5 − 9 21 104 80
− 45 ##
− 9 − 24 ##5 − 9 21 104 80
− 45 − 120 #
− 9 − 24 ##(^5) − 9 21 104 80
− 45 − 120 #
− 9 − 24 − 16 #
5 − 9 21 104 80
− 45 − 120 − 80
− 9 − 24 − 16 #
5 − 9 21 104 80
− 45 − 120 − 80
− 9 − 24 − 16 0
We get a remainder of 0, so we know that x= 5 is a real root of the original cubic.
- We can write the cubic presented in Prob. 8 in binomial-trinomial form on the basis of
the results of the synthetic division. Because x= 5 is a real root of the cubic, we know
that it has a binomial factor of (x− 5). In the trinomial factor, the coefficient of x^2 is
−9, the coefficient of x is −24, and the stand-alone constant is −16, because those three
numbers appear, in that order, in the bottom row before the remainder 0. We can now
write down the entire binomial-trinomial cubic equation:
(x− 5)(− 9 x^2 − 24 x− 16) = 0To figure out whether there are any other roots besides x= 5, we must find the discrimi-
nant of the trinomial factor. If we let a 2 =−9,b 2 =−24, and c=−16, we can find the
discriminant,d, as follows:d=b 22 − 4 a 2 c= (−24)^2 − 4 × (−9)× (−16)
= 576 − 576
= 0
Chapter 25 691