Solve the cubic equation:

$$3x^3+6x^2+4x+1=0$$

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

$$\Delta=0.00034293552812071$$

$$\begin{cases} x_1=-1 \\ x_2=-\dfrac{1}{2}+\dfrac{\sqrt{3}}{6}i \\ x_3=-\dfrac{1}{2}-\dfrac{\sqrt{3}}{6}i \end{cases}$$

In decimals,

$$\begin{cases} x_1=-1 \\ x_2=-0.5+0.28867513459481i \\ x_3=-0.5-0.28867513459481i \end{cases}$$

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 $$3$$,

$$1, 3$$

$$-1, -3$$

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

$$1, -1, \dfrac{1}{3}, -\dfrac{1}{3},$$

Substitute the fractions to the function $$f(x) = 3x³ + 6x² + 4x + 1$$ 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 $$3x³ + 6x² + 4x + 1$$ 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.

$$f(-1) = 3(-1)³ + 6(-1)² + 4(-1) + 1 = 0$$

then, $$x = -1$$ is one of roots of the cubic equaiton $$3x³ + 6x² + 4x + 1 = 0$$. And, the 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 $$3x³ + 6x² + 4x + 1$$ by $$x + 1$$

 3x² +3x +1 x + 1 3x³ +6x² +4x +1 3x³ +3x² 3x² +4x 3x² +3x x +1 x +1 0

Now we get another factor of the cubic equation $$3x² + 3x + 1$$

Solve the quadratic equation: $$3x² + 3x + 1 = 0$$

Given $$a =3, b=3, c=1$$,

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{-3\pm\sqrt{3^2-4\cdot 3\cdot 1}}{2 \cdot 3}\\ & =\dfrac{-3\pm\sqrt{-3}}{6}\\ & =-\dfrac{1}{2}\pm\dfrac{\sqrt{3}}{6}i\\ \end{aligned}

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

That is,

$$\begin{cases} t_2 =-\dfrac{1}{2}+\dfrac{\sqrt{3}}{6}i \\ t_3=-\dfrac{1}{2}-\dfrac{\sqrt{3}}{6}i \end{cases}$$

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

$$p = \dfrac{3\cdot 3\cdot 4-6^2}{3\cdot 3^2}=0$$

$$q = \dfrac{2\cdot 6^3-9\cdot3\cdot 6\cdot 4+27\cdot 3^2\cdot1}{27\cdot 3^3}=\dfrac{1}{27}$$

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 $$x=t-\dfrac{2}{3}$$

The cubic equation $$3x³ + 6x² + 4x + 1=0$$ is transformed to

$$t^3 +\dfrac{1}{27}=0$$

For the equation $$t^3 +\dfrac{1}{27}$$, we have $$p=0$$ and $$q = \dfrac{1}{27}$$

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{\Big(\dfrac{1}{27}\Big)^2}{4}+0\\ & =\dfrac{1}{2916}\\ & =0.00034293552812071\\ \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}{54}+\sqrt{\dfrac{1}{2916}}}+\sqrt[3]{-\dfrac{1}{54}-\sqrt{\dfrac{1}{2916}}}\\ & =\sqrt[3]{-\dfrac{1}{54}+\dfrac{1}{54}}+\sqrt[3]{-\dfrac{1}{54}-\dfrac{1}{54}}\\ & =\sqrt[3]{\dfrac{-1}{2\cdot 3^3}+\dfrac{1}{2\cdot 3^3}}+\sqrt[3]{\dfrac{-1}{2\cdot 3^3}-\dfrac{1}{2\cdot 3^3}}\\ & =\dfrac{1}{3}\Big(\sqrt[3]{\dfrac{-1}{2}+\dfrac{1}{2}}+\sqrt[3]{\dfrac{-1}{2}-\dfrac{1}{2}}\Big)\\ & =-\dfrac{1}{3}\\ \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,

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

\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(-\dfrac{1}{3}\Big)\Big]+\dfrac{\sqrt{3}}{2}\Big[0-\Big(-\dfrac{1}{3}\Big)\Big]i\\ & =\dfrac{1}{6}+\dfrac{\sqrt{3}}{6}\\ \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(-\dfrac{1}{3}\Big)\Big]-\dfrac{\sqrt{3}}{2}\Big[0-\Big(-\dfrac{1}{3}\Big)\Big]i\\ & =\dfrac{1}{6}-\dfrac{\sqrt{3}}{6}\\ \end{aligned}

Roots of the general cubic equation

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

$$x_1 = t_1-\dfrac{2}{3}$$

$$x_2 = t_2-\dfrac{2}{3}$$

$$x_3 = t_3-\dfrac{2}{3}$$

3. Summary

In summary, we have tried the method of factorization, cubic root formula to explore the solutions of the equation. The cubic equation $$3x³ + 6x² + 4x + 1=0$$ is found to have one real root and two complex roots. Exact values and approximations are given below.

$$\begin{cases} x_1=-\dfrac{1}{3}-\dfrac{2}{3} \\ x_2=\dfrac{1}{6}+\dfrac{\sqrt{3}}{6}-\dfrac{2}{3} \\ x_3=\dfrac{1}{6}-\dfrac{\sqrt{3}}{6}-\dfrac{2}{3} \end{cases}$$

in decimal notation,

$$\begin{cases} x_1=-1 \\ x_2=-0.5+0.28867513459481i \\ x_3=-0.5-0.28867513459481i \end{cases}$$

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

$$\begin{cases} x_1=-1 \\ x_2=-\dfrac{1}{2}+\dfrac{\sqrt{3}}{6}i \\ x_3=-\dfrac{1}{2}-\dfrac{\sqrt{3}}{6}i \end{cases}$$

The decimal results by using factorization method are

$$\begin{cases} x_1=-1 \\ x_2=-0.5+0.28867513459481i \\ x_3=-0.5-0.28867513459481i \end{cases}$$

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.

$$-\dfrac{1}{3}-\dfrac{2}{3}=-1$$

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

$$=\sqrt{\dfrac{1}{2}}$$

5. Graph for the function $$f(x) = 3x³ + 6x² + 4x + 1$$

Since the discriminat is greater than zero, the curve of the cubic function $$f(x) = 3x³ + 6x² + 4x + 1$$ has one intersection point with the x-axis.

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