We can rewrite the cubic equation x3−100x=0 into one linear equation and one quadratic equation by factoring out x.x(x2−100)=0.
By using the Zero Product Property, we can now rewrite this equation into the two simpler equations
x=0ochx2−100=0.
From the first equation we learn that x=0 is a solution. Let's go on and solve the second equation.
The equation has three solutions and they are
x=0,x=-10andx=10.
b
In this case, it is a good idea to begin by combining the like terms we have on the left-hand side of the equation. After that, we solve the equation using the same method we practiced in Part A.
We will now use the Zero Product Property and rewrite the equation into
x=0and2x2−162=0.
As before, we find that one solution is x=0. By solving the second equation we will find an additional two solutions.
We have found the three solutions of the equation and they are
x=0,x=-9andx=9.
c
This time, we have a quartic equation, x4−8x=0. Just as before, it is possible to factor out an x. Then, we get
x(x3−8)=0,
and the Zero Product Property gives us the two equations
x=0andx3−8=0.
From the first we learn, as previously, that x=0 is a solution. The second equation is a cubic equation. Let's solve it.
