Several methods are developed for the solution of the cubic equations (polynomial equations of the third degree). The most known of them are based on application of formulas of Viete and Cardan. But except these ways there is simpler algorithm of finding of roots of the cubic equation.
Instruction
1. Consider the cubic equation of a type of Ax³+Bx²+Cx+D=0 where A≠0. Find an equation root a trial and error method. Take into account that one of the third degrees equation roots is always a divider of the free member.
2. Find all dividers of coefficient D, that is all integers (positive and negative) into which the free member of D is divided without the rest. Set up them serially in the initial equation to the place of variable x. Find that number of x1 at which the equation addresses in right equality. It will also be one of roots of the cubic equation. In total at the cubic equation three roots (both material, and complex).
3. Divide a polynomial into Ax³+Bx²+Cx+D on a binomial (x-x1). As a result of division the square polynomial of ax²+bx+c will turn out, the rest will be equal to zero.
4. Equate the received polynomial to zero: ax²+bx+c=0. Find roots of this quadratic equation on formulas x2= (-b+ √ (b²−4ac)) / (2a), x3= (-b− √ (b²−4ac)) / (2a). They will also be roots of the initial cubic equation.
5. Review an example. Let the equation of the third degree 2x³−11x²+12x+9=0 be given. A=2≠0, and free member D=9. Find all dividers of coefficient of D: 1,-1, 3,-3, 9,-9. Substitute these dividers 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 roots of this cubic equation x1=3. Now divide both parts of the initial equation into a binomial (x−3). As a result the quadratic equation turns out: 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 the valid roots x1=x2=3 and x3=-0.5.