Welcome to Further Pure Mathematics 1: Roots of Polynomial Equations!
Hello! This chapter might seem intimidating because we move beyond simple quadratic equations into cubics and quartics, but don't worry—it’s actually a beautiful puzzle! We will be learning how the roots (solutions) of a polynomial are secretly linked to the coefficients (the numbers) used in the equation. Mastering these relationships allows you to solve complicated problems without ever having to find the actual roots!
Why is this important? In Further Mathematics, we often deal with roots that are complex or messy. Using these relationships bypasses the need for complicated arithmetic, making calculations elegant and efficient. This topic is mandatory for Paper 1 (Further Pure Mathematics 1).
1. Understanding Polynomial Structure
1.1 What is a Polynomial?
A polynomial is an expression built from variables and coefficients, involving only operations of addition, subtraction, multiplication, and non-negative integer exponents of the variables.
- The degree is the highest power of the variable.
- A root is a value of \(x\) that makes the polynomial equal to zero. If a polynomial has degree \(n\), it has exactly \(n\) roots (counting multiplicity and complex roots).
In this chapter, we focus solely on equations of degree 2, 3, and 4, as specified by the syllabus.
To use the root relationships correctly, the polynomial must be normalised so that the coefficient of the highest power of \(x\) is \(a\):
- Degree 2 (Quadratic): \(ax^2 + bx + c = 0\)
- Degree 3 (Cubic): \(ax^3 + bx^2 + cx + d = 0\)
- Degree 4 (Quartic): \(ax^4 + bx^3 + cx^2 + dx + e = 0\)
2. The Relations Between Roots and Coefficients
The core of this chapter lies in using Vieta's formulas, which connect the roots (\(\alpha, \beta, \gamma, \dots\)) to the ratios of the coefficients (\(b/a, c/a, \dots\)).
2.1 Degree 2 (Quadratic)
Equation: \(ax^2 + bx + c = 0\). Roots: \(\alpha, \beta\).
The relations are:
1. Sum of roots (S1):
\( \sum \alpha = \alpha + \beta = - \frac{b}{a} \)
2. Product of roots (S2):
\( \alpha\beta = \frac{c}{a} \)
Example: If \(2x^2 - 6x + 5 = 0\), then \(a=2, b=-6, c=5\).
\(\alpha + \beta = -(-6)/2 = 3\).
\(\alpha\beta = 5/2\).
2.2 Degree 3 (Cubic)
Equation: \(ax^3 + bx^2 + cx + d = 0\). Roots: \(\alpha, \beta, \gamma\).
Here we have three main symmetric functions:
1. Sum of roots (S1):
\( \sum \alpha = \alpha + \beta + \gamma = - \frac{b}{a} \)
2. Sum of products taken two at a time (S2):
\( \sum \alpha\beta = \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} \)
3. Product of roots (S3):
\( \alpha\beta\gamma = - \frac{d}{a} \)
! Memory Aid: The Sign Switch !
The sign alternates, starting with negative, then positive, then negative, as you move down the coefficients (excluding \(a\)):
b is negative, c is positive, d is negative...
This pattern holds true for all higher degrees as well!
2.3 Degree 4 (Quartic)
Equation: \(ax^4 + bx^3 + cx^2 + dx + e = 0\). Roots: \(\alpha, \beta, \gamma, \delta\).
1. S1 (Sum of roots):
\( \sum \alpha = \alpha + \beta + \gamma + \delta = - \frac{b}{a} \)
2. S2 (Sum of products of two):
\( \sum \alpha\beta = \alpha\beta + \dots = \frac{c}{a} \)
3. S3 (Sum of products of three):
\( \sum \alpha\beta\gamma = \alpha\beta\gamma + \alpha\beta\delta + \alpha\gamma\delta + \beta\gamma\delta = - \frac{d}{a} \)
4. S4 (Product of roots):
\( \alpha\beta\gamma\delta = \frac{e}{a} \)
Key Takeaway: Vieta's Formula Pattern
The sum of the products of the roots taken \(k\) at a time is given by:
$$ S_k = \frac{(-1)^k \times (\text{Coefficient of } x^{n-k})}{(\text{Coefficient of } x^n)} $$
Always remember to divide by the leading coefficient \(a\), and check the sign based on \(k\)!
3. Evaluating Symmetric Functions of the Roots
Often, you will be asked to find the value of an expression involving the roots, like \(\sum \alpha^2\) or \(\sum \frac{1}{\alpha}\). These are called symmetric functions because the expression remains the same even if you swap the roots around.
The key trick is to express the desired complicated symmetric function purely in terms of the basic relations (\(S_1, S_2, S_3, \dots\)).
3.1 The Classic Example: Sum of Squares (\(\sum \alpha^2\))
If we have a cubic equation with roots \(\alpha, \beta, \gamma\), we want to find \(\alpha^2 + \beta^2 + \gamma^2\).
We know the identity based on S1:
$$ (\alpha + \beta + \gamma)^2 = \alpha^2 + \beta^2 + \gamma^2 + 2(\alpha\beta + \beta\gamma + \gamma\alpha) $$
Using our summation notation, this is written as:
$$ ( \sum \alpha )^2 = \sum \alpha^2 + 2 \sum \alpha\beta $$
Rearranging to find the sum of squares:
$$ \sum \alpha^2 = ( \sum \alpha )^2 - 2 \sum \alpha\beta $$
$$ \sum \alpha^2 = (S_1)^2 - 2(S_2) $$
Step-by-Step Process for Symmetric Functions:
- Identify the basic relations (S1, S2, S3): Read the coefficients \(a, b, c, \dots\) and calculate S1, S2, S3.
- Relate the required sum to the basics: Use algebraic identities to express the desired sum in terms of S1, S2, etc.
- Substitute and Calculate: Plug the values from Step 1 into the identity derived in Step 2.
Common Mistake Alert!
Students often forget the importance of the leading coefficient \(a\). Make sure you calculate \(S_1 = -b/a\), \(S_2 = c/a\), etc., before using them in identities like \((S_1)^2 - 2S_2\). Do not use the raw coefficients \(b\) and \(c\)!
3.2 Other Useful Symmetric Functions
- Sum of Reciprocals:
If we have a cubic, we want \(\sum \frac{1}{\alpha} = \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}\).
Find a common denominator:
\( \sum \frac{1}{\alpha} = \frac{\beta\gamma + \alpha\gamma + \alpha\beta}{\alpha\beta\gamma} = \frac{S_2}{S_3} \) - Sum of Cubes (\(\sum \alpha^3\)):
This is usually derived by substituting a root back into the original equation: \(a\alpha^3 + b\alpha^2 + c\alpha + d = 0\). Rearrange and sum over all roots. Questions involving \(\sum \alpha^3\) often involve these substitutions.
Polynomials with coefficients that are rational numbers but whose roots are irrational or complex have a deep connection to symmetry, studied in an advanced field called Galois Theory.
4. Forming New Equations via Substitution
This section is powerful because it allows us to create a new polynomial whose roots are simply related to the roots of the original polynomial, without knowing the original roots themselves.
Suppose the original equation \(f(x) = 0\) has roots \(\alpha, \beta, \dots\). We want a new equation \(g(y) = 0\) whose roots are \(y = h(\alpha)\).
4.1 The Substitution Technique (The "Root Translator")
If the new root is \(y\), we need to express the old root \(x\) in terms of \(y\). This means finding the inverse relation: \(x = h^{-1}(y)\).
Step-by-Step Process:
- State the relationship: Let \(y\) be a root of the new equation. Write down \(y\) in terms of the old root \(x\). Example: If the new roots are the squares of the old roots, \(y = x^2\).
- Find \(x\) in terms of \(y\): Rearrange the relationship to make \(x\) the subject. Example: If \(y = x^2\), then \(x = \pm \sqrt{y}\).
- Substitute into the original equation: Replace every \(x\) in the original equation \(f(x) = 0\) with the expression for \(x\) from Step 2.
- Simplify and Eliminate Radicals: Manipulate the new equation to eliminate any square roots or fractional powers (like \(\sqrt{y}\) or \(y^{1/2}\)) to obtain a standard polynomial in \(y\).
4.2 Common Substitutions (Syllabus Focus)
The syllabus requires you to handle "easiest cases" where substitutions are not given, such as reciprocals, squares, or simple linear functions.
A. Reciprocal Roots (\(y = 1/x\))
If the new roots are the reciprocals of the old roots, then \(y = 1/x\). Therefore, \(x = 1/y\).
Substitute \(x = 1/y\) into the equation \(ax^3 + bx^2 + cx + d = 0\):
$$ a\left(\frac{1}{y}\right)^3 + b\left(\frac{1}{y}\right)^2 + c\left(\frac{1}{y}\right) + d = 0 $$
Multiply by \(y^3\) to clear denominators:
$$ a + by + cy^2 + dy^3 = 0 $$
The new equation is \(dy^3 + cy^2 + by + a = 0\). Notice how the coefficients are reversed!
B. Squared Roots (\(y = x^2\))
If the new roots are the squares of the old roots, then \(y = x^2\). Therefore, \(x = \sqrt{y}\).
Substitute \(x = \sqrt{y}\) into the equation \(ax^3 + bx^2 + cx + d = 0\):
$$ a(\sqrt{y})^3 + b(\sqrt{y})^2 + c(\sqrt{y}) + d = 0 $$
Separate terms with \(\sqrt{y}\) (odd powers of \(x\)) from those without (even powers of \(x\)):
$$ (ay\sqrt{y} + c\sqrt{y}) + (by + d) = 0 $$
$$ \sqrt{y}(ay + c) = -(by + d) $$
To eliminate the square root, square both sides:
$$ y(ay + c)^2 = (by + d)^2 $$
Expanding and simplifying gives the new polynomial in \(y\).
C. Simple Linear Transformation (\(y = kx + c\))
Example: New roots are twice the old roots plus one (\(y = 2x + 1\)).
1. Relationship: \(y = 2x + 1\)
2. Inverse: \(x = \frac{y-1}{2}\)
3. Substitute this expression for \(x\) into the original equation \(f(x) = 0\). Since the transformation is linear, no squaring or difficult algebraic manipulation is usually needed, just careful expansion.
Key Takeaway: Forming New Equations
Always start by defining the relationship \(y\) in terms of \(x\), then rearrange it to find \(x\) in terms of \(y\). Substitute \(x = f^{-1}(y)\) into the original equation \(f(x) = 0\), and simplify the result into a polynomial in \(y\).
Chapter Summary Review
You have successfully covered the core concepts of Roots of Polynomial Equations (Syllabus 1.1):
1. Vieta's Formulas: Used the alternating sign pattern to relate roots (\(\sum \alpha\), \(\sum \alpha\beta\), \(\alpha\beta\gamma\)) to coefficients of polynomials up to degree 4.
2. Symmetric Functions: Learned how to express complex sums (like \(\sum \alpha^2\)) entirely in terms of the basic Vieta sums (S1, S2, S3), allowing calculation without finding the roots.
3. Forming New Equations: Mastered the substitution technique, using \(x = f^{-1}(y)\) to create new polynomials whose roots are related to the original roots (e.g., reciprocals or squares).
Keep practising those algebraic manipulations, especially for the substitution methods. You've got this!