Algebra Know-It-ALL

430 Cubic Equations in Real Numbers

That’s (x+ 3/2). In the trinomial factor, the coefficient of x^2 is 6, the coefficient of x is 4, and the stand-

alone constant is 2. We know this because 6, 4, and 2 appear, in that order, in the bottom row before the

remainder 0. Here’s the binomial-trinomial cubic:

(x+ 3/2)(6x^2 + 4 x+ 2) = 0

Let’s examine the trinomial factor. If we let a 2 = 6, b 2 = 4, and c= 2, we can find the discriminant, d, as

follows:

d=b 22 − 4 a 2 c

= 42 − 4 × 6 × 2

= 16 − 48

=− 32

If we set the trinomial factor equal to 0 to get a quadratic equation, then that quadratic has no real roots

becaused < 0. The only way the original cubic polynomial can attain the value 0 is if either the binomial

factor is 0, the trinomial factor is 0, or both. The binomial becomes 0 if and only if x=−3/2. We’ve just

discovered that no real number x can make the trinomial become 0. It follows that x=−3/2 is the only

real root of the original cubic equation.

Practice Exercises

This is an open-book quiz. You may (and should) refer to the text as you solve these problems.

Don’t hurry! You’ll find worked-out answers in App. C. The solutions in the appendix may

not represent the only way a problem can be figured out. If you think you can solve a particu-

lar problem in a quicker or better way than you see there, by all means try it!

1. Multiply out the following equation to obtain a polynomial cubic:
   (ax+b)^3 = 0
   where x is the variable, and a and b are real numbers with a≠ 0.


2. Multiply out the following equation to get a polynomial cubic:

(31/2x− 12 1/2)^3 = 0

3. In the chapter text, we solved this cubic in binomial factor form:

(3x+ 2)(5x+ 6)(− 7 x− 1) = 0

We found that the real roots are

x=−2/3 or x=−6/5 or x=−1/7

Multiply this equation out to get it into the polynomial standard form.