The solution set is X= {−4, 4, −36}. The multiplicity of each root is the same as the power
to which its binomial is raised in the original equation. Therefore, the root x=−4 has
multiplicity 1, the root x= 4 has multiplicity 2, and the root x=−36 has multiplicity 3.
The degree of the original equation is the sum of the exponents attached to the factors,
which is 1 + 2 + 3 = 6.- For reference, the polynomial equation is
2 x^5 − 3 x^3 − 2 x+ 2 = 0We have many options! The largest absolute value of any coefficient or constant is 3, so
we can try 3 for the upper bound and −3 for the lower bound. If either or both of these
fail, we can try values farther from 0. The coefficients and constant, in order of decreasing
powers of x, are 2, 0, −3, 0, −2, and 2. (The coefficients of x^4 and x^2 are both equal to 0.)
Here’s the synthetic division array for the “test root” 3:(^320) − 3 0 − 2 2
When we go through the synthetic division process, we end up with
320 − 3 0 − 2 2
6 18 45 135 399
2 6 15 45 133 401
None of the numbers in the bottom row are negative. This tells us that 3 is an upper
bound for the real roots. To try the “test root” −3, we set up the array
320 − 3 0 − 2 2
The synthetic division process leads us to
− 3 20 − 3 0 − 2 2
− 6 − 18 − 45 135 − 399
(^2) − 6 15 − 45 133 − 397
The numbers in the bottom row alternate in sign. This indicates that −3 is a lower bound
for the real roots.
- Here is an outline of the process for finding the rational roots of the equation
2 x^5 − 3 x^3 − 2 x+ 2 = 0Chapter 26 697