We can morph the mixed products to get
(31/2× 9 1/2)x^3 − (31/2× 144 1/2)x^2 + (31/2× 144 1/2)x
− (121/2× 9 1/2)x^2 + (121/2× 144 1/2)x− 12 1/2× 144 1/2
Applying the product of powers rule to each of these terms, we obtain
(3× 9)1/2x^3 − (3 × 144)1/2x^2 + (3 × 144)1/2x
− (12 × 9)1/2x^2 + (12 × 144)1/2x− (12 × 144)1/2
This simplifies to
27 1/2x^3 − 432 1/2x^2 + 432 1/2x
− 108 1/2x^2 + 1,7281/2x− 1,7281/2
Consolidating the terms for each power of x and then returning the expression to the left
side of the complete equation, we get
27 1/2x^3 − (4321/2+ 108 1/2)x^2
+ (4321/2+ 1,7281/2)x− 1,7281/2= 0
That’s as “simple” as we can get this cubic in polynomial standard form. All of the coefficients,
and the stand-alone constant, are irrational. This is the same equation that we solved in the
first “challenge” in the chapter text, finding a single real root of 2. For extra credit, find a
calculator that displays a lot of digits (such as the scientific calculator program in a personal
computer), substitute 2 for x in the result we just got, and verify that it works out!
- Here again, for reference, is the binomial factor form of the equation that we have been
told to multiply out:
(3x+ 2)(5x+ 6)(− 7 x− 1) = 0
Let’s multiply the second two factors first. We get
(3x+ 2)(− 35 x^2 − 5 x− 42 x− 6) = 0
which consolidates to
(3x+ 2)(− 35 x^2 − 47 x− 6) = 0
Now we can multiply the binomial by the trinomial, obtaining
− 105 x^3 − 141 x^2 − 18 x− 70 x^2 − 94 x− 12 = 0
which consolidates to
− 105 x^3 − 211 x^2 − 112 x− 12 = 0
Chapter 25 687
