Solve the cubic equation:

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

Quick Answer

Since the discriminant $$ \Delta=0$$, the cubic equation has three real roots, of which two are equal.

$$ \Delta=0$$

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

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 $$f(x) = x³ - 3x + 2$$ and find the one that makes $$f(x) = 0$$.

According to factor theorem, $$f(n) = 0$$, if and only if the polynomial $$x³ - 3x + 2$$ 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.

$$f(1) = 1³ - 3*1 + 2 = 0$$

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 $$x³ - 3x + 2$$ by $$x - 1$$

+x-2
x - 10x²-3x2
-x²
-3x
-x
-2x2
-2x2
0

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

Solve the quadratic equation: $$x² + x - 2 = 0$$

Find factors of trinomial

Create a table of all factors for constant of $$-2$$

$$\begin{array}{|c|} \hline1\times( -2)\\ \hline(-1)\times 2\\ \hline\end{array}$$

Create another table of all factors for coeffcient of quadratic term $$1$$

$$\begin{array}{|c|} \hline1\times1\\ \hline(-1)\times (-1)\\ \hline\end{array}$$

Determine the roots by factorizing

Cross multiply the pairs from constant and leading term to find pairs of which the sum of products is equal to the coefficient of linear term 1

$$\begin{array}{|c|} \hline1\times1 + 1\times(-2) = -1\\ \hline1\times(-2) + 1\times1 = -1\\ \hline1\times(-1) + 1\times2 = 1\\ \hline\end{array}$$

The pair $$1\times(-1) + 1\times2 = 1$$ is found to satisfy the rule.

Then, the quadratic equaiton is factored as

$$(x + 2)(x - 1) = 0$$

which results in another two roots

$$x_2 = -2$$

$$x_3 = 1$$

That is,

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

2. Cardano's solution

Let $$x=u-v$$

Cube both sides and extract common factor from two middle terms after expanding the bracket.

$$\begin{aligned} \\x^3&=(u-v)^3\\ & =u^3-3u^2v+3uv^2-v^3\\ & =-3uv(u-v)+u^3-v^3\\ \end{aligned}$$

Since $$u-v=x$$, substitution gives a linear term for the equation. Rearrange terms.

$$x^3+3uvx-u^3+v^3=0$$

Compare the cubic equation with the original one (1)

$$\begin{cases} 3uv=-3\quad\text{or}\quad v=-\dfrac{1}{u}\\ v^3-u^3=2\\ \end{cases}$$

$$v=-\dfrac{1}{u}$$ gives relationship between the two variables. Substitute the value of $$v$$ to the second equation

$$\Big(-\dfrac{1}{u}\Big)^3-u^3=2$$

Simplifying gives,

$$u^3+1\dfrac{1}{u^3}+2=0$$2

Let $$m=u^3$$, then the equation is transformed to a quadratic equation in terms of $$m$$. Once the value of $$m$$ is determined, $$v^3$$ could be determined by $$v^3=2+u^3$$.

$$m^2+2m+1=0$$

Sovling the quadratic euqation will give two roots (some may be equal). Here we only cosider one case with positive sign before the square root radical since the negative case will produce the same result.

$$\begin{aligned} \\u^3=m&=-1+\dfrac{1}{2}\sqrt{\Big(2^2\Big)-4\cdot 1}\\ & =-1+\dfrac{1}{2}\sqrt{4-4}\\ & =-1\\ \end{aligned}$$

$$\begin{aligned} \\v^3&=2+u^3\\ & =2-1\\ & =1\\ \end{aligned}$$

Taking cubic root offor $$u^3$$ and $$v^3$$ gives 3 roots.

$$\begin{cases} u_1=-1\\ u_2=\dfrac{-1+i\sqrt{3}}{2}\cdot -1\\ u_3=\dfrac{-1-i\sqrt{3}}{2}\cdot -1\\ \end{cases}$$

$$\begin{cases} v_1=1\\ v_2=\dfrac{-1-i\sqrt{3}}{2}\cdot 1\\ v_3=\dfrac{-1+i\sqrt{3}}{2}\cdot 1\\ \end{cases}$$

Subtracting $$v$$ from $$u$$ yields the roots of the equation.

$$\begin{aligned} \\x_1&=u_1-v_1\\ & =-1-1\\ & =-2\\ \end{aligned}$$

$$\begin{aligned} \\x_2&=u_2-v_2\\ & =\dfrac{-1+i\sqrt{3}}{2}\cdot -1-\dfrac{-1-i\sqrt{3}}{2}\cdot 1\\ & =1\\ \end{aligned}$$

$$\begin{aligned} \\x_3&=u_3-v_3\\ & =\dfrac{-1-i\sqrt{3}}{2}\cdot -1-\dfrac{-1+i\sqrt{3}}{2}\cdot 1\\ & =1\\ \end{aligned}$$

3. Vieta's Substitution

In Cardano' solution, $$x$$ is defined as the difference of $$u$$ and $$v$$. If we substitute the value of $$v$$ (4) into (2), we get the equation. $$x=u+\dfrac{1}{u}$$. And then substitute the equation to the cubic equation $$x^3-3x+2=0$$. This method is called Vieta's Substitution for solving a cubic equation, which simplied the Cardano' solution. The substitution expression can be obtained by the following formula directly.

$$x=u-\dfrac{p}{3u}$$

Substitute the expression $$x=u+\dfrac{1}{u}$$ to the cubic equation

$$\Big(u+\dfrac{1}{u}\Big)^3-3\Big(u+\dfrac{1}{u}\Big)+2=0$$

Expand brackets and cancel the like terms

$$u^3+\cancel{3u^2\dfrac{1}{u}}+\cancel{3u\dfrac{1}{u^2}}+1\dfrac{1}{u^3}-\cancel{3u}-\cancel{3\dfrac{1}{u}}+2=0$$

Then we get the same equation as (2)

$$u^3+1\dfrac{1}{u^3}+2=0$$

The rest of the steps will be the same as those of Cardano's solution

4. Euler's Solution

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

Move the linear term and constant of (1) to its right hand side. We get the following form of the equation.

$$x^3=3x-2 $$3

Let the root of the cubic equation be the sum of two cubic roots

$$x=\sqrt[3]{r_1}+\sqrt[3]{r_2} $$4

in which $$r_1$$ and $$r_2$$ are two roots of a quadratic equation

$$z^2-\alpha z+ β=0 $$5

Using Vieta's Formula, the following equations are established.

$$r_1+r_2 = \alpha \quad \text{and} \quad r_1r_2 = β $$

To determine $$\alpha$$, $$β$$, cube both sides of the equation (4)

$$x^3=3\sqrt[3]{r_1r_2}(\sqrt[3]{r_1}+\sqrt[3]{r_2})+r_1+r_2 $$

Substituting, the equation is simplified to

$$x^3=3\sqrt[3]{β}x+\alpha $$

Compare the cubic equation with (3), the following equations are established

$$\begin{cases} 3\sqrt[3]{β}=3\\ \alpha=-2\\ \end{cases}$$

Solving for $$β$$ gives

$$β=1 $$

So the quadratic equation (5) is determined as

$$z^2+2z+1=0$$6

Solving the quadratic equation yields

$$\begin{cases} r_1=-1\approx-1\\ r_2=-1\approx-1\\ \end{cases}$$

Therefore, one of the roots of the cubic equation could be obtained from (4).

$$x_1=\sqrt[3]{-1}+\sqrt[3]{-1} $$

in decimals,

$$x_1=-2 $$

However, since the cube root of a quantity has triple values,

The other two roots could be determined as,

$$x_2=\dfrac{-1+i\sqrt{3}}{2}\sqrt[3]{-1}+\dfrac{-1-i\sqrt{3}}{2}\sqrt[3]{-1} $$

$$x_3=\dfrac{-1-i\sqrt{3}}{2}\sqrt[3]{-1}+\dfrac{-1+i\sqrt{3}}{2}\sqrt[3]{-1} $$

Combining the real and imaginary parts results in the same result as that obtained by Cardano's solution.

5. Cubic Root Formula

The equation $$x³ - 3x + 2=0$$ has no quadratic term, compared with the general cubic equation. We can use the root formula to calculate the roots direvtly.

$$x^3 +px+q=0$$

For the equation $$$$, we have $$p=-3$$ and $$q = 2$$

Calculate the discriminant

The nature of the roots are determined by the sign of the discriminant.

Since $$p$$ is negative, the discriminant will be less than zero if the absolute value of $$p$$ is large enough.

$$\begin{aligned} \\\Delta&=\dfrac{q^2}{4}+\dfrac{p^3}{27}\\ & =\dfrac{2^2}{4}+\dfrac{(-3)^3}{27}\\ & =1-1\\ & =0\\ \end{aligned}$$

5.1 Use the root formula directly

If $$\Delta = 0$$, then the cubic equation has three real roots, of which two are equal.

Substitute the values of p and q to the root formula, then we get,

$$\begin{aligned} \\t_1&=\dfrac{3q}{p}\\ & =\dfrac{3\cdot 2}{-3}\\ & =-2\\ \end{aligned}$$

$$\begin{aligned} \\t_2=t_3&=-\dfrac{3q}{2p}\\ & =\dfrac{3\cdot 2}{2\cdot -3}\\ & =1\\ \end{aligned}$$

6. Summary

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

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

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

7. Graph for the function $$f(x) = x³ - 3x + 2$$

Since the discriminat is equal to zero, the curve of the cubic function $$f(x) = x³ - 3x + 2$$ has one intersection point with the x-axis.

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