# Determining the Number of Real Roots of a Cubic Equation

## Determining the Number of Real Roots of a Cubic Equation: A Comprehensive Guide

Cubic equations are polynomial equations of the form ax^3 + bx^2 + cx + d = 0. Understanding the number of real roots or solutions these equations possess is a fundamental problem in algebra. One powerful technique for solving this is the “Sums and Products” method. In this guide, we will explore this method in detail to determine the number of real roots of a cubic equation while optimizing for the search term “number of real roots of a cubic equation.”

### Understanding Cubic Equations

Cubic Equations: These are polynomial equations of degree three and can have up to three real roots. Determining how many real roots exist in a given cubic equation is essential in various mathematical and scientific contexts.

The ability to determine the number of real roots of a cubic equation using the Sums and Products method is a valuable skill in mathematics and its applications. By following the steps outlined in this guide, you can confidently analyze cubic equations and understand their real root count, contributing to problem-solving across various domains.

Let $\alpha, \beta$ and $\gamma$ be the roots of the polynomial $P(x) = 2x^3 + 4x^2 + 5x + 6$.

## Part 1

Find $\alpha + \beta + \gamma$.

$\alpha + \beta + \gamma = \displaystyle -\frac{4}{2} = -2$

## Part 2

Find $\alpha \beta + \beta \gamma + \alpha \gamma$.

$\alpha \beta + \beta \gamma + \alpha \gamma = \displaystyle \frac{5}{2}$

## Part 3

Find $\alpha \beta \gamma$.

$\alpha \beta \gamma = \displaystyle – \frac{6}{2} = -3$

## Part 4

Find $\alpha^2 \beta \gamma + \alpha \beta^2 \gamma + \alpha \beta \gamma^2$.

\begin{align} \alpha^2 \beta \gamma + \alpha \beta^2 \gamma + \alpha \beta \gamma^2 &= \alpha \beta \gamma (\alpha + \beta + \gamma) \\ &= -3 \times -2 \\ &= 6 \end{align}

## Part 5

Find $\alpha^2 + \beta^2 + \gamma^2$.

\displaystyle \begin{align} \alpha^2 + \beta^2 + \gamma^2 &= (\alpha + \beta + \gamma)^2 – 2(\alpha \beta + \beta \gamma + \alpha \gamma) \\ &= (-2)^2 – 2 \times \frac{5}{2} \\ &= 4-5 \\ &= -1 \end{align}

## Part 6

Determine how many of the roots (zeros) of $P(x)$ are real.

Since $P(x)$ has real coefficients, then any complex root occurs in conjugate pairs, in the form $a \pm bi$.
Therefore $P(x)$ has either 3 real roots or 1 real root and 2 complex roots in a conjugate pair.
As $\alpha^2 + \beta^2 + \gamma^2 = -1 \lt 0$, then $\alpha, \beta$ and $\gamma$ cannot be all real.
Thus $P(s)$ must have 1 real root and a pair of complex conjugate roots.
That is, $P(x)$ has only one real root.

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