Algebra Know-It-ALL

420 Cubic Equations in Real Numbers

What are the real roots?

The process of finding the real roots of a cubic in the binomial-trinomial form is straightforward,

as long as all the coefficients and constants are real numbers. First, we “manufacture” a first-degree

equation from the binomial, setting it equal to 0. In the general form above, that would be

a 1 x+b 1 = 0

which solves to

x=−b 1 /a 1

This will always give us one real root for the cubic. After that, we set the trinomial equal to 0,

obtaining a quadratic equation. In the general form shown above, we get

a 2 x^2 +b 2 x+c= 0

We can find the real roots of this equation, if any exist, using techniques we’ve already learned

for solving quadratics. (I like to use the quadratic formula, because it always works! Also, if the

root or roots are complex but not real, the quadratic formula will produce them.) Expressed

for the above general equation, the quadratic formula is

x= [−b 2 ± (b 22 − 4 a 2 c)1/2] / (2a 2 )

Are you confused?

You’ve learned that a quadratic equation can have two different real roots, or only one real root, or none

at all. How about cubics? You’ve already seen an example of a cubic with three real roots. You’re about to

see that a cubic equation in the binomial-trinomial form with real coefficients and a real constant can have

two real roots. Then you’ll discover that a cubic in the binomial-trinomial form with real coefficients and

a real constant can have only one real root, along with two others that are complex.

“Okay,” you say. Then you ask, “How about no real roots?” The answer: Any cubic in the binomial-

trinomial form with real coefficients and a real constant always has at least one real root. That’s because a

first-degree equation can always be created from the binomial factor, and that equation always has a real

solution. We can take this statement further: A cubic equation, no matter what the form, has at least one

real root if all the coefficients and constants are real numbers.

Here’s a challenge!

Find the real roots of the following cubic equation using the method described in this section, and state

the real solution set X.

(3x+ 5)(16x^2 − 56 x+ 49) = 0

Solution

First, we can construct a first-degree equation by setting the binomial factor equal to 0. That gives us

3 x+ 5 = 0