Algebra Know-It-ALL

428 Cubic Equations in Real Numbers

What are the other roots?

We can write the binomial-trinomial form of a cubic equation straightaway, once we have

performed synthetic division on the original cubic and managed to come up with a remainder

of 0. Take a close look at the numbers in the bottom line of the last step in the synthetic divi-

sion process for the “test root” of 3. Those numbers are 1, 3, 2, and 0. The first three of these,

in the order shown, are the coefficients and the stand-alone constant in the trinomial factor

of the cubic equation.

We know, in the above example, that (x− 3) is the binomial factor. The factor theorem tells

us so. So the trinomial factor is (x^2 + 3 x+ 2). We can therefore write the binomial-trinomial

version of the original cubic as

(x− 3)(x^2 + 3 x+ 2) = 0

To be certain of this, and to get a little complementary credit, you can multiply the left side

of the equation out and see that it produces the original polynomial.

The other two real roots, if they exist, can be found by solving the quadratic equation

x^2 + 3 x+ 2 = 0

The roots of this equation turn out to be x=−1 and x=−2. You can verify this fact as another

complementary-credit exercise. That gives us three real roots for the original cubic:

x= 3 or x=−1 or x=− 2

and a real-number solution set X= {3,−1,−2}. For some more complementary credit, you can

substitute each of these roots for x in the original cubic to demonstrate that they are correct.

Are you confused?

Synthetic division sometimes works out nicely, but in some cases it does not. What if the real roots of a

cubic are all complicated fractions—or worse, all irrational numbers? What if all the methods to “crack

a cubic” that we’ve seen in this chapter fail us? There are other schemes available, some of which will be

covered in the next chapter.

Here’s a challenge!

Consider the following cubic in polynomial standard form:

6 x^3 + 13 x^2 + 8 x+ 3

Show by synthetic division that −3/2 is a real root of this equation. Then, from the results of that process,

show that −3/2 is the only real root.

Solution

First, let’s set up the synthetic division array with the “test root,” x=−3/2, and the coefficients in the

top row.