# Solve the cubic equation:

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

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

$$\Delta=0.25$$

$$\begin{cases} x_1=-2 \\ x_2=-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i \\ x_3=-\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i \end{cases}$$

In decimals,

$$\begin{cases} x_1=-2 \\ x_2=-0.5+0.86602540378444i \\ x_3=-0.5-0.86602540378444i \end{cases}$$

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 $$f(x) = x³ + 3x² + 3x + 2$$ and find the one that makes $$f(x) = 0$$.

According to factor theorem, $$f(n) = 0$$, if and only if the polynomial $$x³ + 3x² + 3x + 2$$ 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.

$$f(-2) = (-2)³ + 3(-2)² + 3(-2) + 2 = 28$$

then we get the first root

$$x_1 = -2$$

And the cubic equation can be factored as

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

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

### Long division

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

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

Now we get another factor of the cubic equation $$x² + x + 1$$

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

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

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{-1\pm\sqrt{1^2-4\cdot 1\cdot 1}}{2 \cdot 1}\\ & =\dfrac{-1\pm\sqrt{-3}}{2}\\ & =-\dfrac{1}{2}\pm\dfrac{\sqrt{3}}{2}i\\ \end{aligned}

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

That is,

$$\begin{cases} t_2 =-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i \\ t_3=-\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i \end{cases}$$

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

$$p = \dfrac{3\cdot 1\cdot 3-3^2}{3\cdot 1^2}=0$$

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

### 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 $$x³ + 3x² + 3x + 2=0$$ is transformed to

$$t^3 +1=0$$

For the equation $$t^3 +1$$, we have $$p=0$$ and $$q = 1$$

### Calculate the discriminant

The nature of the roots are determined by the sign of the discriminant.

\begin{aligned} \\\Delta&=\dfrac{q^2}{4}+\dfrac{p^3}{27}\\ & =\dfrac{1^2}{4}+0\\ & =\dfrac{1}{4}\\ & =0.25\\ \end{aligned}

### 2.1 Use the root formula directly

If the discriminant is greater than zero, we can use the root formula to determine the roots of the cubic equation.

$$t_{1,2,3} =\begin{cases} \sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } } +\sqrt[3]{-\dfrac{q}{2} -\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }}& \\ ω\cdotp \sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } } + \overline{ω} \sqrt[3]{-\dfrac{q}{2} -\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }}&\ \\ \overline{ω}\cdotp \sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } } + ω\cdotp \sqrt[3]{-\dfrac{q}{2} -\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }} \end{cases}$$

in which, $$ω = \dfrac{-1+i\sqrt{3}}{2}$$ and $$\overline{ω} =\dfrac{-1-i\sqrt{3}}{2}$$

Substitute the values of $$p, q$$ and $$\Delta$$ which we have calculated. Then,

\begin{aligned} \\t_1&=\sqrt[3]{-\dfrac{1}{2}+\sqrt{\dfrac{1}{4}}}+\sqrt[3]{-\dfrac{1}{2}-\sqrt{\dfrac{1}{4}}}\\ & =\sqrt[3]{-\dfrac{1}{2}+\dfrac{1}{2}}+\sqrt[3]{-\dfrac{1}{2}-\dfrac{1}{2}}\\ & =-1\\ \end{aligned}

If we denote

$$R = -\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }$$

$$\overline{R} = -\dfrac{q}{2} -\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }$$

then,

$$\sqrt[3]{R} = 0$$, $$\sqrt[3]{\overline{R}} =-1$$

\begin{aligned} \\t_2&= ω\cdotp \sqrt[3]{R}+ \overline{ω} \sqrt[3]{\overline{R} }\\ & =\dfrac{-\sqrt[3]{R}-\sqrt[3]{\overline{R} }}{2} +\dfrac{\sqrt{3}( \sqrt[3]{R} - \sqrt[3]{\overline{R} }) }{2} i\\ & =\dfrac{1}{2}\Big[0-\Big(-1\Big)\Big]+\dfrac{\sqrt{3}}{2}\Big[0-\Big(-1\Big)\Big]i\\ & =\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\\ \end{aligned}

\begin{aligned} \\t_3&= \overline{ω}\cdotp \sqrt[3]{R}+ ω\cdotp \sqrt[3]{\overline{R}}\\ & =\dfrac{-\sqrt[3]{R}-\sqrt[3]{\overline{R} }}{2} +\dfrac{\sqrt{3}(- \sqrt[3]{R} + \sqrt[3]{\overline{R} }) }{2}i \\ & =\dfrac{1}{2}\Big[0-\Big(-1\Big)\Big]-\dfrac{\sqrt{3}}{2}\Big[0-\Big(-1\Big)\Big]i\\ & =\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}\\ \end{aligned}

## Roots of the general cubic equation

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

$$x_1 = t_1-1$$

$$x_2 = t_2-1$$

$$x_3 = t_3-1$$

## 3. Summary

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

$$\begin{cases} x_1=-1-1 \\ x_2=\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}-1 \\ x_3=\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}-1 \end{cases}$$

in decimal notation,

$$\begin{cases} x_1=-2 \\ x_2=-0.5+0.86602540378444i \\ x_3=-0.5-0.86602540378444i \end{cases}$$

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

$$\begin{cases} x_1=-2 \\ x_2=-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i \\ x_3=-\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i \end{cases}$$

The decimal results by using factorization method are

$$\begin{cases} x_1=-2 \\ x_2=-0.5+0.86602540378444i \\ x_3=-0.5-0.86602540378444i \end{cases}$$

## 4. Math problems derived by the cubic equation

It is found that the methods of factorizaiton and using cubic root formula yield the roots for the cubic equation in different forms. However, their deicmal results show that they are equal in values. Comparing the real roots, we get the equation.

$$-1-1=-2$$

Similarly, comparison of the imaginary parts of complex roots yields the following equation.

$$=\sqrt{\dfrac{3}{2}}$$

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

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

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