Understanding of conjugate pairs in polynomial equations is essential. This topic explores the principle that non-real roots of polynomials with real coefficients occur in conjugate pairs, particularly focusing on cubic and quartic equations.

## Introduction to Conjugate Pairs

**Complex Roots:**In any polynomial with real coefficients, if a complex number is a root, its conjugate will also be a root.**Conjugate Pair:**A pair of complex numbers of the form $a + bi$ and $a - bi$.**Polynomial Equations:**These are equations of the form $ax^n + bx^{n-1} + \ldots + k = 0$, where $a, b, \ldots, k$ are real coefficients.

## Application in Cubic and Quartic Equations

### Cubic Equations

**Example:** Consider the cubic equation $x^3 - 6x^2 + 11x - 6 = 0$.

**Roots:** This equation has three real roots: 1, 2, and 3.

**Conjugate Pairs:** Since all roots are real, there are no conjugate pairs in this case.

Image courtesy of Cuemath

### Example

**Equation: **$x^3 - 3x^2 + 5x - 3 = 0$

**Solution:**

**1. Trial and Error Method:**

- Test simple numbers $0, 1, -1, etc.$ to find roots.
- Here, $x = 1$ works (equation equals 0), so 1 is a root.

**2. Polynomial Division:**

- With a known root $x = 1$, divide the cubic equation by $(x - 1)$ to get a quadratic equation.

**3. Solving the Quadratic Equation:**

- Use the quadratic formula to find the other two roots.

**Roots Found:**

- Real Root: $x = 1$
- Complex Roots: $x = 1 - \sqrt{2}i$ and $x = 1 + \sqrt{2}i$

### Quartic Equations

**Example:**For a quartic equation $x^4 - 10x^3 + 35x^2 - 50x + 24 = 0$.**Roots:**The roots of this equation are 1, 2, 3, and 4.**Conjugate Pairs:**Similar to the cubic example, this equation also has only real roots, hence no conjugate pairs.

The graph provides a visual representation of the function in the range $x = -2$ to $x = 6$.

### Example

**Equation: **$x^4 - 2x^3 - 3x^2 + 4x - 2 = 0$

**Solution:**

**1. Trial and Error Method:**

- Test simple values for roots. This equation has no obvious roots from this method.

**2. Algebraic Techniques:**

- For quartic equations, consider factoring, completing the square, or specific quartic formulas.
- These methods can be intricate, involving multiple steps.

**3. Finding Real Roots:**

- The equation simplifies to a form where roots involve square roots.

**Roots Found:**

- $x = \frac{1}{2} - \frac{\sqrt{9 - 4\sqrt{6}}}{2}$
- $x = \frac{1}{2} + \frac{\sqrt{9 - 4\sqrt{6}}}{2}$
- $x = \frac{1}{2} - \frac{\sqrt{9 + 4\sqrt{6}}}{2}$
- $x = \frac{1}{2} + \frac{\sqrt{9 + 4\sqrt{6}}}{2}$

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.