Algebra Know-It-ALL

418 Cubic Equations in Real Numbers

That’s a reasonable question. Dividing through by x can “knock the equation down” from third degree to

second degree—or so we might be tempted to believe. Let’s try it and see what happens! We get

(x^3 + 14 x^2 + 48 x ) / x= 0/x

which apparently simplifies to

x^3 /x+ 14 x^2 /x+ 48 x/x= 0

and further to

x^2 + 14 x+ 48 = 0

This is a quadratic that can be factored into

(x+ 6)(x+ 8) = 0

The roots x=−6 or x=−8 come out of this process. We conclude that the solution set X for the original

cubic must be {−6,−8}. That’s easy, isn’t it? Not so fast! It’s also wrong, because it’s incomplete. Let’s find

out how to get it right.

Here’s a challenge!

Solve the following cubic equation in a way that works properly:

x^3 + 14 x^2 + 48 x= 0

Solution

We’ve seen this equation in its binomial factor form. Suppose that we could look at the above equation and

see the binomial factor version immediately:

x(x+ 6)(x+ 8) = 0

To find the first root, we can take literally the simple first-degree equation

x= 0

To find the second root, we take the first-degree equation

x+ 6 = 0

which resolves to x=−6. To find the third root, we take the first-degree equation

x+ 8 = 0

which resolves to x=−8. The real roots of the cubic are therefore

x= 0 or x=−6 or x=− 8