Article: Cubic equations take the form ax3+bx2+cx+d=0{\displaystyle ax^{3}+bx^{2}+cx+d=0}. However, the only essential requirement is x3{\displaystyle x^{3}}, which means the other elements need not be present to have a cubic equation.  If your equation does contain a constant (a d{\displaystyle d} value), you'll need to use another solving method. If a=0{\displaystyle a=0}, you do not have a cubic equation. Since your equation doesn't have a constant, every term in the equation has an x{\displaystyle x} variable in it. This means that one x{\displaystyle x} can be factored out of the equation to simplify it. Do this and re-write your equation in the form x(ax2+bx+c){\displaystyle x(ax^{2}+bx+c)}.  For example, let's say that your starting cubic equation is 3x3−2x2+14x=0{\displaystyle 3x^{3}-2x^{2}+14x=0}  Factoring a single x{\displaystyle x} out of this equation, you get x(3x2−2x+14)=0{\displaystyle x(3x^{2}-2x+14)=0} In many cases, you will be able to factor the quadratic equation (ax2+bx+c{\displaystyle ax^{2}+bx+c}) that results when you factor the x{\displaystyle x} out. For example, if you are given x3+5x2−14x=0{\displaystyle x^{3}+5x^{2}-14x=0}, then you can do the following:  Factor out the x{\displaystyle x}: x(x2+5x−14)=0{\displaystyle x(x^{2}+5x-14)=0}  Factor the quadratic in parentheses: x(x+7)(x−2)=0{\displaystyle x(x+7)(x-2)=0}  Set each of these factors equal to0{\displaystyle 0}. Your solutions are x=0,x=−7,x=2{\displaystyle x=0,x=-7,x=2}. You can find the values for which this quadratic equation equals 0{\displaystyle 0} by plugging a{\displaystyle a}, b{\displaystyle b}, and c{\displaystyle c} into the quadratic formula (−b±b2−4ac2a{\displaystyle {\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}}). Do this to find two of the answers to your cubic equation.  In the example, plug your a{\displaystyle a}, b{\displaystyle b}, and c{\displaystyle c} values (3{\displaystyle 3}, −2{\displaystyle -2}, and 14{\displaystyle 14}, respectively) into the quadratic equation as follows:  −b±b2−4ac2a{\displaystyle {\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}} −(−2)±((−2)2−4(3)(14)2(3){\displaystyle {\frac {-(-2)\pm {\sqrt {((-2)^{2}-4(3)(14)}}}{2(3)}}} 2±4−(12)(14)6{\displaystyle {\frac {2\pm {\sqrt {4-(12)(14)}}}{6}}} 2±(4−1686{\displaystyle {\frac {2\pm {\sqrt {(4-168}}}{6}}} 2±−1646{\displaystyle {\frac {2\pm {\sqrt {-164}}}{6}}}   Answer 1:  2+−1646{\displaystyle {\frac {2+{\sqrt {-164}}}{6}}} 2+12.8i6{\displaystyle {\frac {2+12.8i}{6}}}   Answer 2: 2−12.8i6{\displaystyle {\frac {2-12.8i}{6}}} While quadratic equations have two solutions, cubics have three. You already have two of these — they're the answers you found for the "quadratic" portion of the problem in parentheses. In cases where your equation is eligible for this "factoring" method of solving, your third answer will always be 0{\displaystyle 0}.  Factoring your equation into the form x(ax2+bx+c)=0{\displaystyle x(ax^{2}+bx+c)=0} splits it into two factors: one factor is the x{\displaystyle x} variable on the left, and the other is the quadratic portion in parentheses. If either of these factors equals 0{\displaystyle 0}, the entire equation will equal 0{\displaystyle 0}. Thus, the two answers to the quadratic portion in parentheses, which will make that factors equal 0{\displaystyle 0}, are answers to the cubic, as is 0{\displaystyle 0} itself, which will make the left factor equal 0{\displaystyle 0}.
What is a summary of what this article is about?
Check whether your cubic contains a constant (a d{\displaystyle d} value). Factor an x{\displaystyle x} out of the equation. Factor the resulting quadratic equation, if possible. Solve the portion in parentheses with the quadratic formula if you can’t factor it manually. Use zero and the quadratic answers as your cubic's answers.