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.
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).
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.
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.
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 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.
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.
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.
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.
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
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.