- Here’s the general binomial-trinomial equation once again:
(a 1 x+b 1 )(a 2 x^2 +b 2 x+c)= 0
Using the product of sums rule, we can rewrite this as
a 1 a 2 x^3 +a 1 b 2 x^2 +a 1 cx+b 1 a 2 x^2 +b 1 b 2 x+b 1 c= 0
When we bring the terms for x^2 next to each other, and then do the same thing with the
terms for x, we get
a 1 a 2 x^3 +a 1 b 2 x^2 +b 1 a 2 x^2 +a 1 cx+b 1 b 2 x+b 1 c= 0
Let’s use the commutative law for multiplication on the third term to get “a before b” in
the interest of elegance! That gives us
a 1 a 2 x^3 +a 1 b 2 x^2 +a 2 b 1 x^2 +a 1 cx+b 1 b 2 x+b 1 c= 0
Finally, we can use the distributive law for multiplication over addition to consolidate the
coefficients for x^2 and x, giving us the equation in true polynomial standard form:
a 1 a 2 x^3 + (a 1 b 2 +a 2 b 1 )x^2 + (a 1 c+b 1 b 2 )x+b 1 c= 0
- If the coefficients and constants are real numbers, and if a 1 and a 2 are both nonzero,
then the cubic
(a 1 x+b 1 )(a 2 x^2 +b 2 x+c)= 0
has at least one real root, which is x=−b 1 /a 1. That’s the root that we get when we create
a first-degree equation from the binomial term by setting it equal to 0. The cubic might
have no more real roots (therefore one real root in total), one more (two in total), or two
more (three in total). To find out which of these situations is the true case, we can look at
the discriminant d for the quadratic
a 2 x^2 +b 2 x+c= 0
In this notation,
d=b 22 − 4 a 2 c
If d > 0, then the quadratic has two real roots, so the original cubic has three. If d= 0,
then the quadratic has one real root with multiplicity 2, so the original cubic has two real
roots, one of which has multiplicity 2. If d < 0, then the quadratic has no real roots, so
the original cubic has only one.
- For reference, here’s the equation again:
(3x+ 5)(16x^2 − 56 x+ 49) = 0
Chapter 25 689
