Solve the cubic equation:

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

Quick Answer

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+2x2
x + 0.52x³+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,

$$t_1 = 0$$

and a quadratic equation.

$$4t² + 3 = 0$$

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,

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

Another is a quadratic equation.

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

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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