How to solve cubic equations...
Answers
Answer:
A cubic equation is an equation which can be represented in the form ax^3+bx^2+cx+d=0ax
3
+bx
2
+cx+d=0, where a,b,c,da,b,c,d are complex numbers and aa is non-zero. By the fundamental theorem of algebra, cubic equation always has 33 roots, some of which might be equal.
Step-by-step explanation:
For a cubic equation ax^3+bx^2+cx+d=0ax
3
+bx
2
+cx+d=0, let p,q,p,q, and rr be its roots, then the following holds:
Root expression Equals to
p+q+rp+q+r -\frac{b}{a}−
a
b
pq+qr+rppq+qr+rp \ \ \ \frac{c}{a}
a
c
pqrpqr -\frac{d}{a}−
a
d
This is a special case of Vieta's formulas.
Step-by-step explanation:
1
Check whether your cubic contains a constant (a d value). Cubic equations take the form ax^{3}+bx^{2}+cx+d=0. However, the only essential requirement is x^{3}, which means the other elements need not be present to have a cubic equation.[1]
If your equation does contain a constant (a d value), you'll need to use another solving method.
If a=0, you do not have a cubic equation.[2]
2
Factor an x out of the equation. Since your equation doesn't have a constant, every term in the equation has an x variable in it. This means that one x can be factored out of the equation to simplify it. Do this and re-write your equation in the form x(ax^{2}+bx+c).[3]
For example, let's say that your starting cubic equation is 3x^{3}-2x^{2}+14x=0
Factoring a single x out of this equation, you get x(3x^{2}-2x+14)=0
3
Factor the resulting quadratic equation, if possible. In many cases, you will be able to factor the quadratic equation (ax^{2}+bx+c) that results when you factor the x out. For example, if you are given x^{3}+5x^{2}-14x=0, then you can do the following:[4]
Factor out the x: x(x^{2}+5x-14)=0
Factor the quadratic in parentheses: x(x+7)(x-2)=0
Set each of these factors equal to{\displaystyle 0}. Your solutions are x=0,x=-7,x=2.
4
Solve the portion in parentheses with the quadratic formula if you can’t factor it manually. You can find the values for which this quadratic equation equals {\displaystyle 0} by plugging a, b, and c into the quadratic formula ({\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}). Do this to find two of the answers to your cubic equation.[5]
In the example, plug your a, b, and c values (3, -2, and 14, respectively) into the quadratic equation as follows:
{\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}
{\frac {-(-2)\pm {\sqrt {((-2)^{2}-4(3)(14)}}}{2(3)}}
{\frac {2\pm {\sqrt {4-(12)(14)}}}{6}}
{\frac {2\pm {\sqrt {(4-168}}}{6}}
{\frac {2\pm {\sqrt {-164}}}{6}}
Answer 1:
{\frac {2+{\sqrt {-164}}}{6}}
{\frac {2+12.8i}{6}}
Answer 2:
{\frac {2-12.8i}{6}}
5
Use zero and the quadratic answers as your cubic's answers. 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 {\displaystyle 0}.[6]
Factoring your equation into the form x(ax^{2}+bx+c)=0 splits it into two factors: one factor is the x variable on the left, and the other is the quadratic portion in parentheses. If either of these factors equals {\displaystyle 0}, the entire equation will equal {\displaystyle 0}.
Thus, the two answers to the quadratic portion in parentheses, which will make that factors equal {\displaystyle 0}, are answers to the cubic, as is {\displaystyle 0} itself, which will make the left factor equal {\displaystyle 0}.