﻿ Solve x^3+3x^2+4x+2=0 | Uniteasy.com

# Solve the cubic equation:

## $$x^3+3x^2+4x+2=0$$

Since the discriminant $$\Delta >0$$, the cubic equation has one real root and two conjugate complex roots.

$$\Delta=0.037037037037037$$

$$\begin{cases} x_1=-1 \\ x_2=-1+1i \\ x_3=-1-1i \end{cases}$$

Detailed Steps on Solution

A cubic equation has at least one real root. If the coefficient of leading term is 1, one of solutions could be a factor of the constant term.

## 1. Factorization Method

Find all possible factors for constant

$$1, 2$$

$$-1, -2$$

Substitute the factors to the function $$f(x) = x³ + 3x² + 4x + 2$$ and find the one that makes $$f(x) = 0$$.

According to factor theorem, $$f(n) = 0$$, if and only if the polynomial $$x³ + 3x² + 4x + 2$$ has a factor $$x-n$$, that is, $$x=n$$ is a root of the equation.

Fortunetely, one of the numbers is found to make the equation equal.

$$f(-1) = (-1)³ + 3(-1)² + 4(-1) + 2 = 10$$

then we get the first root

$$x_1 = -1$$

And the cubic equation can be factored as

$$(x +1)(ax^2+bx+c) = 0$$

Next we can use either long division or synthetic division to determine the expression of trinomial

### Long division

Divide the polynomial $$x³ + 3x² + 4x + 2$$ by $$x + 1$$

 x² +2x 2 x + 1 x³ +3x² +4x 2 x³ +x² 2x² +4x 2x² +2x 2x 2 2x 2 0

Now we get another factor of the cubic equation $$x² + 2x + 2$$

## Solve the quadratic equation: $$x² + 2x + 2 = 0$$

Given $$a =1, b=2, c=2$$,

Use the root solution formula for a quadratic equation, the roots of the equation are given as

\begin{aligned} \\t&=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}\\ & =\dfrac{-2\pm\sqrt{2^2-4\cdot 1\cdot 2}}{2 \cdot 1}\\ & =\dfrac{-2\pm2\sqrt{-1}}{2}\\ & =-1\pm1i\\ \end{aligned}

Since the discriminat is less than zero, we get another two complex roots:

That is,

$$\begin{cases} t_2 =-1+1i \\ t_3=-1-1i \end{cases}$$

Another method to find the roots of a general cubic equation is simplifing it to a depressed form. This method is applicable to all cases, especially to those difficult to find factors.

## 2. Convert to depressed cubic equation

The idea is to convert general form of cubic equation

$$ax^3+bx^2+cx+d = 0$$

to the form without quadratic term.

$$t^3+pt+q = 0$$

By substituting $$x$$ with $$t - \dfrac{b}{3a}$$, the general cubic equation could be transformed to

$$t^3+\dfrac{3ac-b^2}{3a^2}t+\dfrac{2b^3-9abc+27a^2d}{27a^3} = 0$$

Compare with the depressed cubic equation. Then,

$$p = \dfrac{3ac-b^2}{3a^2}$$

$$q = \dfrac{2b^3-9abc+27a^2d}{27a^3}$$

Substitute the values of coefficients, $$p, q$$ is obtained as

$$p = \dfrac{3\cdot 1\cdot 4-3^2}{3\cdot 1^2}=1$$

$$q = \dfrac{2\cdot 3^3-9\cdot1\cdot 3\cdot 4+27\cdot 1^2\cdot2}{27\cdot 1^3}=0$$

### Use the substitution to transform

Let $$p$$ and $$q$$ being the coefficient of the linean and constant terms, the depressed cubic equation is expressed as.

$$t^3 +pt+q=0$$

Let $$x=t-1$$

The cubic equation $$x³ + 3x² + 4x + 2=0$$ is transformed to

$$t^3 +t=0$$

For this equation, we have $$p=1$$ and $$q = 0$$

## Factor out common factor

Since each term contains a common factor, we can factor out the common factor $$x$$ out of each term.

$$(t² + 1)t = 0$$

In order for the equation to hold true, either of the factors is equal to 0. Then, we get the first root,

$$t_1 = 0$$

$$t² + 1 = 0$$

And then, the problem turns to solving a quadratic equation.

## Solve the quadratic equation: $$x² + 1 = 0$$

$$t^2 = -1$$

Solving for $$t$$.

Then,

\begin{aligned} \\ t&=\pm i\\ \end{aligned}

That is,

$$\begin{cases} t_2 =i \\ t_3=-i \end{cases}$$

Since $$x = t - \dfrac{b}{3a}$$, substituting the values of $$t$$, $$a$$ and $$b$$ gives

$$x_1 = t_1-1=-1$$

$$x_2 = t_2-0=-1+$$

$$x_3 = t_3-0=-1$$

## 3. Summary

In summary, we have tried the method of factorization, cubic root formula to explore the solutions of the equation. The cubic equation $$x³ + 3x² + 4x + 2=0$$ is found to have one real root and two complex roots. Exact values and approximations are given below.

$$\begin{cases} x_1=-1 \\ x_2=-1+ \\ x_3=-1 \end{cases}$$

Using the method of factorization, the roots are derived to the following forms

$$\begin{cases} x_1=-1 \\ x_2=-1+1i \\ x_3=-1-1i \end{cases}$$

The decimal results by using factorization method are

$$\begin{cases} x_1=-1 \\ x_2=-1+1i \\ x_3=-1-1i \end{cases}$$

## 4. Graph for the function $$f(x) = x³ + 3x² + 4x + 2$$

Since the discriminat is greater than zero, the curve of the cubic function $$f(x) = x³ + 3x² + 4x + 2$$ has one intersection point with the x-axis.

Scroll to Top