3 Ways To Solve 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. [2] X Research source

For example, let’s say that your starting cubic equation is 3x3+5x2−14x=0{\displaystyle 3x^{3}+5x^{2}-14x=0} Factoring a single x{\displaystyle x} out of this equation, you get x(3x2+5x−14)=0{\displaystyle x(3x^{2}+5x-14)=0}

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 also factor it by grouping.

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}}}

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}.

Take, for example, 2x3+9x2+13x=−6{\displaystyle 2x^{3}+9x^{2}+13x=-6}. In this case, getting a 0{\displaystyle 0} on the right side of the equals sign requires you to add 6{\displaystyle 6} to both sides. In the new equation, 2x3+9x2+13x+6=0{\displaystyle 2x^{3}+9x^{2}+13x+6=0}. Since d=6{\displaystyle d=6}, you can’t use the quadratic equation method.

For example, since you can make 6 by multiplying 6×1{\displaystyle 6\times 1} and 2×3{\displaystyle 2\times 3}, that means 1, 2, 3, and 6 are the factors of 6. In the sample problem, a=2{\displaystyle a=2} and d=6{\displaystyle d=6}. The factors of 2 are 1 and 2. The factors of 6 are 1, 2, 3, and 6.

In the sample equation, taking the factors of d{\displaystyle d} (1, 2, 3 and 6) over the factors of a{\displaystyle a} (1 and 2) gives this list: 6{\displaystyle 6}, −6{\displaystyle -6}, 3{\displaystyle 3}, −3{\displaystyle -3}, 2{\displaystyle 2}, −2{\displaystyle -2}, 1{\displaystyle 1} and −1{\displaystyle -1}. Your cubic equation’s solutions are somewhere in this list.

2(1)3+9(1)2+13(1)+6{\displaystyle 2(1)^{3}+9(1)^{2}+13(1)+6}, or 2+9+13+6{\displaystyle 2+9+13+6}, which clearly does not equal 0{\displaystyle 0}. So, move on to the next value on your list. If you plug in −1{\displaystyle -1}, you get (−2)+9+(−13)+6{\displaystyle (-2)+9+(-13)+6}, which does equal 0{\displaystyle 0}. This means −1{\displaystyle -1} is one of your integer solutions.

Synthetic division is a complex topic that’s beyond the scope of describing fully here. However, here’s a sample of how to find one of the solutions to your cubic equation with synthetic division: -1 | 2 9 13 6 __| -2-7-6 __| 2 7 6 0 Since you got a final remainder of 0{\displaystyle 0}, you know that one of your cubic’s integer solutions is −1{\displaystyle -1}.

For the sample equation x3−3x2+3x−1{\displaystyle x^{3}-3x^{2}+3x-1}, write a=1{\displaystyle a=1}, b=−3{\displaystyle b=-3}, c=3{\displaystyle c=3}, and d=−1{\displaystyle d=-1}. Don’t forget that when an x{\displaystyle x} variable doesn’t have a coefficient, it’s implicitly assumed that its coefficient is 1{\displaystyle 1}.

A discriminant is simply a number that gives us information about the roots of a polynomial (you may already know the quadratic discriminant: b2−4ac{\displaystyle b^{2}-4ac}). In your sample problem, solve as follows: b2−3ac{\displaystyle b^{2}-3ac} (−3)2−3(1)(3){\displaystyle (-3)^{2}-3(1)(3)} 9−3(1)(3){\displaystyle 9-3(1)(3)} 9−9=0=Δ0{\displaystyle 9-9=0=\Delta _{0}}

In the example, solve as follows: 2(−3)3−9(1)(−3)(3)+27(1)2(−1){\displaystyle 2(-3)^{3}-9(1)(-3)(3)+27(1)^{2}(-1)} 2(−27)−9(−9)+27(−1){\displaystyle 2(-27)-9(-9)+27(-1)} −54+81−27{\displaystyle -54+81-27} 81−81=0=Δ1{\displaystyle 81-81=0=\Delta _{1}}

A cubic equation always has at least one real solution, because the graph will always cross the x-axis at least once. In the example, since both Δ0{\displaystyle \Delta _{0}} and Δ1{\displaystyle \Delta _{1}} =0{\displaystyle =0}, finding Δ{\displaystyle \Delta } is relatively easy. Solve as follows: (Δ12−4Δ03)÷(−27a2){\displaystyle (\Delta _{1}^{2}-4\Delta _{0}^{3})\div (-27a^{2})} ((0)2−4(0)3)÷(−27(1)2){\displaystyle ((0)^{2}-4(0)^{3})\div (-27(1)^{2})} 0−0÷27{\displaystyle 0-0\div 27} 0=Δ{\displaystyle 0=\Delta }, so the equation has one or two answers.

In your example, find C{\displaystyle C} as follows: 3(Δ12−4Δ03)+Δ1÷2{\displaystyle ^{3}{\sqrt {{\sqrt {(\Delta _{1}^{2}-4\Delta _{0}^{3})+\Delta _{1}}}\div 2}}} 3(02−4(0)3)+(0)÷2{\displaystyle ^{3}{\sqrt {{\sqrt {(0^{2}-4(0)^{3})+(0)}}\div 2}}} 3(0−0)+0÷2{\displaystyle ^{3}{\sqrt {{\sqrt {(0-0)+0}}\div 2}}} 0=C{\displaystyle 0=C}

You can solve the example by checking the answer when n is equal to 1, 2, and 3. The answers you get from these tests are the possible answers to the cubic equation — any that give an answer of 0 when plugged into the equation are correct. For example, since plugging 1 into x3−3x2+3x−1{\displaystyle x^{3}-3x^{2}+3x-1} gives an answer of 0, 1 is one of the answers to your cubic equation.