# Solve the cubic equation:

## $$2x^3+6x^2+5x+1=0 $$

**Quick Answer**

Since the discriminant $$ \Delta <0$$, the cubic equation has three distinct real roots.

$$ \Delta=-0.0046296296296296$$

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

**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 $$f(x) = 2x³ + 6x² + 5x + 1$$ 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 $$2x³ + 6x² + 5x + 1$$ 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.

$$f(-1) = 2(-1)³ + 6(-1)² + 5(-1) + 1 = 0$$

then, $$x = -1$$ is one of roots of the cubic equaiton $$2x³ + 6x² + 5x + 1 = 0$$. 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 $$2x³ + 6x² + 5x + 1$$ by $$x + 1$$

2x² | +4x | +1 | |||

x + 1 | 2x³ | +6x² | +5x | +1 | |

2x³ | +2x² | ||||

4x² | +5x | ||||

4x² | +4x | ||||

x | +1 | ||||

x | +1 | ||||

0 |

Now we get another factor of the cubic equation $$2x² + 4x + 1$$

## Solve the quadratic equation: $$2x² + 4x + 1 = 0$$

Given $$a =2, b=4, c=1$$,

Use the root solution formula for a quadratic equation, the roots of the equation are given as

$$\begin{aligned} \\t&=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}\\ & =\dfrac{-4\pm\sqrt{4^2-4\cdot 2\cdot 1}}{2 \cdot 2}\\ & =\dfrac{-4\pm2\sqrt{2}}{4}\\ & =-1\pm\dfrac{\sqrt{2}}{2}\\ \end{aligned}$$

Since the discriminat is greater than zero, we get another two real roots:That is,

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

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

$$p = \dfrac{3\cdot 2\cdot 5-6^2}{3\cdot 2^2}=-\dfrac{1}{2}$$

$$q = \dfrac{2\cdot 6^3-9\cdot2\cdot 6\cdot 5+27\cdot 2^2\cdot1}{27\cdot 2^3}=0$$

### 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 $$x=t-1$$

The cubic equation $$2x³ + 6x² + 5x + 1=0$$ is transformed to

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

For this equation, we have $$p=-\dfrac{1}{2}$$ 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.

$$(2t² - 1)t = 0$$

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.

$$2t² - 1 = 0$$

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

## Solve the quadratic equation: $$2x² - 1 = 0$$

$$2t^2 = 1$$

Solving for $$t$$.

Then,

$$\begin{aligned} \\ t&=\pm \sqrt{\dfrac{1}{2}}\\ &=\pm \sqrt{\dfrac{1^2}{2}}\\ &=\pm \dfrac{1}{\sqrt{2}}\\ \end{aligned}$$

That is,

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

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

$$x_1 = t_1-1=-1$$

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

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

## 3. Summary

In summary, we have tried the method of factorization, cubic root formula to explore the solutions of the equation. The cubic equation $$2x³ + 6x² + 5x + 1=0$$ is found to have three real roots . Exact values and approximations are given below.

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

Convert to decimals,

$$\begin{cases} x_1=-1 \\ x_2=-0.29289321881345 \\ x_3=-1.7071067811865 \end{cases}$$

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

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

## 4. Graph for the function $$f(x) = 2x³ + 6x² + 5x + 1$$

Since the discriminat is less than zero, the curve of the cubic function $$f(x) = 2x³ + 6x² + 5x + 1$$ has 3 intersection points with horizontal axis.