Algebra Know-It-ALL

Part Three 521

If we substitute the quantity (x− 3)^2 for the trinomial in the original equation, we get

[(x− 3)^2 ]^2 = 0

which can be simplified to

(x− 3)^4 = 0

We can solve by setting the binomial equal to 0:

x− 3 = 0

This first-degree equation resolves to x= 3. This is the only real root of the original higher-
degree equation, and it has multiplicity 4. The real solution set is X= {3}.

Question 26-6

What is the binomial factor form of an nth-degree equation in the variable x?

Answer 26-6

Suppose that a 1 ,a 2 ,a 3 , ..., an are nonzero real numbers, and b 1 ,b 2 , b 3 , ..., bn are real stand-alone
constants. Let x be the variable in the following equation:

(a 1 x+b 1 )(a 2 x+b 2 )(a 3 x+b 3 ) ··· (anx+bn)= 0

This is the binomial factor form for an nth-degree equation in the variable x.

Question 26-7

How many real roots does the equation stated in Answer 25-5 have? What is that root, or what
are they? What is the real solution set X?

Answer 26-7

There are n real roots. They are the solutions to the equations we obtain when we set the
binomials equal to 0:

a 1 x+b 1 = 0

a 2 x+b 2 = 0

a 3 x+b 3 = 0

↓

anx+bn= 0

When we subtract the “b-constant” from both sides in each of these equations and then divide
through by the “a-coefficient” (which is okay because we know that a 1 ,a 2 ,a 3 , ..., an are all
nonzero), we get these roots:

x=−b 1 /a 1 or x=−b 2 /a 2 or x=−b 3 /a 3

... or x=−bn/an