Solve the cubic equation:

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

Quick Answer

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 $$x + 1$$

+2x2
x + 1+3x²+4x2
+x²
2x²+4x
2x²+2x
2x2
2x2
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,

$$t_1 = 0$$

and a quadratic equation.

$$t² + 1 = 0$$

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.

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