# Solve the cubic equation:

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

**Quick Answer**

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

$$ \Delta=-0.125$$

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

**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 $$f(x) = 2x³ - 6x² + 3x + 1$$ 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 $$2x³ - 6x² + 3x + 1$$ 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.

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

then, $$x = 1$$ is one of roots of the cubic equaiton $$2x³ - 6x² + 3x + 1 = 0$$. 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 $$2x³ - 6x² + 3x + 1$$ by $$x - 1$$

2x² | -4x | -1 | |||

x - 1 | 2x³ | -6x² | +3x | +1 | |

2x³ | -2x² | ||||

-4x² | +3x | ||||

-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

$$\begin{aligned} \\t&=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}\\ & =\dfrac{-(-4)\pm\sqrt{(-4)^2-4\cdot 2\cdot (-1)}}{2 \cdot 2}\\ & =\dfrac{4\pm2\sqrt{6}}{4}\\ & =1\pm\dfrac{\sqrt{6}}{2}\\ \end{aligned}$$

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

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

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

$$p = \dfrac{3\cdot 2\cdot 3-(-6)^2}{3\cdot 2^2}=-\dfrac{3}{2}$$

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

### 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 $$x=t+1$$

The cubic equation $$2x³ - 6x² + 3x + 1=0$$ is transformed to

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

For this equation, we have $$p=-\dfrac{3}{2}$$ 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.

$$(2t² - 3)t = 0$$

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.

$$2t² - 3 = 0$$

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

## Solve the quadratic equation: $$2x² - 3 = 0$$

$$2t^2 = 3$$

Solving for $$t$$.

Then,

$$\begin{aligned} \\ t&=\pm \sqrt{\dfrac{3}{2}}\\ \end{aligned}$$

That is,

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

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

$$x_1 = t_1+1=1$$

$$x_2 = t_2+1=1+\sqrt{\dfrac{3}{2}}$$

$$x_3 = t_3+1=1-\sqrt{\dfrac{3}{2}}$$

## 3. Summary

In summary, we have tried the method of factorization, cubic root formula to explore the solutions of the equation. The cubic equation $$2x³ - 6x² + 3x + 1=0$$ is found to have three real roots . Exact values and approximations are given below.

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

Convert to decimals,

$$\begin{cases} x_1=1 \\ x_2=2.2247448713916 \\ x_3=-0.22474487139159 \end{cases}$$

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

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

## 4. Graph for the function $$f(x) = 2x³ - 6x² + 3x + 1$$

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