
- Polar Coordinates
- Polar Coordinates Functions
- Polar Curves
- Population Change
- Power Series
- Radius of Convergence
- Ratio Test
- Removable Discontinuity
- Riemann Sum
- Rolle's Theorem
- Root Test
- Second Derivative Test
- Separable Equations
- Separation of Variables
- Simpson's Rule
- Solid of Revolution
- Solutions to Differential Equations
- Surface Area of Revolution
- Symmetry of Functions
- Tangent Lines
- Taylor Polynomials
- Taylor Series
- Techniques of Integration
- The Fundamental Theorem of Calculus
- The Mean Value Theorem
- The Power Rule
- The Squeeze Theorem
- The Trapezoidal Rule
- Theorems of Continuity
- Trigonometric Substitution
- Vector Valued Function
- Vectors in Calculus
- Vectors in Space
- Washer Method
- Decision Maths
- Geometry
- 2 Dimensional Figures
- 3 Dimensional Vectors
- 3-Dimensional Figures
- Altitude
- Angles in Circles
- Arc Measures
- Area and Volume
- Area of Circles
- Area of Circular Sector
- Area of Parallelograms
- Area of Plane Figures
- Area of Rectangles
- Area of Regular Polygons
- Area of Rhombus
- Area of Trapezoid
- Area of a Kite
- Composition
- Congruence Transformations
- Congruent Triangles
- Convexity in Polygons
- Coordinate Systems
- Dilations
- Distance and Midpoints
- Equation of Circles
- Equilateral Triangles
- Figures
- Fundamentals of Geometry
- Geometric Inequalities
- Geometric Mean
- Geometric Probability
- Glide Reflections
- HL ASA and AAS
- Identity Map
- Inscribed Angles
- Isometry
- Isosceles Triangles
- Law of Cosines
- Law of Sines
- Linear Measure and Precision
- Median
- Parallel Lines Theorem
- Parallelograms
- Perpendicular Bisector
- Plane Geometry
- Polygons
- Projections
- Properties of Chords
- Proportionality Theorems
- Pythagoras Theorem
- Rectangle
- Reflection in Geometry
- Regular Polygon
- Rhombuses
- Right Triangles
- Rotations
- SSS and SAS
- Segment Length
- Similarity
- Similarity Transformations
- Special quadrilaterals
- Squares
- Surface Area of Cone
- Surface Area of Cylinder
- Surface Area of Prism
- Surface Area of Sphere
- Surface Area of a Solid
- Surface of Pyramids
- Symmetry
- Translations
- Trapezoids
- Triangle Inequalities
- Triangles
- Using Similar Polygons
- Vector Addition
- Vector Product
- Volume of Cone
- Volume of Cylinder
- Volume of Pyramid
- Volume of Solid
- Volume of Sphere
- Volume of prisms
- Mechanics Maths
- Acceleration and Time
- Acceleration and Velocity
- Angular Speed
- Assumptions
- Calculus Kinematics
- Coefficient of Friction
- Connected Particles
- Conservation of Mechanical Energy
- Constant Acceleration
- Constant Acceleration Equations
- Converting Units
- Elastic Strings and Springs
- Force as a Vector
- Kinematics
- Newton's First Law
- Newton's Law of Gravitation
- Newton's Second Law
- Newton's Third Law
- Power
- Projectiles
- Pulleys
- Resolving Forces
- Statics and Dynamics
- Tension in Strings
- Variable Acceleration
- Work Done by a Constant Force
- Probability and Statistics
- Bar Graphs
- Basic Probability
- Charts and Diagrams
- Conditional Probabilities
- Continuous and Discrete Data
- Frequency, Frequency Tables and Levels of Measurement
- Independent Events Probability
- Line Graphs
- Mean Median and Mode
- Mutually Exclusive Probabilities
- Probability Rules
- Probability of Combined Events
- Quartiles and Interquartile Range
- Systematic Listing
- Pure Maths
- ASA Theorem
- Absolute Value Equations and Inequalities
- Addition and Subtraction of Rational Expressions
- Addition, Subtraction, Multiplication and Division
- Algebra
- Algebraic Fractions
- Algebraic Notation
- Algebraic Representation
- Analyzing Graphs of Polynomials
- Angle Measure
- Angles
- Angles in Polygons
- Approximation and Estimation
- Area and Circumference of a Circle
- Area and Perimeter of Quadrilaterals
- Area of Triangles
- Argand Diagram
- Arithmetic Sequences
- Average Rate of Change
- Bijective Functions
- Binomial Expansion
- Binomial Theorem
- Chain Rule
- Circle Theorems
- Circles
- Circles Maths
- Combination of Functions
- Combinatorics
- Common Factors
- Common Multiples
- Completing the Square
- Completing the Squares
- Complex Numbers
- Composite Functions
- Composition of Functions
- Compound Interest
- Compound Units
- Conic Sections
- Construction and Loci
- Converting Metrics
- Convexity and Concavity
- Coordinate Geometry
- Coordinates in Four Quadrants
- Cubic Function Graph
- Cubic Polynomial Graphs
- Data transformations
- De Moivre's Theorem
- Deductive Reasoning
- Definite Integrals
- Deriving Equations
- Determinant of Inverse Matrix
- Determinants
- Differential Equations
- Differentiation
- Differentiation Rules
- Differentiation from First Principles
- Differentiation of Hyperbolic Functions
- Direct and Inverse proportions
- Disjoint and Overlapping Events
- Disproof by Counterexample
- Distance from a Point to a Line
- Divisibility Tests
- Double Angle and Half Angle Formulas
- Drawing Conclusions from Examples
- Ellipse
- Equation of Line in 3D
- Equation of a Perpendicular Bisector
- Equation of a circle
- Equations
- Equations and Identities
- Equations and Inequalities
- Estimation in Real Life
- Euclidean Algorithm
- Evaluating and Graphing Polynomials
- Even Functions
- Exponential Form of Complex Numbers
- Exponential Rules
- Exponentials and Logarithms
- Expression Math
- Expressions and Formulas
- Faces Edges and Vertices
- Factorials
- Factoring Polynomials
- Factoring Quadratic Equations
- Factorising expressions
- Factors
- Finding Maxima and Minima Using Derivatives
- Finding Rational Zeros
- Finding the Area
- Forms of Quadratic Functions
- Fractional Powers
- Fractional Ratio
- Fractions
- Fractions and Decimals
- Fractions and Factors
- Fractions in Expressions and Equations
- Fractions, Decimals and Percentages
- Function Basics
- Functional Analysis
- Functions
- Fundamental Counting Principle
- Fundamental Theorem of Algebra
- Generating Terms of a Sequence
- Geometric Sequence
- Gradient and Intercept
- Graphical Representation
- Graphing Rational Functions
- Graphing Trigonometric Functions
- Graphs
- Graphs and Differentiation
- Graphs of Common Functions
- Graphs of Exponents and Logarithms
- Graphs of Trigonometric Functions
- Greatest Common Divisor
- Growth and Decay
- Growth of Functions
- Highest Common Factor
- Hyperbolas
- Imaginary Unit and Polar Bijection
- Implicit differentiation
- Inductive Reasoning
- Inequalities Maths
- Infinite geometric series
- Injective functions
- Instantaneous Rate of Change
- Integers
- Integrating Polynomials
- Integrating Trig Functions
- Integrating e^x and 1/x
- Integration
- Integration Using Partial Fractions
- Integration by Parts
- Integration by Substitution
- Integration of Hyperbolic Functions
- Interest
- Inverse Hyperbolic Functions
- Inverse Matrices
- Inverse and Joint Variation
- Inverse functions
- Iterative Methods
- Law of Cosines in Algebra
- Law of Sines in Algebra
- Laws of Logs
- Limits of Accuracy
- Linear Expressions
- Linear Systems
- Linear Transformations of Matrices
- Location of Roots
- Logarithm Base
- Logic
- Lower and Upper Bounds
- Lowest Common Denominator
- Lowest Common Multiple
- Math formula
- Matrices
- Matrix Addition and Subtraction
- Matrix Determinant
- Matrix Multiplication
- Metric and Imperial Units
- Misleading Graphs
- Mixed Expressions
- Modulus Functions
- Modulus and Phase
- Multiples of Pi
- Multiplication and Division of Fractions
- Multiplicative Relationship
- Multiplying and Dividing Rational Expressions
- Natural Logarithm
- Natural Numbers
- Notation
- Number
- Number Line
- Number Systems
- Numerical Methods
- Odd functions
- Open Sentences and Identities
- Operation with Complex Numbers
- Operations with Decimals
- Operations with Matrices
- Operations with Polynomials
- Order of Operations
- Parabola
- Parallel Lines
- Parametric Differentiation
- Parametric Equations
- Parametric Integration
- Partial Fractions
- Pascal's Triangle
- Percentage
- Percentage Increase and Decrease
- Percentage as fraction or decimals
- Perimeter of a Triangle
- Permutations and Combinations
- Perpendicular Lines
- Points Lines and Planes
- Polynomial Graphs
- Polynomials
- Powers Roots And Radicals
- Powers and Exponents
- Powers and Roots
- Prime Factorization
- Prime Numbers
- Problem-solving Models and Strategies
- Product Rule
- Proof
- Proof and Mathematical Induction
- Proof by Contradiction
- Proof by Deduction
- Proof by Exhaustion
- Proof by Induction
- Properties of Exponents
- Proportion
- Proving an Identity
- Pythagorean Identities
- Quadratic Equations
- Quadratic Function Graphs
- Quadratic Graphs
- Quadratic functions
- Quadrilaterals
- Quotient Rule
- Radians
- Radical Functions
- Rates of Change
- Ratio
- Ratio Fractions
- Rational Exponents
- Rational Expressions
- Rational Functions
- Rational Numbers and Fractions
- Ratios as Fractions
- Real Numbers
- Reciprocal Graphs
- Recurrence Relation
- Recursion and Special Sequences
- Remainder and Factor Theorems
- Representation of Complex Numbers
- Rewriting Formulas and Equations
- Roots of Complex Numbers
- Roots of Polynomials
- Roots of Unity
- Rounding
- SAS Theorem
- SSS Theorem
- Scalar Triple Product
- Scale Drawings and Maps
- Scale Factors
- Scientific Notation
- Second Order Recurrence Relation
- Sector of a Circle
- Segment of a Circle
- Sequences
- Sequences and Series
- Series Maths
- Sets Math
- Similar Triangles
- Similar and Congruent Shapes
- Simple Interest
- Simplifying Fractions
- Simplifying Radicals
- Simultaneous Equations
- Sine and Cosine Rules
- Small Angle Approximation
- Solving Linear Equations
- Solving Linear Systems
- Solving Quadratic Equations
- Solving Radical Inequalities
- Solving Rational Equations
- Solving Simultaneous Equations Using Matrices
- Solving Systems of Inequalities
- Solving Trigonometric Equations
- Solving and Graphing Quadratic Equations
- Solving and Graphing Quadratic Inequalities
- Special Products
- Standard Form
- Standard Integrals
- Standard Unit
- Straight Line Graphs
- Substraction and addition of fractions
- Sum and Difference of Angles Formulas
- Sum of Natural Numbers
- Surds
- Surjective functions
- Tables and Graphs
- Tangent of a Circle
- The Quadratic Formula and the Discriminant
- Transformations
- Transformations of Graphs
- Translations of Trigonometric Functions
- Triangle Rules
- Triangle trigonometry
- Trigonometric Functions
- Trigonometric Functions of General Angles
- Trigonometric Identities
- Trigonometric Ratios
- Trigonometry
- Turning Points
- Types of Functions
- Types of Numbers
- Types of Triangles
- Unit Circle
- Units
- Variables in Algebra
- Vectors
- Verifying Trigonometric Identities
- Writing Equations
- Writing Linear Equations
- Statistics
- Bias in Experiments
- Binomial Distribution
- Binomial Hypothesis Test
- Bivariate Data
- Box Plots
- Categorical Data
- Categorical Variables
- Central Limit Theorem
- Chi Square Test for Goodness of Fit
- Chi Square Test for Homogeneity
- Chi Square Test for Independence
- Chi-Square Distribution
- Combining Random Variables
- Comparing Data
- Comparing Two Means Hypothesis Testing
- Conditional Probability
- Conducting a Study
- Conducting a Survey
- Conducting an Experiment
- Confidence Interval for Population Mean
- Confidence Interval for Population Proportion
- Confidence Interval for Slope of Regression Line
- Confidence Interval for the Difference of Two Means
- Confidence Intervals
- Correlation Math
- Cumulative Distribution Function
- Cumulative Frequency
- Data Analysis
- Data Interpretation
- Degrees of Freedom
- Discrete Random Variable
- Distributions
- Dot Plot
- Empirical Rule
- Errors in Hypothesis Testing
- Estimator Bias
- Events (Probability)
- Frequency Polygons
- Generalization and Conclusions
- Geometric Distribution
- Histograms
- Hypothesis Test for Correlation
- Hypothesis Test for Regression Slope
- Hypothesis Test of Two Population Proportions
- Hypothesis Testing
- Inference for Distributions of Categorical Data
- Inferences in Statistics
- Large Data Set
- Least Squares Linear Regression
- Linear Interpolation
- Linear Regression
- Measures of Central Tendency
- Methods of Data Collection
- Normal Distribution
- Normal Distribution Hypothesis Test
- Normal Distribution Percentile
- Paired T-Test
- Point Estimation
- Probability
- Probability Calculations
- Probability Density Function
- Probability Distribution
- Probability Generating Function
- Quantitative Variables
- Quartiles
- Random Variables
- Randomized Block Design
- Residual Sum of Squares
- Residuals
- Sample Mean
- Sample Proportion
- Sampling
- Sampling Distribution
- Scatter Graphs
- Single Variable Data
- Skewness
- Spearman's Rank Correlation Coefficient
- Standard Deviation
- Standard Error
- Standard Normal Distribution
- Statistical Graphs
- Statistical Measures
- Stem and Leaf Graph
- Sum of Independent Random Variables
- Survey Bias
- T-distribution
- Transforming Random Variables
- Tree Diagram
- Two Categorical Variables
- Two Quantitative Variables
- Type I Error
- Type II Error
- Types of Data in Statistics
- Variance for Binomial Distribution
- Venn Diagrams

Tree roots, square roots, root beer. There are many things in life that deal with roots of some kind, but here we'll be talking about roots of polynomials.

Polynomials have many uses, from modelling the curves of a rollercoaster to predicting growth patterns in economics. The roots of a polynomial are solutions where the value of the polynomial is equal to zero, and in this article we'll be covering different methods of calculating these roots using summations and recurrence relations.

This section covers the general equations for each of the three types of polynomials discussed in this article.

The **quadratic equation** takes the general form \(ax^{2} + bx + c = 0\).

It is a polynomial where the highest power of \(x\) is 2.

It is often further simplified to

\(x^{2} + \frac{b}{a}x + \frac{c}{a} = 0\),

known as the monic form of the polynomial, because the leading coefficient is equal to 1.

This equation can also be written as

\((x - x_{1})(x - x_{2}) = 0\),

where \(x_{1}\) and \(x_{2}\) are the two roots of the equation.

A **cubic equation** takes the general form \(ax^{3} + bx^{2} + cx + d = 0\).

It is a polynomial in which the highest power of \(x\) is 3.

As with the quadratic equation, is often simplified to

\(x^{3} + \frac{b}{a}x^{2} + \frac{c}{a}x + \frac{d}{a} = 0\).

It is also written as

\((x - x_{1})(x - x_{2})(x - x_{3}) = 0\),

where \(x_{1}, \; x_{2} \text{ and } x_{3}\) represent the roots of the equation.

**Quartic equations** take the general form \(ax^{4} + bx^{3} + cx^{2} + dx + e = 0\).

They are polynomials in which the highest power of \(x\) is 4.

Likewise, they can be simplified to

\(x^{4} + \frac{b}{a}x^{3} + \frac{c}{a}x^{2} + \frac{d}{a}x + \frac{e}{a} = 0\).

They are also written as

\((x - x_{1})(x - x_{2})(x - x_{3})(x - x_{4}) = 0\),

where \(x_{1}, \; x_{2}, \; x_{3} \text{ and } x_{4}\) represent the roots of equation.

Take the quadratic equation, \(ax^{2} + bx + c = 0\). Divide the equation by the coefficient of \(x^{2}, a\) to rewrite the equation as \(x^{2} + \frac{b}{a}x + \frac{c}{a} = 0\).

Now let us take \(\alpha\) and \(\beta\) to be the roots of our equation. We can thus also write the equation as

\((x - \alpha)(x - \beta) = 0\).

If we multiply this out, get

\(x^{2} - \alpha x - \beta x + \alpha\beta = 0\).

Simplify the equation further to get

\(x^{2} - (\alpha + \beta)x + \alpha\beta\).

We now compare this with our other equation. What do you notice?

\[x^{2} + \frac{b}{a}x + \frac{c}{a} = 0\qquad \text{and} \qquadx^{2} - (\alpha + \beta)x + \alpha\beta = 0\]

From this comparison, we can see that:

the

**sum of the roots**, \(\alpha + \beta\), is equal to \(-\frac{b}{a}\);

the

**product of the roots**, \(\alpha\beta\), is equal to \(\frac{c}{a}\).

We can use these properties to calculate many other results, but a new notation needs to be defined first.

\(\alpha + \beta\) can be written as \(\Sigma{\alpha}\);

\(\alpha\beta\) can be written as \(\Sigma{\alpha\beta}\).

This form is called **summation notation**.

How would we determine \(\alpha^{2} + \beta^{2}\), using what we have listed above?

First, let us square \(\alpha + \beta\) to get the two squared terms. We know that \((\alpha + \beta)^{2} = \alpha^{2} + \beta^{2} + 2\alpha\beta\) so to get the final answer of \(\alpha^{2} + \beta^{2}\), we simply subtract \(2\alpha\beta\).

In summation notation, this calculation will look like this:

\[\alpha^{2} + \beta^{2} = (\Sigma{\alpha})^{2} - 2\Sigma{\alpha\beta},\]

where \(\alpha^{2} + \beta^{2}\) is denoted as \(\Sigma{\alpha^{2}}\).

Note that \(\Sigma{\alpha^{2}}\) and \((\Sigma{\alpha})^{2}\) mean different things. \((\Sigma{\alpha})^{2} = (\alpha + \beta)^{2}\) whereas \( \Sigma{\alpha^{2}} = \alpha^{2} + \beta^{2} \).

We can determine \(\frac{1}{\alpha} + \frac{1}{\beta}\), denoted as \(\Sigma{\frac{1}{\alpha}}\), using similar methods.

Combine the fractions to get \(\frac{\alpha + \beta}{\alpha\beta}\) which we can rewrite in summation notation to get \(\frac{\Sigma{\alpha}}{\Sigma{\alpha\beta}}\).

Another way of working with roots of polynomials is to make use of recurrence relations. The relation used for each type of polynomial is summarized in the table below:

Type of polynomial | Recurrence relation used |

Quadratic equations | \[S_{n} = \alpha^{n} + \beta^{n} \] |

Cubic equations | \[S_{n} = \alpha^{n} + \beta^{n} + \gamma^{n} \] |

Quartic equations | \[S_{n} = \alpha^{n} + \beta^{n} + \gamma^{n} + \delta^{n} \] |

Let's look at an example right away.

Consider the quadratic equation \(x^{2} + 6x + 8 = 0\).

We know that \(\alpha + \beta = -6\). We can also view this as \(S_{1}\). Using \(S_{1}\), we can calculate \(\alpha^{2} + \beta^{2}\) with a different method.

Since \(\alpha\) and \(\beta\) are roots of the equation, we can say that \(\alpha^{2} + 6\alpha + 8 = 0 \; \text{and} \; \beta^{2} + 6\beta + 8 = 0\). We add these two together to get \((\alpha^{2} + \beta^{2}) + 6({\alpha + \beta}) + 16 = 0\), which can also be written as \(S_{2} + 6S_{1} + 16 = 0\).

Remember that \(S_{n} = \alpha^{n} + \beta^{n}\), so \(S_{1} = \alpha + \beta\) and \(S_{2} = \alpha^{2} + \beta^{2}\).

By substituting \(S_{1}\) back into the equation, we can calculate that \(\alpha^{2} + \beta^{2} = S_{2} = 20\).

This can also be found by using the summation method mentioned above, \(\alpha^{2} + \beta^{2} = (\Sigma{\alpha})^{2} - 2\Sigma{\alpha\beta}\).

Let's look at another example.

Assuming that \(3x^{2} + 4x + 12 = 0\) has roots \(\alpha\) and \(\beta\), find the value of \(S_{3}\).

**Solution**

First, substitute \(\alpha \; \text{and}\; \beta\) into the equation to obtain the following:

\[\begin{align}3\alpha^{2} + 4\alpha + 12 & = 0 \\\newline 3\beta^{2} + 4\beta + 12 & = 0 \end{align}\]

Add the two equations together to get:\[3(\alpha^{2} + \beta^{2}) + 4(\alpha + \beta) + 24 = 0\]

Which is the same as:\[3S_{2} + 4S_{1} + 24 = 0\]

We know that \(S_{1} = \alpha + \beta = -\frac{b}{a}\)

\[\therefore \qquad S_{1} = -\frac{4}{3}\]

Substitute \(S_{1}\) back into the equation to calculate that \(S_{2} = -\frac{56}{9}\).

Using \(S_{n} = \alpha^{n} + \beta^{n}\), we know that \(S_{3}\) must thus equal \(\alpha^{3} +\beta^{3}\). Returning to the original equation, \(3x^{2} + 4x + 12 = 0\), we can see that there are no cubes present, so we multiply the whole equation by \(x\) to get \(3x^{3} + 4x^{2} + 12x = 0\).

As we did earlier, substitute \(\alpha \; \text{and} \; \beta\) in the equation to get:

\[\begin{align}3\alpha^{3} + 4\alpha^{2} + 12\alpha & = 0 \\\newline 3\beta^{3} + 4\beta^{2} + 12\beta & = 0 \end{align}\]

Add the two equations together and apply the recurrence relation, \(S_{n} = \alpha^{n} + \beta^{n}\) to result in:

\[3S_{3} + 4S_{2} + 12S_{1} = 0\]

Since we already know the values of \(S_{1} \; \text{and} \; S_{2}\), we just substitute them back into the equation to calculate the value of \(S_{3}\).

\[\begin{align}3S_{3} + 4\left(-\frac{56}{9} \right) + 12\left(-\frac{4}{3} \right) & = 0 \\\newline S_{3} & = \frac{368}{27}\end{align}\]

**Alternative method**

Alternatively, this can also be solved using summation notation.

We know that \(S_{3} = \alpha^{3} + \beta^{3}\) and that \(\alpha^{3} + \beta^{3}\ = (\alpha + \beta)^{3} - 3\alpha\beta(\alpha + \beta)\)

In summation form this gives us:

\[\Sigma{\alpha^{3}} = (\Sigma{\alpha})^{3} - 3\Sigma{\alpha\beta}\Sigma{\alpha}\]

The sum of the roots, \(\Sigma{\alpha}\), is \(-\frac{4}{3}\) and the product of the roots, \(\Sigma{\alpha\beta}\), is \(4\).

Substitute these values into the summation form above to get:

\[\begin{align}\Sigma{\alpha^{3}} & = \left(-\frac{4}{3} \right)^{3} - 3(4)\left(-\frac{4}{3} \right) = \frac{368}{27}\end{align}\]

Recurrence relations can also be used to find values such \(S_{-1}\). This can be done similarly to how we found \(S_{3}\) in the previous example, but there is a multitude of methods one can use to calculate it.

The following example explains two of these different methods in detail:

You are given the quadratic equation \(3x^{2} + 5x - 8 = 0\) with roots \(\alpha\) and \(\beta\). Calculate the value of \(S_{-1}\).

**Solution - Method 1**

First, calculate the values of \(S_{1}\).

\[ \begin{align}S_{1} & = -\frac{b}{a} \\ & = -\frac{5}{3} \end{align} \]

Next, divide by \(x\) to obtain the following:

\[ 3x + 5 - \frac{8}{x} = 0\]

Substitute the roots into the equation and apply your recurrence relation to get:

\[ 3S_{1} + 10 - 8S_{-1} = 0\]

You have already calculated the value of \(S_{1}\), so substitute that value back into the equation and simplify to calculate \(S_{-1}\).

\[ \begin{align} 3 \left(-\frac{5}{3} \right) + 10 - 8S_{-1} & = 0 \\-8S_{-1} & = -5 \\ S_{-1} & = \frac{5}{8} \end{align} \]

**Method 2**

\(S_{-1}\) is equivalent to \(\frac{1}{\alpha} + \frac{1}{\beta}\).

Remember that \(S_{n} = \alpha^{n} + \beta^{n}\) for quadratic equations. It therefore follows that \(S_{-1} = \alpha^{-1} + \beta^{-1} = \frac{1}{\alpha} + \frac{1}{\beta}\).

It was established in the previous section that

\[\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\Sigma{\alpha}}{\Sigma{\alpha}{\beta}} \]

You know that \( \Sigma{\alpha} = -\frac{b}{a} \text{ and } \Sigma{\alpha\beta} = \frac{c}{a} \). Substitute these into the equations to get the following:

\[ S_{-1} = -\frac{b}{c}\]

From there, all you need to do is substitute the values for \(b\) and \(c\) in and simplify.

We thus get that \(S_{-1} = \frac{5}{8}\).

Cubic equations are very similar to quadratic equations, except the roots will now be represented by \(\alpha,\;\beta \; \text{and} \; \gamma\).

We start off with \(ax^{3} + bx^{2} + cx + d = 0\). As with quadratic equations, we divide by \(a\) to get \(x^{3} + \frac{b}{a}x^{2} + \frac{c}{a}x + \frac{d}{a} = 0\).

We've established that we will be using \(\alpha, \; \beta \; \text{and} \; \gamma\) to represent the roots of the cubic equation. We thus get

\((x - \alpha)(x - \beta)(x - \gamma) = 0\),

which, when expanded, gives us:

\[x^{3} - (\alpha + \beta + \gamma)x^{2} + (\alpha\beta + \alpha\gamma + \beta\gamma)x -\alpha\beta\gamma = 0.\]

When we compare this to \(x^{3} + \frac{b}{a}x^{2} + \frac{c}{a}x + \frac{d}{a} = 0\), we can see that

\(\alpha + \beta + \gamma = -\frac{b}{a}\), which is denoted by \(\Sigma{\alpha}\) or \(S_{1}\);

\(\alpha\beta + \alpha\gamma + \beta\gamma = \frac{c}{a}\) which we denote with \(\Sigma{\alpha\beta}\);

\(\alpha\beta\gamma = -\frac{d}{a}\), which we write as \(\Sigma{\alpha\beta\gamma}\).

We know from quadratics that \(\Sigma{\alpha^{2}} = (\Sigma{\alpha})^{2} - 2\Sigma{\alpha\beta}\). This also holds true for cubic equations where

\((\Sigma{\alpha})^{2} = (\alpha + \beta + \gamma)^{2} = \alpha^{2} + \beta^{2} + \gamma^{2} + 2\alpha\beta + 2\alpha\gamma + 2\beta\gamma\),

and \(\Sigma{\alpha^{2}}\) therefore equals \((\Sigma{\alpha})^{2} - 2\Sigma{\alpha\beta}\).

We can now use \(S_{n} = \alpha^{n} + \beta{n} + \gamma^{n}\) to represent the roots of the cubic equations. For example, \(S_{3}\) will be used to represent \(\alpha^{3} + \beta^{3} + \gamma^{3}\), e.t.c.

All the results that have been derived from quadratic equations hold true for cubic equations, but the calculations get more complicated as we work with higher powers. It is easier to use recurrence relations rather than summation forms to find \(S_{n}\) from powers of 3 and higher.

Refer to the following example that shows this:

Find \(S_{3}\) for the equation \(6x^{3} - 15x^{2} - 17x + 6 = 0\).

**Solution - Using Summation Form**

We know that \(S_{3} = (\Sigma{\alpha})^{3} - 3\Sigma{\alpha\beta}\Sigma{\alpha} + 3\Sigma{\alpha\beta\gamma}\).

First, let us calculate the values of \(\Sigma{\alpha}, \; \Sigma{\alpha\beta} \; \text{and} \; \Sigma{\alpha\beta\gamma}\):

\[\begin{align}\Sigma{\alpha} & = -\frac{b}{a} \\& = -\frac{-15}{6} \\& = \frac{5}{2}\end{align}\]

\[\begin{align}\Sigma{\alpha\beta} & = \frac{c}{a} \\& = -\frac{17}{6}\end{align}\]

\[\begin{align}\Sigma{\alpha\beta\gamma} & = -\frac{d}{a} \\& = -\frac{6}{6} \\& = -1\end{align}\]

Now that we have these values, we can substitute them back into the summation form to calculate \(S_{3}\).

\[\begin{align}S_{3} & = (\Sigma{\alpha})^{3} - 3\Sigma{\alpha\beta}\Sigma{\alpha} + 3\Sigma{\alpha\beta\gamma} \\& = \left(\frac{5}{2} \right)^{3} - 3\left(-\frac{17}{6} \right) \left(\frac{5}{2} + 3(-1) \right) \\& = \frac{271}{8}\end{align}\]

\[\therefore \qquad S_{3} = \frac{169}{8} \]

As you can see, this method does work, but it's easy to make mistakes, and it gets messy quickly. There is a more efficient way of solving this problem using recurrence relations.

**Alternative Solution - Using Recurrence Relations**

\(\alpha, \; \beta \text{ and } \gamma\) all satisfy our cubic equation, so we know that:

\[\begin{align}6\alpha^{3} - 15\alpha^{2} - 17\alpha + 6 & = 0 \\6\beta^{3} - 15\beta^{2} - 17\beta + 6 & = 0 \\6\gamma^{3} - 15\gamma^{2} - 17\gamma + 6 & = 0\end{align}\]

When we add these three equations, we get:

\[6(\alpha^{3} + \beta^{3} + \gamma^{3}) - 15(\alpha^{2} + \beta^{2} + \gamma^{2}) - 17(\alpha + \beta + \gamma) + 18 = 0\]

Which is the same as:

\[6S_{3} - 15S_{2} - 17S_{1} + 18 = 0\]

We know that:

\[ S_{1} = -\frac{b}{a} = \frac{5}{2} \]

and,

\[ \begin{align}S_{2} & = (\Sigma{\alpha})^{2} - 2\Sigma{\alpha\beta} \\& = \left(\frac{5}{2}\right)^{2} - 2\left(-\frac{17}{6}\right) \\& = \frac{143}{12} \\\end{align}\]

\[\begin{align}\therefore \qquad 6S_{3} - 15S_{2} - 17S_{1} + 18 & = 0 \qquad \text{Substitute } S_{1} \text{ and } S_{2} \text{ in} \\6S_{3} - 15 \left( \frac{143}{12}\right) - 17 \left(\frac{5}{2} \right) + 18 & = 0 \\S_{3} & = \frac{271}{8}\end{align}\]

For quartic equations, we will be using \(\alpha, \; \beta, \; \gamma \text{ and } \delta\) to represent the roots of the equation.

As with quadratic and cubic equations, the sum of the roots, \(\alpha + \beta + \gamma + \delta\), is denoted by \(\Sigma{\alpha}\) and \(S_{1}\) and is equivalent to \(-\frac{b}{a}\).

It is further given that, for quartic equations in the form \(ax^{4} + bx^{3} + cx^{2} + dx + e = 0\).

When working with quartic equations, make sure to use the recurrence formula wherever possible, especially when working with the sum of the cubes, \((\alpha^{3} + \beta^{3} + \gamma^{3} + \delta^{3})\).

The best method for determining the sum of the cubes of a general quartic equation is to calculate \(S_{1}\) first and then to determine \(S_{2}\) and \(S_{-1}\). Once we have determined those values, we can then use \(aS_{4} + bS_{3} + cS_{2} + dS_{1} + 4e = 0\) and divide by \(x\) to calculate \(S_{3}\).

For quartic equations in the form \(ax^{4} + bx^{3} + cx^{2} + dx + e = 0\), \[ \begin{gather*}\Sigma{\alpha} = \alpha + \beta + \gamma + \delta = -\frac{b}{a} \\\Sigma{\alpha\beta} = \alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta = \frac{c}{a} \\\Sigma{\alpha\beta\gamma} = \alpha\beta\gamma + \alpha\beta\delta + \alpha\gamma\delta + \beta\gamma\delta = -\frac{d}{a} \end{gather*}\]

and,

\[\Sigma{\alpha\beta\gamma\delta} = \alpha\beta\gamma\delta = \frac{e}{a}\]

These formulas come from expanding \((x - \alpha)(x - \beta)(x - \gamma)(x - \delta) = 0\) and comparing it to \(x^{4} + \frac{b}{a}x^{3} + \frac{c}{a}x^{2} + \frac{d}{a}x + e = 0\).

\[ \begin{align} (x + \alpha)(x + \beta)(x + \gamma)(x + \delta) & = 0 \\ x^{4} - (\alpha + \beta + \delta + \gamma)x^{3} + (\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma +\beta\delta + \gamma\delta)x^{2} - (\alpha\beta\gamma + \alpha\beta\delta + \alpha\delta\gamma\beta\gamma\delta)x + \alpha\beta\gamma\delta & = 0 \\x^{4} + \Sigma{\alpha}x^{3} + \Sigma{\alpha\beta}x^{2} + \Sigma{\alpha\beta\gamma}x +\Sigma{\alpha\beta\gamma\delta} & = 0 \end{align} \]

\(\therefore \qquad\) when we compare the two equations, we can conclude the following:

\[ \begin{gather*} \Sigma{\alpha} = -\frac{b}{a} \\ \hspace{1cm} \\ \Sigma{\alpha\beta} = \frac{c}{a} \\ \hspace{1cm} \\ \Sigma{\alpha\beta\gamma} = -\frac{d}{a} \\ \hspace{1cm} \\ \Sigma{\alpha\beta\gamma\delta} = \frac{e}{a} \end{gather*} \]The full workings of the expansion of \((x - \alpha)(x - \beta)(x - \gamma)(x - \delta) = 0\) are not shown here as it is not necessary for you to know them. It is only necessary for you to know how to apply the formulas that result from the expansion.

For recurrence relations involving quartic equations, use the following formulas:

\[ \begin{gather*} S_{n} = \alpha^{n} + \beta^{n} + \gamma^{n} + \delta^{n} \\ \hspace{1cm} \\S_{-1} = \Sigma{\frac{1}{\alpha}} = \frac{\Sigma{\alpha\beta\gamma}}{\Sigma{\alpha\beta\gamma\delta}} \\\end{gather*} \]

As with quadratic and cubic equations, \(S_{2} = (\Sigma{\alpha})^{2} - 2\Sigma{\alpha\beta}\).

You are given the quartic equation \(3x^{4} - x^{3} + 2x + 7 = 0\) with roots \(\alpha, \; \beta, \; \gamma \text{ and } \delta\). Calculate the values of \(S_{3}\) and \(S_{4}\).

**Solution**

First, calculate \(S_{1}\):

\[ S_{1} = -\frac{b}{a} = \frac{1}{3} \]

Next, calculate the value of \(S_{2}\):

\[\begin{align}S_{2} & = \left( \Sigma{\alpha} \right)^{2} – 2\Sigma{\alpha\beta} \\& = \left ( \frac{1}{3} \right )^{2} – 2\left( \frac{0}{2} \right) \\& = \frac{1}{9}\end{align}\]

Divide the original equation by \(x\) to get \(3x^{3} - x^{2} + 2 + \frac{7}{x} = 0\) and use this to calculate the value of \(S_{-1}\).

You know from quadratic equations that \(S_{-1} = \Sigma{\frac{1}{\alpha}}\). It therefore

Suppose that you are given a quadratic equation with roots \(\alpha\) and \(\beta\). You are asked to find another quadratic that has roots \(3\alpha\) and \(3\beta\). How would you go about solving this?

The following example explains the two methods you could use in detail.

\(6x^{2} - x - 12 = 0\) has roots \(\alpha\) and \(\beta\). Find the quadratic equation with roots \(3\alpha\) and \(3\beta\).

**Solution - Method 1**

Start by considering the quadratic \((y - 3\alpha)(y - 3\beta)\ = 0\). This equation has roots \(3\alpha\) and \(3\beta\), which is what you want.

Expand it to get \(y^{2} - 3(\alpha + \beta)y + 9\alpha\beta = 0\).

Now compare this to the original equation. From the original, we know that \(\alpha + \beta = \frac{1}{6}\) and that \(\alpha\beta = -2\).

Input these values into the new equation to get the final result of \(y^{2} - \frac{1}{2}y - 18 = 0 \), which can also be multiplied by 2 to give \(2y^{2} - y - 36 = 0\).

**Method 2**

The second method of determining the new equation works by relating the roots of the different equations.

We know that the roots of the new equation (\(y\)) are three times that of the original.

The relation is thus shown as:

\[y =3x\]

Rearrange the equation so that \(x\) is the subject of the formula.

\[x = \frac{y}{3}\]

Next, substitute for \(x\) in the original equation and simplify

\[\begin{align}6\left(\frac{y}{3} \right)^{2} - \frac{y}{3} - 12 & = 0 \\\frac{6}{9}y^{2} - \frac{y}{3} - 12 & = 0 \\2y^{2} - y - 36 & = 0\end{align}\]

A more complicated use of substitution involves reciprocal functions. The following example details how one would go about using reciprocal functions in substitution:

Consider the given cubic equation, \(x^{3} - 2x^{2} - 6x + 4 = 0\) with roots \(\alpha, \; \beta \text{ and } \gamma\). Find the cubic equation with roots \(\frac{1}{\alpha}, \; \frac{1}{\beta}, \; \frac{1}{\gamma}\).

**Solution**

First, relate the roots to each other and make \(x\) the subject of the formula.

\[\begin{align} y & = \frac{1}{x} \\x & = \frac{1}{y}\end{align}\]

Next, substitute for \(x\) in the original equation and simplify.

\[\begin{align}\left(\frac{1}{y} \right)^{3} - 2\left(\frac{1}{y} \right)^{2} - 6\left(\frac{1}{y} \right) + 4 & = 0 \\\hspace{1cm} \\\frac{1}{y^{3}} - \frac{2}{y^{2}} - \frac{6}{y} + 4 & = 0 \\\hspace{1cm} \\ 4y^{3} - 6y^{2} - 2y + 1 & = 0 \qquad \text{Multiply by } y^{3}\end{align}\]

The cubic equation with roots \(\alpha, \; \beta \text{ and } \gamma\) is thus \(4y^{3} - 6y^{2} - 2y + 1 = 0\).

There are two methods one can use when substituting with powers of roots.

You are given the quadratic equation \(2x^{2} - 4x + 7 = 0\) with roots \(\alpha \text{ and } \beta\). Find the quadratic equation with roots \(\alpha^{2}\) and \(\beta^{2}\).

**Solution - Method 1**

We can state that \(y = x^{2}\) and therefore \(x = \sqrt{y}\).

Substituting this value into the equation results in the following:

\[\begin{split}2((y^{\frac{1}{2}})^{2} - 4y^{\frac{1}{2}} + 7 = 0 \\2y - 4y^{\frac{1}{2}} + 7 = 0\end{split}\]

We can then rearrange the equation and square both sides. This will result in the quadratic equation that is the solution to this question.

\[\begin{align}2y + 7 & = 4y^{\frac{1}{2}} \\4y^{2} + 28y + 49 & = 16y \\4y^{2} + 12y + 49 & = 0 \end{align}\]

**Method 2 **

In this method, we reverse the order we followed to solve the problem in the first method.

First, rearrange the equation. We want to perform an operation so that we get even powers for every term of \(x\). This is done by squaring both sides.

\[\begin{align}-4x & = -2x^{2} - 7 \\16x^{2} & = 4x^{4} + 28x^{2} + 49 \qquad \text{Square both sides} \end{align}\]

Now substitute \(y = x^{2}\) in, like we did in the first method.

\[\begin{align}16y & = 4y^{2} + 28y + 49 \\4y^{2} + 12y + 49 & = 0 \end{align}\]

As you can see, this method results in the same answer as the first method.

For both methods, you must ensure that you are using the correct the powers of \(x\).

Another use of substitution methods is determining the values of recurrence relations, such as \(S_{6}\).

Suppose you were asked to determine the value of \(S_{6}\) for the cubic equation \(x^{3} + 3x^{2} - 1 = 0\). Not only is the process for doing this long, it is also easy to make mistakes.

Now, imagine that there was another cubic equation for which \(y= x^{3}\). For the \(x\) cubic, \(S_{n} = \alpha^{n} + \beta^{n} + \gamma^{n}\).

Since the roots of the \(y\) cubic are equal to the roots of the \(x\) cubic cubed, the \(y\) cubic would have \(S_{n} = \alpha^{3n} + \beta^{3n} + \gamma^{3n}\). The power for each root has been tripled, so we only need to find \(S_{2}\) of the \(y\) cubic to determine the value of \(S_{6}\) for the \(x\) cubic.

The following example demonstrates how one would do this:

Consider the cubic equation \(2x^{3} - 5x + 1 = 0\) with roots \(\alpha, \; \beta \text{ and } \gamma\). Determine the value of \(S_{6}\) using the substitution of \(y = x^{3}\).

**Solution**

Label the cubic and rearrange it as follows:

\[\begin{equation}\tag{1}2x^{3} + 1 = 5x \end{equation}\]

Cube both sides of the cubic and then simplify to get:

\[\begin{align}8x^{9} + 6x^{6} + 6x^{3} + 1 & = 125x^{3} \\8x^{9} + 6x^{6} - 119x^{3} + 1 & = 0\end{align}\]

Next, substitute \(y\) in for \(x^{3}\) and label this new equation.

\[\begin{equation}\tag{2}8y^{3} + 6y^{2} - 119y + 1 = 0\end{equation}\]

Determine \(S_{1}\) for equation 2.

\[\begin{align}S_{1} & = -\frac{b}{a} \\\hspace{1cm} \\ & = -\frac{6}{8} \\\hspace{1cm} \\ & = -\frac{3}{4}\end{align}\]

Finally, determine \(S_{2}\) using the \(S_{1}\) value you just calculated, and state the value of \(S_{6}\) for Equation 1.

\[\begin{align} S_{2} & = (S_{1})^{2} - \frac{c}{a} \\\hspace{1cm} \\ & = \left(-\frac{3}{4} \right)^{2} - \left(-\frac{119}{8} \right) \\\hspace{1cm} \\& = \frac{247}{16} \\ \end{align} \]

\[ \begin{split}\therefore \qquad S_{2} \text{ for Equation } 2 = \frac{247}{16} \\\hspace{1cm} \\ \therefore \qquad S_{6} \text{ for Equation 1 } = \frac{247}{16}\end{split}\]

This section summarises the formulas derived in previous sections.

For quadratic equations in the form \(ax^{2} + bx + c = 0\):

\[ \begin{gather*} \Sigma{\alpha} = \alpha + \beta = -\frac{b}{a} \\ \hspace{1cm} \\\Sigma{\alpha^{2}} = \alpha^{2} + \beta^{2} \\ \hspace{1cm} \\\Sigma{\alpha\beta} = \alpha\beta = \frac{c}{a} \\ \hspace{1cm} \\S_{n} = \alpha^{n} + \beta^{n} \\\hspace{1cm} \\ S_{-1} = \Sigma{\frac{1}{\alpha}} = \frac{\Sigma{\alpha}}{\Sigma{\alpha\beta}} \\\hspace{1cm} \\ \Sigma{\frac{1}{\alpha}} = \frac{1}{\alpha} + \frac{1}{\beta} \end{gather*} \]

For cubic equations in the form \(ax^{3} + bx^{2} + cx + d = 0\):

\[ \begin{gather*} \Sigma{\alpha} = \alpha + \beta + \gamma = -\frac{b}{a} \\ \hspace{1cm} \\\Sigma{\alpha\beta} = \alpha\beta + \alpha\gamma + \beta\gamma = \frac{c}{a} \\ \hspace{1cm} \\\Sigma{\alpha\beta\gamma} = \alpha\beta\gamma = -\frac{d}{a} \\ \hspace{1cm} \\S_{n} = \alpha^{n} + \beta^{n} + \gamma^{n} \\\hspace{1cm} \\ S_{-1} = \Sigma{\frac{1}{\alpha}} = \frac{\Sigma{\alpha\beta}}{\Sigma{\alpha\beta\gamma}} \\ \hspace{1cm} \\\Sigma{\frac{1}{\alpha}} = \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}\end{gather*} \]

For quartic equations in the form \(ax^{4} + bx^{3} + cx^{2} + dx + e = 0\):

\[ \begin{gather*}\Sigma{\alpha} = \alpha + \beta + \gamma + \delta = -\frac{b}{a} \\\hspace{1cm} \\\Sigma{\alpha\beta} = \alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta= \frac{c}{a} \\\hspace{1cm} \\\Sigma{\alpha\beta\gamma} = \alpha\beta\gamma + \alpha\beta\delta + \alpha\gamma\delta + \beta\gamma\delta = -\frac{d}{a} \\ \hspace{1cm} \\\Sigma{\alpha\beta\gamma\delta} = \alpha\beta\gamma\delta = \frac{e}{a} \\ \hspace{1cm} \\S_{n} = \alpha^{n} + \beta^{n} + \gamma^{n} + \delta^{n} \\ \hspace{1cm} \\ S_{-1} = \Sigma{\frac{1}{\alpha}} = \frac{\Sigma{\alpha\beta\gamma}}{\Sigma{\alpha\beta\gamma\delta}} \\ \hspace{1cm} \\\Sigma{\frac{1}{\alpha}} = \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} + \frac{1}{\delta} \\\end{gather*} \]

The formulas you need to know for recurrence notation are as follows:

\[ \begin{gather*}S_{1} = \Sigma{\alpha} \\ \hspace{1cm} \\ S_{2} = \Sigma{\alpha^{2}} = (\Sigma{\alpha})^{2} - 2\Sigma{\alpha\beta} \\ \hspace{1cm} \\ S_{-1} = \Sigma{\frac{1}{\alpha}}\end{gather*} \]

In this section, we will work through a multitude of practice questions involving roots of polynomials.

You are given the cubic equation \(3x^{3} + mx^{2} + nx + p = 0\) with \(S_{1} = -5, \; S_{2} = \frac{79}{3} \text{ and } S_{-1} = \frac{1}{2}\). Find the values of the constants \(m, \; n \text{ and } p\).

**Solution**

Calculate the value of \(m\). You already know that \(S_{1} = \alpha + \beta = -\frac{b}{a}\), so make use of this relationship to find \(m\).

\[ \begin{align}S_{1} & = -\frac{b}{a} \qquad \quad \text{ Substitute in the values you know.} \\ -5 & = -\frac{m}{3} \\ \therefore \qquad m & = 15\end{align} \]

Similarly, when calculating \(n\), you know that \(S_{2} = (\Sigma{\alpha})^{2} - 2\Sigma{\alpha\beta}\) and also that \(\Sigma{\alpha} = S_{1} \text { and } \Sigma{\alpha\beta} = \frac{c}{a}\). Substitute the values you know in and simplify to solve for \(n\).

\[ \begin{align} S_{2} & = (\Sigma{\alpha})^{2} - 2\Sigma{\alpha}{\beta} \\ \frac{79}{3} & = (-5)^{2} - 2\frac{n}{3} \\ \frac{4}{3} & = -2\frac{n}{3} \\ n & = -2 \end{align} \]

Finally, to calculate \(p\), you use \(S_{-1} = \frac{\Sigma{\alpha\beta}}{\Sigma{\alpha\beta\gamma}}\) (there are other ways of doing this, but this method is the easiest).

\[ \begin{align}S_{-1} & = \frac{\Sigma{\alpha}{\beta}}{\Sigma{\alpha\beta\gamma}} \\ \frac{1}{2} & = -\frac{c}{d} \\ \frac{1}{2} & = -\frac{n}{p} \\ \frac{1}{2} & = -\frac{-2}{p} \\ p & = 4\end{align} \]

Let's see another example.

You are given the quartic equation \(x^{4} + 3x^{3} - 8x^{2} + 5 = 0\) with roots \(\alpha, \beta, \; \gamma \text{ and } \delta\). Show that \(S_{4} = -22S_{3} + 1\).

**Solution**

First, you must calculate the values of \(S_{1}, \; S_{2}\) and \(S_{-1}\): \[ \begin{align}S_{1} & = \alpha + \beta + \gamma + \delta \\ & = -\frac{b}{a} \\ & = -3 \end{align} \]

\[ \begin{align} S_{2} & = (\Sigma{\alpha})^{2} - 2\Sigma{\alpha\beta} \\ & = (-3)^{2} - 2(8) \\ & = -7 \end{align} \]

\[ \begin{align} S_{-1} & = \frac{\Sigma{\alpha\beta\gamma}}{\Sigma{\alpha\beta\gamma\delta}} \\ & = -\frac{d}{e} \\ & = -\frac{0}{5} \\ & = 0\end{align} \]

Next, you need to calculate the value of \(S_{3}\). This is done by dividing the original equation by \(x\) so that it looks like this \(x^{3} + 3x^{2} - 8x + \frac{5}{x} = 0\) and then substituting each of the four roots, \(\alpha, \; \beta, \; \gamma \text{ and } \delta\) into the equation and adding the four resulting equations together to get the following:

\[(\alpha^{3} + \beta^{3} + \gamma^{3} + \delta^{3}) + 3(\alpha^{2} + \beta^{2} + \gamma^{2} + \delta^{2} ) - 8(\alpha + \beta + \gamma + \delta) + 5(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} + \frac{1}{\delta}) = 0\]

Apply the recurrence relation \(S_{n} = \alpha^{n} + \beta^{n} + \gamma^{n} + \delta^{n}\) and your equation should now look like this:

\[ S_{3} + 3S_{2} - 8S_{1} + 5S_{-1} = 0\]

Substitute \(S_{1}, \; S_{2} \text{ and } \S_{-1}\) into the equation and solve for \(S_{3}\):

\[ \begin{align} S_{3} + 3(-7) - 8(-3) + 5(0) & = 0 \\ S_{3} & = -3 \\ \end{align} \]

To solve for \(S_{4}\), we repeat a similar process to solving for \(S_{3}\), except for dividing by \(x\). Substitute each of the roots, \(\alpha, \; \beta, \; \gamma \text{ and } \delta\) into the original equation and add the four resulting equations to get the following:

\[ (\alpha^{4} + \beta^{4} + \gamma^{4} + \delta^{4}) + 3(\alpha^{3} + \beta^{3} + \gamma^{3} + \delta^{3}) - 8(\alpha^{2} + \beta^{2} + \gamma^{2} + \delta^{2}) + 20 = 0\]

Apply the recurrence relation and you get:

\[ S_{4} + 3S_{3} - 8S_{2} + 20 = 0\]Finally, substitute the values for \(S_{3}\) and \(S_{2}\) in and solve for \(S_{4}\):

\[ \begin{align}S_{4} + 3(-3) - 8(-7) + 20 & = 0 \\S_{4} & = 67\end{align} \]

\(\therefore \qquad -22S_{3} + 1 = -22(-3) + 1 = 67 = S_{4}\)

This final example covers a more complicated problem involving substitution.

The cubic equation \(x^{3} + 4x^{2} - 7x - 1 = 0\) has roots \(\alpha, \beta\) and \(\gamma\). Find the cubic equations with roots \(\frac{2\alpha - 3}{\alpha}, \; \frac{2\beta - 3}{\beta} \text{ and } \frac{2\gamma - 3}{\gamma}\).

**Solution**

We know that the roots of this new cubic equation are:

\[ y = \frac{2x - 3}{x}\]

Make \(x\) the subject of the formula:

\[ \begin{align} xy & = 2x - 3 \\ xy - 2x & = -3 \\ x(y - 2) & = -3 \\ x & = -\frac{3}{y-2} \end{align} \]

Substitute this value for \(x\) back into the original equation and simplify to obtain the new cubic.

\[ \begin{align} \left( -\frac{3}{y - 2} \right)^{3} + 4\left(-\frac{3}{y-2}\right)^{2} - 7\left(-\frac{3}{y-2}\right) + 1 & = 0 \\ \hspace{1cm} \\ -\frac{27}{(y-2)^{3}} + \frac{36}{(y-2)^{2}} + \frac{21}{y-2} + 1 & = 0 \\ \hspace{1cm} \\ -27 + 36(y-2) + 21(y-2)^{2} + (y-2)^{3} & = 0 \\ \hspace{1cm} \\-27 + 36y - 72 + 21y^{2} - 84y + 84 + y^{3} - 6y^{2} + 12y - 8 & = 0 \\ \hspace{1cm} \\y^{3} + 15y^{2} - 36y - 23 & = 0\end{align}\]

The cubic equation with roots \(\frac{2\alpha - 3}{\alpha}, \; \frac{2\beta - 3}{\beta} \text{ and } \frac{2\gamma - 3}{\gamma}\) is thus:

\[ y^{3} + 15y^{2} - 36y - 23 = 0 \]

- We use \(\alpha\) and \(\beta\) to represent the roots of a quadratic equation, \(\alpha, \; \beta \text{ and } \gamma\) to represent the roots of a cubic equation, and \(\alpha, \; \beta, \; \gamma \text{ and } \delta\) to represent the roots of a quartic equation.
- The sum of the roots of a polynomial equation can be calculated using \(\Sigma{\alpha} = -\frac{b}{a}\).
- It is better to use recurrence relations when working with powers of 3 and higher.
- Substitution can be used to find a polynomial by relating the roots of the polynomial you want to find with those of a known polynomial, followed by substituting that relation into the known polynomial and simplifying.
- The sum of the inverse roots of a polynomial, \(\Sigma{1}{\alpha}\), is always equal to the negative of the coefficient of the linear term divided by the constant term.

Considering a, b, c, d, e the coefficients, you analyse the sign of the polynomial's discriminant:

- for quadratic equations, if Δ = b
^{2}– 4ac < 0, then the roots are complex; - for cubic equations, if Δ = b
^{2}c^{2}– 4ac^{3}– 4b^{3}d – 27a^{2}d^{2}+ 18abcd < 0, then there's one real root and two complex conjugate roots; - for quartic equations, if Δ = 256a
^{3}e^{3}– 192a^{2}bde^{2}– 128a^{2}c^{2}e^{2}+ 144a^{2}cd^{2}e – 27a^{2}d^{4}+ 144ab^{2}ce^{2}– 6ab^{2}d^{2}e – 80abc^{2}de + 18abcd^{3}+ 16ac^{4}e – 4ac^{3}d^{2}– 27b^{4}e^{2}+ 18b^{3}cde – 4b^{3}d^{3}– 4b^{2}c^{3}e + b^{2}c^{2}d^{2}> 0, there are two real and two complex roots, and if Δ < 0, all roots are real or complex.

Considering a, b, c, d, e the coefficients, you analyse the sign of the polynomial's discriminant:

- for quadratic equations, if Δ = b
^{2}– 4ac > 0 or Δ = 0, then the roots are real; - for cubic equations, if Δ = b
^{2}c^{2}– 4ac^{3}– 4b^{3}d – 27a^{2}d^{2}+ 18abcd > 0 or Δ= 0, then all the roots are real; - for quartic equations, if Δ = 256a
^{3}e^{3}– 192a^{2}bde^{2}– 128a^{2}c^{2}e^{2}+ 144a^{2}cd^{2}e – 27a^{2}d^{4}+ 144ab^{2}ce^{2}– 6ab^{2}d^{2}e – 80abc^{2}de + 18abcd^{3}+ 16ac^{4}e – 4ac^{3}d^{2}– 27b^{4}e^{2}+ 18b^{3}cde – 4b^{3}d^{3}– 4b^{2}c^{3}e + b^{2}c^{2}d^{2}> 0, there are two real and two complex roots, and if Δ < 0, all roots are real or complex.

To find the roots of a cubic polynomial, you can:

- use factorisation, reducing it to three linear equations or one linear and one quadratic equation;
- use the division method;
- use the graphical method.

More about Roots of Polynomials

Be perfectly prepared on time with an individual plan.

Test your knowledge with gamified quizzes.

Create and find flashcards in record time.

Create beautiful notes faster than ever before.

Have all your study materials in one place.

Upload unlimited documents and save them online.

Identify your study strength and weaknesses.

Set individual study goals and earn points reaching them.

Stop procrastinating with our study reminders.

Earn points, unlock badges and level up while studying.

Create flashcards in notes completely automatically.

Create the most beautiful study materials using our templates.

Sign up to highlight and take notes. It’s 100% free.

Over 10 million students from across the world are already learning smarter.

Get Started for Free