Algebra Know-It-ALL

Are you confused?

If you wonder how the two trinomials in the above situation can be multiplied out, remember the rule
you derived when you solved Practice Exercise 10 at the end of Chap. 9. If you’ve forgotten how that rule
works, review it now. It can be extrapolated to all real, imaginary, and complex numbers.

Here’s a challenge!

Consider a general quadratic equation in polynomial standard form:

ax^2 +bx+c= 0

Suppose that b is not equal to 0. Also suppose that the discriminant, d, is negative, where

d=b^2 − 4 ac

Write the general quadratic equation in factored form.

Solution

Let’s begin with the quadratic formula we obtained earlier in this chapter for cases of this sort:

x= [−b±j(|d|1/2)] / (2a)

We can break the numerator apart and write this as the sum of two fractions with a common denominator:

x=−b/(2a)±j(|d|1/2)/(2a)

The roots can be expressed separately like this:

x=−b/(2a)+j(|d|1/2)/(2a) or x=−b/(2a)−j(|d|1/2)/(2a)

This suggests that the factored form of the equation ought to be

{x− [−b/(2a)+j(|d|1/2)/(2a)]}{x− [−b/(2a)−j(|d|1/2)/(2a)]}= 0

Here’s another challenge!

Multiply out the factored equation we’ve just obtained, verifying that it’s equivalent to the original general
quadratic in polynomial standard form.

Solution

We can simplify things if we temporarily rename two of the terms in the binomial. We’ll give them their
“legitimate” names back later. Let

p=−b/(2a)

and

q= (|d|1/2)/(2a)

Conjugate Roots in Factors 393