Algebra Know-It-ALL

where neither p nor r equals 0. If either factor (that is, either binomial) happens to equal 0,
then the entire expression on the left side of the equals sign becomes 0. The equation then
reduces to 0 = 0, indicating a root! The roots of the above quadratic can therefore be found by
solving these two first-degree equations:

px+q= 0

and

rx+s= 0

In the first equation, we can subtract q from each side, getting

px=−q

Dividing through by p, which we have said is nonzero, we get

x=−q/p

In the second equation, we can subtract s from each side to obtain

rx=−s

Then we divide through by r, which we have restricted to nonzero values, getting

x=−s/r

The roots of the quadratic are x=−q/p or x=−s/r. If X is the set of solutions, called the solu-
tion set, then

X= {−q/p,−s/r}

The solution set for an equation with multiple roots is the set containing all of those roots.
As you continue studying algebra, you’ll come across equations with solution sets containing
three, four, or more elements. The solution set for a true quadratic, however, never has more
than two elements.

Here’s a challenge!

Consider the following quadratic equation in standard form:

x^2 − 2 x− 15 = 0

Put this equation into binomial factor form. Then find the solution set X. Here’s a hint: The coefficients
and constants in the binomial factor form are integers.

Binomial Factor Form 369