Part Three 517Question 25-5
What is the binomial-factor form of a cubic equation in the variable x?
Answer 25-5
An equation in binomial-factor form can always be written like this when x is the variable:
(a 1 x+b 1 )(a 2 x+b 2 )(a 3 x+b 3 )= 0where a 1 ,a 2 , and a 3 are nonzero real numbers, and b 1 ,b 2 , and b 3 are real numbers.
Question 25-6
What does the binomial-factor equation from Answer 25-5 look like when multiplied out
into polynomial standard form?
Answer 25-6
Let’s start by multiplying the second and third binomials together. When we do that, we
obtain
(a 1 x+b 1 )(a 2 a 3 x^2 +a 2 b 3 x+b 2 a 3 x+b 2 b 3 )= 0Now let’s multiply the binomial by the polynomial on the left side. That gives us
a 1 a 2 a 3 x^3 +a 1 a 2 b 3 x^2 +a 1 b 2 a 3 x^2 +a 1 b 2 b 3 x+b 1 a 2 a 3 x^2 +b 1 a 2 b 3 x+b 1 b 2 a 3 x+b 1 b 2 b 3 = 0When we group the terms for x^2 and x together and apply the distributive law for multiplica-
tion over addition, we get the polynomial standard form
a 1 a 2 a 3 x^3 + (a 1 a 2 b 3 +a 1 b 2 a 3 +b 1 a 2 a 3 )x^2 + (a 1 b 2 b 3 +b 1 a 2 b 3 +b 1 b 2 a 3 )x+b 1 b 2 b 3 = 0Question 25-7
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 25-7
There are three 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