Instruction

1

Consider the cubic equation Ax3+Bx2+Cx+D=0 where A≠0. Find root of

**equation**by trial and error. Please note that one of the roots of**equations**of the third degree is always a divisor of the free term.2

Find all divisors of the coefficient D, that is, all integers (positive and negative) that the constant D is divisible. Substitute them one by one into the original equation in place of the variable x. Find the number x1, which turns the equation into right equality. It will be one of the roots of the cubic

**equation**. Just at cubic**equation**with three roots (both real and complex).3

Divide the polynomial by Ax3+Bx2+Cx+D on the binomials (x-x1). The result of the division will square polynomial ax2+bx+c, the remainder will be zero.

4

Paranaita the obtained polynomial to zero: ax2+bx+c=0. Find

**the roots**of this square**equation**formula x2=(-b+√(b2−4ac))/(2a), x3=(-b−√(b2−4ac))/(2a). They will also be roots of the original cubic**equation**.5

Let's consider an example. Let the equation of the third degree 2x3−11x2+12x+9=0. A=2≠0 and the constant D=9. Find all divisors of the coefficient D: 1, -1, 3, -3, 9, -9. Substitute these factors in the equation instead of the unknown x. It turns out, 2×13-11×12+12×1+9=12≠0; 2×(-1)3-11×(-1)2+12×(-1)+9=-16≠0; 2×33-11×32+12×3+9=0. Thus, one of the roots of this cubic

**equation**x1=3. Now let's divide both sides of the original**equation**to the binomials (x−3). The result is the quadratic equation: 2x2−5x−3=0, that is, a=2, b=-5, c=-3. Find its**roots**: x2=(5+√((-5)2-4×2×(-3)))/(2×2)=3, x3=(5−√((-5)2-4×2×(-3)))/(2×2)=-0,5. Thus, the cubic equation 2x3−11x2+12x+9=0 has real**roots**x1=x2=3 and x3=-0,5.# Advice 2 : How to solve equation third degree

Equations of the third degree is also called cubic equations. This equation, in which the highest degree when variable x is a cube (3).

Instruction

1

Cubic equation in General form looks like: ax3 + bx2 + cx + d = 0, a does not equal 0; a, b, c, d are real numbers. A universal method of solving equations of the third degree method is Cardano.

2

For starters, here is the equation to the form y3 + py + q = 0. To do this, replace the variable x by y - b/3a. Substitution replacement, see in the picture. To disclose the brackets used two formulas of reduced multiplication: (a-b)3 = a3 - 3a2b + 3ab2 - b3 and (a-b)2 = a2 - 2ab + b2. Then, given similar terms and grouped by powers of the variable y.

3

Now, to get under y3 unit ratio, divide all equation on a. We obtain the following formulas for the coefficients p and q in the equation y3 + py + q = 0.

4

Then computed the special values: Q, α, β, which will allow to calculate the roots of the equation with y.

5

Then the three roots of the equation y3 + py + q = 0 are calculated by the formulas in the figure.

6

If Q > 0, the equation y3 + py + q = 0 has only one real root y1 = α + β (and two complex, evaluate them according to prescribed formulas, if necessary).

If Q = 0, all roots real and at least two of them coincide, with α = β and equal roots: y1 = 2α, y2 = y3 = -α.

If Q < 0 then the roots are real, but you need the ability to extract the root of a negative number.

After finding y1, y2 and y3, substitute them into the substitution x = y - b/3a and find the roots of the original equation.

If Q = 0, all roots real and at least two of them coincide, with α = β and equal roots: y1 = 2α, y2 = y3 = -α.

If Q < 0 then the roots are real, but you need the ability to extract the root of a negative number.

After finding y1, y2 and y3, substitute them into the substitution x = y - b/3a and find the roots of the original equation.

Useful advice

If you can find one of the roots of the cubic equation x1, it can be a cubic polynomial divided by (x - x1) and solve the resulting quadratic equation.

# Advice 3 : How to solve cubic equation

Today, the world knows several ways to solve a cubic equation. The most popular are the formula of Cardan and trigonometric formula of vieta. However, these methods are rather complicated and in practice almost never used. The following is the easiest way to solve the cubic equation.

Instruction

1

So, in order to solve the cubic equation of the form Ах3+Вх2+CX+D=0, you need brute force to find one of the roots of the equation. The root of the cubic equation is always one of the divisors of the free term of the equation. Thus, in the first stage the solution of the equation, you need to find all integers a for which the constant D is divisible.

2

Received integers are alternately substituted into the cubic equation instead of the unknown variable x. The number that draws the equality of the faithful, is the root of the equation.

3

One of the roots of the equation found. For further solutions to apply the method of dividing a polynomial of the binomials. Polynomial Ах3+Вх2+CX+D is divisible, and the binomials x-h where h - the first root of the equation XX). The result of the division will be a square polynomial of the form ах2+bx+C.

4

Equating the obtained polynomial to zero ах2+bx+C =0, you get a quadratic equation, the roots of which and will be the solution to the original cubic equation, i.e. x₂' ₃ ơ=(-b±√(b^2-4ac))/2a

Note

In the first stage equation, namely, finding the root of equation by the method of selection, we should not forget about the whole negative numbers, which can also be a solution of the equation.

# Advice 4 : How to solve equation with a cube

For the solution of cubic equations developed by several mathematical methods. The method of substitution or replacement of cube auxiliary variable, and a number of iterative methods, in particular Newton's method. But the classical solution of the cubic

**equation**is expressed in the use of the formulas of vieta and Cardano. The method of vieta-Cardano is based on the use of the formula of the cube of the sum of the coefficients and is applicable for any kind of cubic**equation**. To search for the roots**of the equation**, the entry must be represented in the form: x3+a*x2+b*x+c=0 where a is not zero.Instruction

1

Make a note of the original cubic equation in the form: x3+a*x2+b*x+c=0. To do this, all the coefficients

**of the equation**divide by the first coefficient with a multiplier of x3 so that it became equal to one.2

Based on the algorithm of the method of vieta, cardan, calculate the values of R and Q according to the respective formulae: Q =(a2-3b)/9, R=(2a3-9ab+27c)/54. Moreover, the coefficients a, b and C are the coefficients of the given

**equation**.3

Compare the obtained values of R and Q. If it is the expression Q3 >R2 , therefore, in the initial equation there are 3 real roots. Calculate them by the formulas of vieta.

4

When Q3 values <= R2 , the solution is one real root x1 and two complex conjugate roots. To define them we need to find intermediate values A and B. Calculate their formulas Cardano.

5

Find the first valid root by the formula x1=(B + A) a/3. For different values of A and b define the complex conjugate roots of the cubic

**equation**by appropriate formulas.6

If the values of A and b was equal, conjugate roots, degenerate in the second real root of the original

**equation**. This is the case when a valid root turns two. Calculate second real root by the formula x2=-A-A/3.