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# Solve the cubic equation:

## $$2x^3+3x^2+3x+1=0$$

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

$$\Delta=0.015625$$

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

In decimals,

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

Detailed Steps on Solution

If a cubic equation has a rational root, the root could be a fracion number with a factor of constant term as the numerator and a factor of the coefficient of leading term as the denumerator.

## 1. Factorization Method

Find all possible factors for constant $$1$$, which are,

$$1$$

$$-1$$

and all possible factors of the coefficient of the leading term $$2$$,

$$1, 2$$

$$-1, -2$$

Dividing factors of constant term by those of the cubic term one by one gives the following fractions.

$$1, -1, \dfrac{1}{2}, -\dfrac{1}{2},$$

Substitute the fractions to the function $$f(x) = 2x³ + 3x² + 3x + 1$$ one by one to find out the one that makes $$f(x) = 0$$.

According to the factor theorem, $$f(n) = 0$$, if and only if the polynomial $$2x³ + 3x² + 3x + 1$$ has a factor $$x-n$$, that is, $$x=n$$ is the root of the equation.

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

$$f(-\dfrac{1}{2}) = 2(-\dfrac{1}{2})³ + 3(-\dfrac{1}{2})² + 3(-\dfrac{1}{2}) + 1 = 0$$

then, $$x = -\dfrac{1}{2}$$ is one of roots of the cubic equaiton $$2x³ + 3x² + 3x + 1 = 0$$. And, the equation can be factored as

$$(x +\dfrac{1}{2})(ax^2+bx+c) = 0$$

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

### Long division

Divide the polynomial $$2x³ + 3x² + 3x + 1$$ by $$x + 0.5$$

 2x² +2x 2 x + 0.5 2x³ +3x² +3x +1 2x³ +x² 2x² +3x 2x² +x 2x +1 2x +1 0

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

Removing common factor $$2$$ from each term gives $$x² + x + 1$$

## 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

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

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

### 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 $$x=t-\dfrac{1}{2}$$

The cubic equation $$2x³ + 3x² + 3x + 1=0$$ is transformed to

$$t^3 +\dfrac{3}{4}t=0$$

For this equation, we have $$p=\dfrac{3}{4}$$ 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.

$$(4t² + 3)t = 0$$

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

$$t_1 = 0$$

$$4t² + 3 = 0$$

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

## Solve the quadratic equation: $$4x² + 3 = 0$$

$$4t^2 = -3$$

Solving for $$t$$.

Then,

\begin{aligned} \\ t&=\pm \sqrt{\dfrac{3}{4}}i\\ &=\pm \sqrt{\dfrac{3}{2^2}}i\\ &=\pm \dfrac{\sqrt{3}}{2}i\\ \end{aligned}

That is,

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

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

$$x_1 = t_1-\dfrac{1}{2}=-\dfrac{1}{2}$$

$$x_2 = t_2-0=-\dfrac{1}{2}+$$

$$x_3 = t_3-0=-\dfrac{1}{2}$$

## 3. Use the sum of cubes identity

Evaluate the coefficients of the cubic equation.

$$2x³ + 3x² + 3x + 1=0$$

The coefficients for the quadratic tand linear terms are equal and multiples of three. And they are threes times of the constant..

Then, we can make the following manipulation.

### 3.1 Factorization

Break up the leading term $$2x^3$$ to $$x^3 +x^3$$. Then,

$$x^3+x^3+3x² + 3x + 1=0$$

Group the last four terms .

$$x^3 +(x^3+3x² + 3x + 1)=0$$

Apply the sum of cubes identity on the second term.

$$x^3+(x+1)^3=0$$

Factorize by using the sum of cubes identity.

$$[x+(x+1)][(x)^2-x\cdot (x+1)+((x+1))^2]=0$$

Then we get two equations. One is a linear equation,

$$x+(x+1)=0$$

$$(x)^2-x\cdot (x+1)+((x+1))^2=0$$

### 3.2 Sove the linear equation

Remove parentheses from the linear equation

$$x+x+1=0$$

Collecting the like terms gives

$$2x+1=0$$

Then the first root is found as,

\begin{aligned} \\x_1&=\dfrac{-1}{2}\\ \end{aligned}

### 3.3 Sove the quadratic equation

$$(x)^2-x\cdot (x+1)+((x+1))^2=0$$

Expand and simplify the equation. We get a quadratic equation with radicals.

$$x^2+x+1=0$$

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,

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

## 4. Summary

In summary, we have tried the method of factorization, cubic root formula, sum of cubes to explore the solutions of the equation. The cubic equation $$2x³ + 3x² + 3x + 1=0$$ is found to have one real root and two complex roots. Exact values and approximations are given below.

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

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

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

The decimal results by using factorization method are

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

## 5. Graph for the function $$f(x) = 2x³ + 3x² + 3x + 1$$

Since the discriminat is greater than zero, the curve of the cubic function $$f(x) = 2x³ + 3x² + 3x + 1$$ has one intersection point with the x-axis.

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