Further Mathematics (9665) | Roots and polynomials

Welcome to Roots and Polynomials!

Hello, and welcome to one of the most powerful and elegant topics in Further Mathematics: the connection between the roots (solutions) of a polynomial equation and its coefficients (the numbers in front of the $x$ terms).

In this chapter, we level up the skills you started in FP1 (with quadratics) and extend them to cubics, quartics, and beyond! This is a fantastic 'detective skill' because it lets you find relationships between roots without ever having to solve the complicated polynomial equation itself. Let's dive in!

Section 1: Reviewing the Foundation (Quadratics)

Don't worry if you need a quick refresh! All higher-order polynomials follow the same fundamental rules established by the simple quadratic equation.

The General Quadratic Equation

A general quadratic equation is written as:
$$\(ax^2 + bx + c = 0\)$$
If the roots are \(\alpha\) and \(\beta\), there are two fundamental relationships, provided $a \neq 0$:

• Sum of the roots: $$\(\alpha + \beta = -\frac{b}{a}\)$$
• Product of the roots: $$\(\alpha \beta = \frac{c}{a}\)$$

Memory Aid: Notice the signs alternate starting with negative: \(-b/a\), then \(+c/a\). This sign pattern is crucial and holds for all polynomials.

Manipulating Expressions

A common requirement is to find the value of expressions involving the roots, like \(\alpha^2 + \beta^2\) or \(\alpha^3 + \beta^3\). You must express these in terms of the basic sum and product only.

Key Identities to Remember:

• $$\(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\)$$
• $$\(\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)\)$$
  Tip: Focus on replacing every \(\alpha\) and \(\beta\) with the group terms \(\alpha+\beta\) or \(\alpha\beta\).

Quick Takeaway (Quadratics)

The relationships \(\alpha + \beta = -b/a\) and \(\alpha \beta = c/a\) are the building blocks. If you can rewrite any expression using these two blocks, you can solve the problem!

Section 2: Extending to Higher Degree Polynomials (FP2 Core)

The elegant pattern we saw with quadratics extends perfectly to cubics and quartics (and beyond!). These are often referred to as the Elementary Symmetric Polynomials of the roots.

2.1: The Cubic Equation (\(n=3\))

Consider the general cubic equation with real coefficients:
$$\(ax^3 + bx^2 + cx + d = 0\)$$
Let the three roots be \(\alpha\), \(\beta\), and \(\gamma\).

The three essential relationships are:

1. Sum of roots (taken one at a time): \(\Sigma \alpha\) $$\(\alpha + \beta + \gamma = -\frac{b}{a}\)$$
2. Sum of roots taken two at a time: \(\Sigma \alpha\beta\) $$\(\alpha\beta + \alpha\gamma + \beta\gamma = +\frac{c}{a}\)$$
3. Product of roots (taken three at a time): \(\Sigma \alpha\beta\gamma\) $$\(\alpha\beta\gamma = -\frac{d}{a}\)$$

The Sign Pattern Trick:
The sign of the relationship always alternates starting with negative:
$$ \begin{array}{c|c|c|c} \text{Term} & x^{n-1} & x^{n-2} & x^{n-3} \\ \hline \text{Coefficient} & b & c & d \\ \hline \text{Sign} & \mathbf{-} & \mathbf{+} & \mathbf{-} \end{array} $$
The coefficient used is always the one corresponding to the term one power lower than the previous step.

2.2: The Quartic Equation (\(n=4\))

Consider the general quartic equation with roots \(\alpha, \beta, \gamma, \delta\):
$$\(ax^4 + bx^3 + cx^2 + dx + e = 0\)$$

The four relationships are:

1. Sum of 1 root: \(\Sigma \alpha\) $$\(\alpha + \beta + \gamma + \delta = -\frac{b}{a}\)$$
2. Sum of 2 roots: \(\Sigma \alpha\beta\) $$\(\alpha\beta + \dots + \gamma\delta = +\frac{c}{a}\)$$
3. Sum of 3 roots: \(\Sigma \alpha\beta\gamma\) $$\(\alpha\beta\gamma + \alpha\beta\delta + \alpha\gamma\delta + \beta\gamma\delta = -\frac{d}{a}\)$$
4. Product of 4 roots: \(\Sigma \alpha\beta\gamma\delta\) $$\(\alpha\beta\gamma\delta = +\frac{e}{a}\)$$

Quick Review: The Generalized Rule

For a polynomial of degree \(n\), the sum of the roots taken \(k\) at a time is given by:

$$\Sigma (\text{roots taken } k \text{ at a time}) = (-1)^k \frac{\text{Coefficient of } x^{n-k}}{a}$$

Don't worry if this general formula seems tricky! Just remember the sequence: $k=1$ is $-b/a$, $k=2$ is $+c/a$, $k=3$ is $-d/a$, and so on, always alternating signs.

Section 3: Complex Roots and the Conjugate Pair Theorem

This is a critical concept in FP2, especially when dealing with cubics and quartics, as it saves you a huge amount of work!

The Theorem

The syllabus states that if a polynomial equation has real coefficients, then any non-real (complex) roots must occur in conjugate pairs.

• If \(\alpha = p + iq\) is a root, then its conjugate \(\alpha^* = p - iq\) must also be a root.

Analogy: Think of complex roots as being magnetic; they can never exist alone! If your polynomial uses only real numbers, the complex roots must pair up to cancel out the imaginary components when multiplied or added, ensuring the final coefficients remain real.

Applying the Theorem

This theorem is essential for factorization:

If a cubic equation has real coefficients, and you are told one root is \(3+i\):

1. The second root must be \(3-i\).
2. Since the cubic only has three roots, the third root (\(\gamma\)) must be real (because if it were complex, its conjugate would give a fourth root, making it a quartic!).

This immediately allows you to find the sum and product of the two complex roots, which combine to form a quadratic factor with real coefficients:
$$(x - \alpha)(x - \alpha^*) = x^2 - (\alpha + \alpha^*)x + \alpha \alpha^*$$

• Sum of the conjugate pair: \((p+iq) + (p-iq) = 2p\) (Real)
• Product of the conjugate pair: \((p+iq)(p-iq) = p^2 + q^2\) (Real)

Once you have found this real quadratic factor, you can use polynomial division or factor comparison to find the remaining factor (which gives the real root).

Did You Know?

Because of the Conjugate Root Theorem, any polynomial with real coefficients can always be factorized into a product of linear factors (corresponding to real roots) and quadratic factors (corresponding to complex conjugate pairs).

Section 4: Forming Equations with Transformed Roots

Often, an exam question asks you to find a new polynomial equation whose roots are a function of the roots of the original equation. For example, if the original roots are \(\alpha, \beta, \gamma\), the new roots might be \(\alpha^2, \beta^2, \gamma^2\) or \(\alpha+2, \beta+2, \gamma+2\).

Method 1: Direct Substitution (The Quickest Way)

The most efficient method is using substitution to relate the old root, \(x\), to the new root, \(y\).

Step-by-Step Process:

1. Let \(x\) be a root of the original equation \(f(x) = 0\).
2. Define the new root, \(y\), in terms of the old root, \(x\). (e.g., \(y = x+2\) or \(y = \frac{1}{x}\)).
3. Rearrange the expression to isolate \(x\). (e.g., if \(y = x^2\), then \(x = \sqrt{y}\)).
4. Substitute this expression for \(x\) back into the original equation \(f(x) = 0\).
5. Simplify the resulting equation in terms of \(y\). This new equation is the required polynomial.

Example: If the original equation is \(x^3 + 4x - 5 = 0\) with roots \(\alpha, \beta, \gamma\), find the equation whose roots are \(1/\alpha, 1/\beta, 1/\gamma\).

• New root: \(y = \frac{1}{x}\)
• Rearrange: \(x = \frac{1}{y}\)
• Substitute into original: $$\(\left(\frac{1}{y}\right)^3 + 4\left(\frac{1}{y}\right) - 5 = 0\)$$
• Multiply by \(y^3\) to clear denominators: $$\(1 + 4y^2 - 5y^3 = 0\)$$
• Rearrange into standard form: $$\(5y^3 - 4y^2 - 1 = 0\)$$

The new equation is \(5x^3 - 4x^2 - 1 = 0\). (Using \(x\) as the variable again).

Method 2: Using the New Sums (The Traditional Way)

If the substitution method is messy (e.g., transforming to \(\alpha^2\), which introduces square roots), you can calculate the new sums of roots directly.

For a new cubic equation \(Y^3 + PY^2 + QY + R = 0\) with roots \(\alpha^2, \beta^2, \gamma^2\):

• $P = -(\alpha^2 + \beta^2 + \gamma^2) = - \Sigma \alpha^2$
• $Q = (\alpha^2\beta^2 + \alpha^2\gamma^2 + \beta^2\gamma^2) = \Sigma (\alpha\beta)^2$
• $R = -(\alpha^2\beta^2\gamma^2) = - (\alpha\beta\gamma)^2$

You must first calculate \(\Sigma \alpha\), \(\Sigma \alpha\beta\), and \(\alpha\beta\gamma\) from the old equation, and then use identities (like those in Section 1) extended to three variables to find $P, Q,$ and $R$.

Advice for struggling students: Method 1 (Direct Substitution) is generally faster and less prone to algebraic error than calculating the new sums using Method 2, unless the question specifically asks for the new sums. Practice the substitution method well!

Key Takeaway (Transformations)

Transforming equations usually involves replacing the original variable $x$ with a function of the new variable $y$. Always rearrange to get $x=$ (a function of $y$) before substituting.