It is the cubic equation x3−9x=0, which should be solved. Let's factor out x. This gives us the equation
We can now use the Zero Product Property to divide it into two equations.
In the first, x already stands by itself, so we know that x=0 is a solution. Let's now solve the other.
We will use the same method to solve the equation 2x3−50x=0. First we factor out x.x(2x2−50)=0
Using the Zero Product Property we can write this as two equations.
From the first equation we learn that x=0 is a solution. Next, we will solve the second equation.
We have found that there are three solutions to the equation and they are
Let's now solve 18x−2x3=0 using the method we practiced in Part A and Part B. If we factor out x, we get
By using the Zero Product Property, we get the equations
Again, we find that x=0 is a solution. Let's go on and solve the second equation.